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Department of Statistics and Operations Research

College of Science, King Saud University

Summer Semester 1426-1427

STAT – 324

Probability and Statistics for Engineers

EXERCISES

A Collection of Questions Selected from

Midterm and Final Examinations' Papers

for the Years from 1422 to 1427

PREPARED BY:

Dr. Abdullah Al-Shiha

1. COMBINATIONS

=The number of combinations of n distinct objects taken r at a time (r objects in each combination)

= The number of different selections of r objects from n distinct objects.

= The number of different ways to select r objects from n distinct objects.

= The number of different ways to divide a set of n distinct objects into 2 subsets; one subset

contains r objects and the other subset contains the rest.

Q1. Compute:

(a)             (b) .

Q2. Show that .

Q3. Compute:

(a) ,           (b) ,           (c)

Q4. A man wants to paint his house in 3 colors. If he can choose 3 colors out of 6 colors, how many different color settings can he make?

(A) 216            (B) 20              (C) 18              (D) 120

Q5. The number of ways in which we can select two students among a group of 5 students is

(A) 120            (B) 10              (C) 60              (D) 20              (E) 110

Q6. The number of ways in which we can select a president and a secretary among a group of 5 students is

(A) 120            (B) 10              (C) 60              (D) 20

2. PROBABILITY, CONDITIONAL PROBABILITY, AND INDPENDENCE

Q1. Let A, B, and C be three events such that: P(A)=0.5, P(B)=0.4, P(CÇAc)=0.6,  P(CÇA)=0.2, and  P(AÈB)=0.9. Then

 (a)  P(C) = (A) 0.1 (B) 0.6 (C) 0.8 (D) 0.2 (E) 0.5 (b)  P(BÇA) = (A) 0.0 (B) 0.9 (C) 0.1 (D) 1.0 (E) 0.3 (c)  P (C|A) = (A) 0.4 (B) 0.8 (C) 0.1 (D) 1.0 (E) 0.7 (d)  P (BcÇAc) = (A) 0.3 (B) 0.1 (C) 0.2 (D) 1.1 (E) 0.8

Q2. Consider the experiment of flipping a balanced coin three times independently.

 (a) The number of points in the sample space is (A) 2 (B) 6 (C) 8 (D) 3 (E) 9 (b) The probability of getting exactly two heads is (A) 0.125 (B) 0.375 (C) 0.667 (D) 0.333 (E) 0.451 (c) The events ‘exactly two heads’ and ‘exactly three heads’ are (A) Independent (B) disjoint (C) equally        likely (D) identical (E) None (d) The events ‘the first coin is head’ and ‘the second and the third coins are tails’ are (A) Independent (B) disjoint (C) equally        likely (D) identical (E) None

Q3.  Suppose that a fair die is thrown twice, then

1. the probability that the sum of numbers of the two dice is less than or equal to 4 is;

(A) 0.1667       (B) 0.6667       (C) 0.8333       (D) 0.1389

1. the probability that at least one of the die shows 4 is;

(A) 0.6667       (B) 0.3056       (C) 0.8333       (D) 0.1389

1. the probability that one die shows one and the sum of the two dice is four is;

(A) 0.0556       (B) 0.6667       (C) 0.3056       (D) 0.1389

1. the event A={the sum of two dice is 4} and the event B={exactly one die shows two} are,

(A) Independent           (B) Dependent  (C) Joint           (D) None of these.

Q4. Assume that , , , and , then

1.             the events A and B are,

(A) Independent           (B) Dependent              (C) Disjoint      (D) None of these.

2.              is equal to,

(A) 0.65           (B) 0.25           (C) 0. 35          (D) 0.14

Q5. If the probability that it will rain tomorrow is 0.23, then the probability that it will not rain tomorrow is:

(A) -0.23          (B) 0.77          (C) -0.77        (D) 0.23

Q6. The probability that a factory will open a branch in Riyadh is 0.7, the probability that it will open a branch in Jeddah is 0.4, and the probability that it will open a branch in either Riyadh or Jeddah or both is 0.8. Then, the probability that it will open a branch:

1) in both cities is:

(A) 0.1             (B) 0.9            (C) 0.3            (D) 0.8

2) in neither city is:

(A) 0.4             (B) 0.7            (C) 0.3            (D) 0.2

Q7. The probability that a lab specimen is contaminated is 0.10. Three independent samples are checked.

1) the probability that none is contaminated is:

(A) 0.0475       (B) 0.001        (C) 0.729        (D) 0. 3

2) the probability that exactly one sample is contaminated is:

(A) 0.243         (B) 0.081        (C) 0.757        (D) 0. 3

Q8. 200 adults are classified according to sex and their level of education in the following table:

 Sex Education Male (M) Female (F) Elementary (E) 28 50 Secondary (S) 38 45 College (C) 22 17

If a person is selected at random from this group, then:

1) the probability that he is a male is:

(A) 0.3182        (B) 0.44          (C) 0.28          (D) 78

2) The probability that the person is male given that the person has a secondary

education is:

(A) 0.4318        (B) 0.4578      (C) 0.19          (D) 0.44

3) The probability that the person does not have a college degree given that the

person is a female is:

(A) 0.8482       (B) 0.1518      (C) 0.475        (D) 0.085

4) Are the events M and E independent? Why?  [P(M)=0.44 ¹ P(M|E)=0.359  Þ dependent]

Q9.  1000 individuals are classified below by sex and smoking habit.

 SEX Male (M) Female (F) SMOKING HABIT Daily  (D) 300 50 Occasionally  (O) 200 50 Not at all   (N) 100 300

A person is selected randomly from this group.

1. Find the probability that the person is female. [P(F)=0.4]
2. Find the probability that the person is female and smokes daily. [P(FÇD)=0.05]
3. Find the probability that the person is female, given that the person smokes daily. [P(F|D)=0.1429]
4. Are the events F and D independent? Why? [P(F)=0.4 ¹ P(F|D)=0.1429  Þ dependent]

Q10. Two engines operate independently, if the probability that an engine will start is 0.4, and the probability that other engine will start is 0.6, then the probability that both will start is:

(A) 1                (B) 0.24           (C) 0.2             (D) 0.5

Q11. If  and , then  equals to;

(A) 0.67           (B) 0.12           (C) 0.75           (D) 0.3

Q12. The probability that a computer system has an electrical failure is 0.15, and the probability that it has a virus is 0.25, and the probability that it has both problems is 0.10, then the probability that the computer system has the electrical failure or the virus is:

(A) 1.15           (B) 0.2             (C) 0.15           (D) 0.30

Q13. From a box containing 4 black balls and 2 green balls, 3 balls are drawn independently in succession, each ball being replaced in the box before the next draw is made. The probability of drawing 2 green balls and 1 black ball is:

(A) 6/27           (B) 2/27           (C) 12/27                     (D) 4/27

Q14. 80 students are enrolled in STAT-324 class. 60 students are from engineering college and the rest are from computer science college. 10% of the engineering college students have taken this course before, and 5% of the computer science college students have taken this course before. If one student from this class is randomly selected, then:

1) the probability that he has taken this course before is:

(A) 0.25           (B) 0.0875      (C) 0.8021      (D) 0. 75

2) If the selected student has taken this course before then the probability that he is from the

computer science college is:

(A) 0.1429       (B) 0.375        (C) 0.80          (D) 0. 25

Q15. Two machines A and B make 80% and 20%, respectively, of the products in a certain factory. It is known that 5% and 10% of the products made by each machine, respectively, are defective. A finished product is randomly selected.

1. Find the probability that the product is defective. [P(D)=0.06]
2. If the product were found to be defective, what is the probability that it was made by machine B. [P(B|D)=0.3333]

Q16. If P(A1)=0.4, P(A1ÇA2)=0.2, and P(A3|A1ÇA2)=0.75, then

(1) P(A2|A1) equals to

(A)        0.00     (B)       0.20     (C)       0.08     (D)       0.50

(2) P(A1ÇA2ÇA3) equals to

(A)        0.06     (B)       0.35     (C)       0.15     (D)       0.08

Q17. If P(A)=0.9, P(B)=0.6, and P(AÇB)=0.5, then:

(1) P(AÇBC) equals to

(A) 0.4                          (B) 0.1                         (C) 0.5                         (D) 0.3

(2) P(ACÇBC) equals to

(A) 0.2                          (B) 0.6                         (C) 0.0                         (D) 0.5

(3) P(B|A) equals to

(A) 0.5556                    (B) 0.8333                   (C) 0.6000                   (D) 0.0

(4) The events A and B are

(A) independent (B) disjoint                   (C) joint                       (D) none

(5) The events A and B are

(A) disjoint                    (B) dependent              (C) independent            (D) none

Q18. Suppose that the experiment is to randomly select with replacement 2 children and register their gender (B=boy, G=girl) from a family having 2 boys and 6 girls.

(1) The number of outcomes (elements of the sample space) of this experiment equals to

(A) 4                            (B) 6                (C) 5    (D)       125

(2) The event that represents registering at most one boy is

(A) {GG, GB, BG}      (B) {GB, BG}              (C)        (D) {GB, BG, BB}

(3) The probability of registering no girls equals to

(A) 0.2500                   (B) 0.0625                   (C) 0.4219       (D) 0.1780

(4) The probability of registering exactly one boy equals to

(A) 0.1406                   (B) 0.3750                   (C) 0.0141       (D) 0.0423

(5) The probability of registering at most one boy equals to

(A) 0.0156                   (B) 0.5000                   (C) 0.4219       (D) 0.9375

3. BAYES RULE:

Q1. 80 students are enrolled in STAT-324 class. 60 students are from engineering college and the rest are from computer science college. 10% of the engineering college students have taken this course before, and 5% of the computer science college students have taken this course before. If one student from this class is randomly selected, then:

1) the probability that he has taken this course before is:

(A) 0.25           (B) 0.0875      (C) 0.8021      (D) 0. 75

2) If the selected student has taken this course before then the probability that he

is from the computer science college is:

(A) 0.1429       (B) 0.375        (C) 0.80          (D) 0. 25

Q2. Two machines A and B make 80% and 20%, respectively, of the products in a certain factory. It is known that 5% and 10% of the products made by each machine, respectively, are defective. A finished product is randomly selected.

(a)    Find the probability that the product is defective. [P(D)=0.06]

(b) If the product were found to be defective, what is the probability that it was made by machine B. [P(B|D)=0.3333]

Q3. Dates' factory has three assembly lines, A, B, and C. Suppose that the assembly lines A, B, and C account for 50%, 30%, and 20% of the total product of the factory. Quality control records show that 4% of the dates packed by line A, 6% of the dates packed by line B, and 12% of the dates packed by line C are improperly sealed. If a pack is randomly selected, then:

 (a) the probability that the pack is from line B and it is improperly sealed is (A) 0.018 (B) 0.30 (C) 0.06 (D) 0.36 (E) 0.53 (b)  the probability that the pack is improperly sealed is (A) 0.62 (B) 0.022 (C) 0.062 (D) 0.22 (E) 0.25 (c)  if it is found that the pack is improperly sealed, what is the probability that it is from line B? (A) 0.0623 (B) 0.0223 (C) 0.6203 (D) 0.2203 (E) 0.2903

Q4. Two brothers, Mohammad and Ahmad own and operate a small restaurant. Mohammad washes 50% of the dishes and Ahmad washes 50% of the dishes. When Mohammad washes a dish, he might break it with probability 0.40. On the other hand, when Ahmad washes a dish, he might break it with probability 0.10. Then,

 (a) the probability that a dish will be broken during washing is: (A) 0.667 (B) 0.25 (C) 0.8 (D) 0.5 (b) If a broken dish was found in the washing machine, the probability that it was washed by Mohammad is: (A) 0.667 (B) 0.25 (C) 0.8 (D) 0.5

Q5. A vocational institute offers two training programs (A) and (B). In the last semester, 100 and 300 trainees were enrolled for programs (A) and (B), respectively. From the past experience it is known that the passing probabilities are 0.9 for program (A) and 0.7 for program (B). Suppose that at the end of the last semester, we selected a trainee at random from this institute.

(1) The probability that the selected trainee passed the program equals to

(A) 0.80                        (B) 0.75           (C) 0.85                       (D) 0.79

(2) If it is known that the selected trainee passed the program, then the probability that he has

been enrolled in program (A) equals to

(A) 0.8              (B) 0.9                         (C) 0.3                         (D) 0.7

4. RANDOM VARIABLES, DISTRIBUTIONS, EXPECTATIONS

AND CHEBYSHEV's THEOREM:

4.1. DISCRETE DISTRIBUTIONS:

Q1. Consider the experiment of flipping a balanced coin three times independently.

Let X= Number of heads – Number of tails.

(a) List the elements of the sample space S.

(b) Assign a value x of X to each sample point.

(c) Find the probability distribution function of X.

(d) Find P( X £ 1 )

(e) Find P( X < 1 )

(f) Find  m=E(X)

(g) Find  s2=Var(X)

Q2. It is known that 20% of the people in a certain human population are female. The experiment is to select a committee consisting of two individuals at random. Let X be a random variable giving the number of females in the committee.

1. List the elements of the sample space S.
2. Assign a value x of X to each sample point.
3. Find the probability distribution function of X.
4. Find the probability that there will be at least one female in the committee.
5. Find the probability that there will be at most one female in the committee.
6. Find  m=E(X)
7. Find  s2=Var(X)

Q3. A box contains 100 cards; 40 of which are labeled with the number 5 and the other cards are labeled with the number 10. Two cards were selected randomly with replacement and the number appeared on each card was observed. Let X be a random variable giving the total sum of the two numbers.

(i) List the elements of the sample space S.

(ii) To each element of S assign a value x of X.

(iii) Find the probability mass function (probability distribution function) of X.

(iv) Find P(X=0).

(v) Find P(X>10).

(vi) Find  m=E(X)

(vii) Find  s2=Var(X)

Q4.  Let X be a random variable with the following probability distribution:

 x -3 6 9 f(x) 0.1 0.5 0.4

1) Find the mean (expected value) of X, m=E(X).

2) Find E(X2).

3) Find the variance of X, Var (X) = .

4) Find the mean of 2X+1, .

5) Find the variance of 2X+1, Var(2X+1)= .

Q5. Which of the following is a probability distribution function:

(A) ; x=0,1,2,3,4             (B) ; x=0,1,2,3,4

(C) ; x=0,1,2,3,4                   (D) ; x=0,1,2,3

Q6. Let the random variable X have a discrete uniform with parameter k=3 and with values 0,1, and 2. The probability distribution function is:

f(x)=P(X=x)=1/3 ;  x=0, 1, 2.

 (1) The mean of X is (A) 1.0 (B) 2.0 (C) 1.5 (D) 0.0 (2) The variance of X is (A) 0.0 (B) 1.0 (C) 0.67 (D) 1.33

Q7. Let X be a discrete random variable with the probability distribution function:

f(x) = kx for x=1, 2, and 3.

(i) Find the value of k.

(ii) Find the cumulative distribution function (CDF), F(x).

(iii) Using the CDF, F(x), find P (0.5 < X £ 2.5).

Q8. Let X be a random variable with cumulative distribution function (CDF) given by:

(a) Find the probability distribution function of X, f(x).

(b) Find P( 1≤X<2). (using both f(x) and F(x))

(c) Find P( X>2). (using both f(x) and F(x))

Q9. Consider the random variable X with the following probability distribution function:

 X 0 1 2 3 f(x) 0.4 c 0.3 0.1

 The value of c is (A) 0.125 (B) 0.2 (C) 0.1 (D) 0.125 (E)   - 0.2

Q10. Consider the random variable X with the following probability distribution function:

 X -1 0 1 2 f(x) 0.2 0.3 0.2 c

Find the following:

(a)    The value of c.

(b)   P( 0 < X ≤ 2 )

(c)     m = E(X)

(d)   E(X2)

(e)    s2 = Var(X)

Q11. Find the value of k that makes the function

for x=0,1,2

serve as a probability distribution function of the discrete random variable X.

Q12. The cumulative distribution function (CDF) of a discrete random variable, X, is given below:

 (a) the is equal to: (A) 3/8 (B) 11/16 (C) 10/16 (D) 5/16 (b) the  is equal to: (A) 20/16 (B) 11/16 (C) 10/16 (D) 5/16

Q13. If a random variable X has a mean of 10 and a variance of 4, then, the random variable

,

 (a) has a mean of: (A) 10 (B) 18 (C) 20 (D) 22 (b) and a standard deviation of: (A) 6 (B) 2 (C) 4 (D) 16

Q14. Let X be the number of typing errors per page committed by a particular typist. The probability distribution function of X is given by:

 x 0 1 2 3 4 3k 3k 2k k k

(1) Find the numerical value of k.

(2) Find the average (mean) number of errors for this typist.

(3) Find the variance of the number of errors for this typist.

(4) Find the cumulative distribution function (CDF) of X.

(5) Find the probability that this typist will commit at least 2 errors per page.

Q15. Suppose that the discrete random variable X has the following probability function: f(-1)=0.05, f(0)=0.25, f(1)=0.25, f(2)=0.45, then:

(1) P(X<1) equals to

(A)        0.30     (B)       0.05     (C)       0.55     (D)       0.50

(2) P(X£1) equals to

(A)        0.05     (B)       0.55     (C)       0.30     (D)       0.45

(3) The mean m=E(X) equals to

(A)        1.1       (B)       0.0       (C)       1.2       (D)       0.5

(4) E(X2) equals to

(A)        2.00     (B)       2.10     (C)       1.50     (D)       0.75

(5) The variance s2=Var(X) equals to

(A)        1.00     (B)       3.31     (C)       0.89     (D)       2.10

(6) If F(x) is the cumulative distribution function (CDF) of X, then F(1) equals to

(A)        0.50     (B)       0.25     (C)       0.45     (D)       0.55

4.2. CONTINUOUS DISTRIBUTIONS:

Q1. If the continuous random variable X has mean μ=16 and variance σ2=5, then  is

(A)   0.0625   (B) 0.5            (C) 0.0            (D) None of these.

Q2. Consider the probability density function:

1) The value of k is:

(A) 1                 (B) 0.5            (C) 1.5            (D) 0.667

2) The probability  is,

(A) 0.4647       (B) 0.3004      (C) 0.1643      (D) 0.4500

3) The expected value of X, is,

(A) 0.6            (B) 1.5            (C) 1                (D) 0.667

[Hint: ]

Q3. Let X be a continuous random variable with the probability density function f(x)=k(x+1) for 0<x<2.

(i) Find the value of k.

(ii) Find P(0 < X £ 1).

(iii) Find the cumulative distribution function (CDF) of X, F(x) .

(iv) Using F(x), find P(0 < X £ 1).

Q4. Let X be a continuous random variable with the probability density function f(x)=3x2/2 for -1<x<1.

1. P(0<X<1) = …………..
2. E(X) = …………………
3. Var(X) = …………….…
4. E(2X+3)= ………………
5. Var(2X+3)= ……………

Q5.  Suppose that the random variable X has the probability density function:

1.      Evaluate k.

2.      Find the cumulative distribution function (CDF) of X, F(x).

3.      Find P(0 < X < 1).

4.      Find P(X = 1) and P(2 < X < 3).

Q6. Let X be a random variable with the probability density function:

1.      Find m = E(X).

2.      Find s2 = Var(X).

3.      Find E (4X+5).

4.      Find Var(4X+5).

Q7. If the random variable X has a uniform distribution on the interval (0,10) with the probability density function given by:

,

1. Find P(X<6).

2. Find the mean of X.

3. Find E(X2).

4. Find the variance of X.

5. Find the cumulative distribution function (CDF) of X, F(x).

6. Use the cumulative distribution function, F(x), to find P(1<X£5).

Q8. Suppose that the failure time (in months) of a specific type of electrical device is distributed with the probability density function:

 (a) the average failure time of such device is: (A) 6.667 (B) 1.00 (C) 2.00 (D) 5.00 (b) the variance of the failure time of such device is: (A) 0 (B) 50 (C) 5.55 (D) 10 (c) P(X>5) = (A) 0.25 (B) 0.55 (C) 0.65 (D) 0.75

Q9. If the cumulative distribution function of the random variable X having the form:

Then

(1)  equals to

(a) 0.555         (b) 0.333          (c) 0.867          (d) none of these.

(2) If , then k equals to

(a) 5     (b) 0.5              (c) 1                 (d) 1.5

Q10. If the diameter of a certain an electrical cable is a continuous random variable X ( in cm ) having the probability density function :

(1)   is:

a) 0.8125    b) 0.1875    c) 0.9844   d) 0.4445

(2)  is:

a) 0.8125    b) 0.1875   c) 0.9844   d) 0.4445

(3) m=E(x) is:

a) 0.667   b) 0.333    c) 0.555    d) none of these.

(4)  =Var(x) is:

a)0.3175    b) 3.175    c) 0.0317    d) 2.3175

(5) For this random variable,  will have an exact value equals to:

a) 0.3175    b) 0.750    c) 0.965    d) 0.250

(6) For this random variable,  will have a lower bound value

according to chebyshev’s theory equals to:

a) 0.3175    b) 0.750    c) 0.965    d) 0.250

(7) If  Y= 3X-1.5,  then E(Y) is:

a) – 0.5     b) – 0.3335     c) 0.5     d) none of these.

and Var(Y) is:

a) 2.8575   b) 0.951    c) 0.2853   d) 6.9525

4.3. CHEBYSHEV’S THEOREM :

Q1. According to Chebyshev’s theorem, for any random variable X with mean m and variance s2, a lower bound for  is,

(A) 0.3175       (B) 0.750         (C) 0.965         (D) 0.250

Note:

Q2. Suppose that X is a random variable with mean μ=12, variance σ2=9, and unknown probability distribution. Using Chebyshev’s theorem,  is at least equal to,

(A) 8/9             (B) 3/4             (C) 1/4             (D) 1/16

Q3. Suppose that E(X)=5 and Var(X)=4. Using Chebyshev's Theorem,

(i) find an approximated value of P(1<X<9).

(ii) find some constants a and b (a<b) such that P(a<X<b) » 15/16.

Q4. Suppose that the random variable X is distributed according to the probability density function given by:

,

Assuming m=5 and s = 2.89,

1. Find the exact value of P(m-1.5s < X <m+1.5s).

2. Using Chebyshev's Theorem, find an approximate value of P(m-1.5s < X <m+1.5s).

3. Compare the values in (1) and (2).

Q5. Suppose that X and Y are two independent random variables with E(X)=30, Var(X)=4, E(Y)=10, and Var(Y)=2. Then:

(1) E(2X-3Y-10) equals to

(A) 40               (B) 20                          (C) 30                          (D) 90

(2) Var(2X-3Y-10) equals to

(A) 34               (B) 24                          (C) 2.0                         (D) 14

(3) Using Chebyshev's theorem, a lower bound of  P(24<X<36) equals to

(A) 0.3333                    (B) 0.6666                   (C) 0.8888                   (D) 0.1111

DISCRETE UNIFORM DISTRIBUTION:

Q1. Let the random variable X have a discrete uniform with parameter k=3 and with values 0,1, and 2. Then:

(a)    P(X=1) is

(A) 1.0                   (B) 1/3             (C) 0.3             (D) 0.1             (E) None

(b)   The mean of X is:

(A) 1.0                   (B) 2.0             (C) 1.5             (D) 0.0             (E) None

(c)    The variance of X is:

(A) 0/3=0.0            (B) 3/3=1.0      (C) 2/3=0.67    (D) 4/3=1.33    (E) None

5. BINOMIAL DISTRIBUTION:

Q1. Suppose that 4 out of 12 buildings in a certain city violate the building code. A building engineer randomly inspects a sample of 3 new buildings in the city.

(a) Find the probability distribution function of the random variable X representing the

number of buildings that violate the building code in the sample.

(b) Find the probability that:

(i) none of the buildings in the sample violating the building code.

(ii) one building in the sample violating the building code.

(iii) at lease one building in the sample violating the building code.

(c) Find the expected number of buildings in the sample that violate the building code (E(X)).

(d) Find s2=Var(X).

Q2. A missile detection system has a probability of 0.90 of detecting a missile attack. If 4 detection systems are installed in the same area and operate independently, then

 (a) The probability that at least two systems detect an attack is (A) 0.9963 (B) 0.9477 (C) 0.0037 (D) 0.0523 (E) 0.5477 (b) The average (mean) number of systems detect an attack is (A) 3.6 (B) 2.0 (C) 0.36 (D) 2.5 (E) 4.0

Q3. Suppose that the probability that a person dies when he or she contracts a certain disease is 0.4. A sample of 10 persons who contracted this disease is randomly chosen.

(1) What is the expected number of persons who will die in this sample?

(2) What is the variance of the number of persons who will die in this sample?

(3) What is the probability that exactly 4 persons will die among this sample?

(4) What is the probability that less than 3 persons will die among this sample?

(5) What is the probability that more than 8 persons will die among this sample?

Q4. Suppose that the percentage of females in a certain population is 50%. A sample of 3 people is selected randomly from this population.

 (a) The probability that no females are selected is (A) 0.000 (B) 0.500 (C) 0.375 (D) 0.125 (b) The probability that at most two females are selected is (A) 0.000 (B) 0.500 (C) 0.875 (D) 0.125 (c) The expected number of females in the sample is (A) 3.0 (B) 1.5 (C) 0.0 (D) 0.50 (d) The variance of the number of females in the sample is (A) 3.75 (B) 2.75 (C) 1.75 (D) 0.75

Q5. 20% of the trainees in a certain program fail to complete the program. If 5 trainees of this program are selected randomly,

(i) Find the probability distribution function of the random variable X, where:

X = number of the trainees who fail to complete the program.

(ii) Find the probability that all trainees fail to complete the program.

(iii) Find the probability that at least one trainee will fail to complete the program.

(iv) How many trainees are expected to fail completing the program?

(v) Find the variance of the number of trainees who fail completing the program.

Q6. In a certain industrial factory, there are 7 workers working independently. The probability of accruing accidents for any worker on a given day is 0.2, and accidents are independent from worker to worker.

 (a) The probability that at most two workers will have accidents during the day is (A) 0.7865 (B) 0.4233 (C) 0.5767 (D) 0.6647 (b) The probability that at least three workers will have accidents during the day is: (A) 0.7865 (B) 0.2135 (C) 0.5767 (D) 0.1039 (c) The expected number workers who will have accidents during the day is (A) 1.4 (B) 0.2135 (C) 2.57 (D) 0. 59

Q7. From a box containing 4 black balls and 2 green balls, 3 balls are drawn independently in succession, each ball being replaced in the box before the next draw is made. The probability of drawing 2 green balls and 1 black ball is:

(A) 6/27           (B) 2/27           (C) 12/27                     (D) 4/27

Q8. The probability that a lab specimen is contaminated is 0.10. Three independent samples are checked.

1) the probability that none is contaminated is:

(A) 0.0475       (B) 0.001        (C) 0.729        (D) 0. 3

2) the probability that exactly one sample is contaminated is:

(A) 0.243         (B) 0.081        (C) 0.757        (D) 0. 3

Q9. If X~Binomial(n,p), E(X)=1, and Var(X)=0.75, find P(X=1).

Q10. Suppose that X~Binomial(3,0.2). Find the cumulative distribution function (CDF) of X.

Q11. A traffic control engineer reports that 75% of the cars passing through a checkpoint are from Riyadh city. If at this checkpoint, five cars are selected at random.

(1) The probability that none of them is from Riyadh city equals to:

(A) 0.00098                 (B) 0.9990       (C) 0.2373       (D) 0.7627

(2) The probability that four of them are from Riyadh city equals to:

(A)       0.3955             (B) 0.6045       (C) 0                (D) 0.1249

(3) The probability that at least four of them are from Riyadh city equals to:

(A)       0.3627             (B) 0.6328       (C) 0.3955       (D) 0.2763

(4) The expected number of cars that are from Riyadh city equals to:

(A) 1                                        (B) 3.75           (C) 3                (D) 0

6. HYPERGEOMETRIC DISTRIBUTION:

Q1. A shipment of 7 television sets contains 2 defective sets. A hotel makes a random purchase of 3 of the sets.

(i) Find the probability distribution function of the random variable X representing the number

of defective sets purchased by the hotel.

(ii) Find the probability that the hotel purchased no defective television sets.

(iii) What is the expected number of defective television sets purchased by the hotel?

(iv) Find the variance of X.

Q2. Suppose that a family has 5 children, 3 of them are girls and the rest are boys. A sample of 2 children is selected randomly and without replacement.

 (a) The probability that no girls are selected is (A) 0.0 (B) 0.3 (C) 0.6 (D) 0.1 (b) The probability that at most one girls are selected is (A) 0.7 (B) 0.3 (C) 0.6 (D) 0.1 (c) The expected number of girls in the sample is (A) 2.2 (B) 1.2 (C) 0.2 (D) 3.2 (d) The variance of the number of girls in the sample is (A) 36.0 (B) 3.6 (C) 0.36 (D) 0.63

Q3. A random committee of size 4 is selected from 2 chemical engineers and 8 industrial engineers.

(i)                  Write a formula for the probability distribution function of the random variable X representing the number of chemical engineers in the committee.

(ii)                Find the probability that there will be no chemical engineers in the committee.

(iii)               Find the probability that there will be at least one chemical engineer in the committee.

(iv)              What is the expected number of chemical engineers in the committee?

(v)                What is the variance of the number of chemical engineers in the committee?

Q4. A box contains 2 red balls and 4 black balls. Suppose that a sample of 3 balls were selected randomly and without replacement. Find,

1.      The probability that there will be 2 red balls in the sample.

2.      The probability that there will be 3 red balls in the sample.

3.      The expected number of the red balls in the sample.

Q5. From a lot of 8 missiles, 3 are selected at random and fired. The lot contains 2 defective missiles that will not fire. Let X be a random variable giving the number of defective missiles selected.

1. Find the probability distribution function of X.
2. What is the probability that at most one missile will not fire?
3. Find m = E(X)  and  s2 = Var(X).

Q6. A particular industrial product is shipped in lots of 20 items. Testing to determine whether an item is defective is costly; hence, the manufacturer samples production rather than using 100% inspection plan. A sampling plan constructed to minimize the number of defectives shipped to consumers calls for sampling 5 items from each lot and rejecting the lot if more than one defective is observed. (If the lot is rejected, each item in the lot is then tested.) If a lot contains 4 defectives, what is the probability that it will be accepted.

Q7. Suppose that X~h(x;100,2,60); i.e., X has a hypergeometric distribution with parameters N=100, n=2, and K=60. Calculate the probabilities P(X=0), P(X=1), and P(X=2) as follows:

(a) exact probabilities using hypergeometric distribution.

(b) approximated probabilities using binomial distribution.

Q8. A particular industrial product is shipped in lots of 1000 items. Testing to determine whether an item is defective is costly; hence, the manufacturer samples production rather than using 100% inspection plan. A sampling plan constructed to minimize the number of defectives shipped to consumers calls for sampling 5 items from each lot and rejecting the lot if more than one defective is observed. (If the lot is rejected, each item in the lot is then tested.) If a lot contains 100 defectives, calculate the probability that the lot will be accepted using:

(a) hypergeometric distribution (exact probability.)

(b) binomial distribution (approximated probability.)

Q9. A shipment of 20 digital voice recorders contains 5 that are defective. If 10 of them are randomly chosen (without replacement) for inspection, then:

(1) The probability that 2 will be defective is:

(A) 0.2140                    (B) 0.9314                   (C) 0.6517                   (D) 0.3483

(2) The probability that at most 1 will be defective is:

(A) 0.9998                    (B) 0.2614                   (C) 0.8483                   (D) 0.1517

(3) The expected number of defective recorders in the sample is:

(A) 1                                         (B) 2                (C) 3.5                         (D) 2.5

(4) The variance of the number of defective recorders in the sample is:

(A) 0.9868                                (B) 2.5             (C) 0.1875                   (D) 1.875

Q10. A box contains 4 red balls and 6 green balls. The experiment is to select 3 balls at random. Find the probability that all balls are red for the following cases:

(1) If selection is without replacement

(A) 0.216                      (B) 0.1667                   (C) 0.6671                   (D) 0.0333

(2) If selection is with replacement

(A) 0.4600                    (B) 0.2000                   (C) 0.4000                   (D) 0.0640

7. POISSON DISTRIBUTION:

Q1. On average, a certain intersection results in 3 traffic accidents per day. Assuming Poisson distribution,

(i) what is the probability that at this intersection:

(1) no accidents will occur in a given day?

(2) More than 3 accidents will occur in a given day?

(3) Exactly 5 accidents will occur in a period of two days?

(ii) what is the average number of traffic accidents in a period of 4 days?

Q2. At a checkout counter, customers arrive at an average of 1.5 per minute. Assuming Poisson distribution, then

 (1) The probability of no arrival in two minutes is (A) 0.0 (B) 0.2231 (C) 0.4463 (D) 0.0498 (E) 0.2498 (2) The variance of the number of arrivals in two minutes is (A) 1.5 (B) 2.25 (C) 3.0 (D) 9.0 (E) 4.5

Q3. Suppose that the number of telephone calls received per day has a Poisson distribution with mean of 4 calls per day.

 (a) The probability that 2 calls will be received in a given day is (A) 0.546525 (B) 0.646525 (C) 0.146525 (D) 0.746525 (b) The expected number of telephone calls received in a given week is (A) 4 (B) 7 (C) 28 (D) 14 (c) The probability that at least 2 calls will be received in a period of 12 hours is (A) 0.59399 (B) 0.19399 (C) 0.09399 (D) 0.29399

Q4. The average number of car accidents at a specific traffic signal is 2 per a week. Assuming Poisson distribution, find the probability that:

(i) there will be no accident in a given week.

(ii) there will be at least two accidents in a period of two weeks.

Q5. The average number of airplane accidents at an airport is two per a year. Assuming Poisson distribution, find

1. the probability that there will be no accident in a year.

2. the average number of airplane accidents at this airport in a period of two years.

3. the probability that there will be at least two accidents in a period of 18 months.

Q6. Suppose that X~Binomial(1000,0.002). By using Poisson approximation, P(X=3) is approximately equal to (choose the nearest number to your answer):

(A) 0.62511     (B) 0.72511     (C) 0.82511     (D) 0.92511     (E) 0.18045

Q7. The probability that a person dies when he or she contracts a certain disease is 0.005. A sample of 1000 persons who contracted this disease is randomly chosen.

(1) What is the expected number of persons who will die in this sample?

(2) What is the probability that exactly 4 persons will die among this sample?

Q8. The number of faults in a fiber optic cable follows a Poisson distribution with an average of 0.6 per 100 feet.

(1) The probability of 2 faults per 100 feet of such cable is:

(A) 0.0988                    (B) 0.9012                   (C) 0.3210                   (D) 0.5

(2) The probability of less than 2 faults per 100 feet of such cable is:

(A) 0.2351                    (B) 0.9769                   (C) 0.8781                   (D) 0.8601

(3) The probability of 4 faults per 200 feet of such cable is:

(A) 0.02602                  (B) 0.1976                   (C) 0.8024                   (D) 0.9739

CONTINOUS UNIFORM DISTRIBUTION:

Q1. If the random variable X has a uniform distribution on the interval (0,10), then

1. P(X<6) equals to

(A) 0.4             (B) 0.6             (C) 0.8                         (D) 0.2             (E) 0.1

2. The mean of X is

(A) 5                (B) 10              (C) 2                            (D) 8                (E) 6

2. The variance X is

(A) 33.33         (B) 28.33         (C) 8.33                       (D) 25              (E) None

Q2. Suppose that the random variable X has the following uniform distribution:

(1)

(A) 0.49                        (B) 0.51                       (C) 0                (D) 3

(2)

(A) 0                                         (B) 1                            (C) 0.5             (D) 0.33

(3)  The variance of X is

(A) 0.00926                  (B) 0.333                     (C) 9                (D) 0.6944

Q3. Suppose that the continuous random variable X has the following probability density function (pdf): f(x)=0.2 for 0<x<5. Then

(1) P(X>1) equals to

(A)        0.4       (B)       0.2                   (C)       0.1                   (D)       0.8

(2) P(X³1) equals to

(A)        0.05     (B)       0.8                   (C)       0.15                 (D)       0.4

(3)  The mean m=E(X) equals to

(A)        2.0       (B)       2.5                   (C)       3.0                   (D)       3.5

(4) E(X2) equals to

(A)        8.3333 (B)       7.3333             (C)       9.3333             (D)       6.3333

(5) Var(X) equals to

(A)        8.3333 (B)       69.444             (C)       5.8333             (D)       2.0833

(6) If F(x) is the cumulative distribution function (CDF) of X, then F(1) equals to

(A)        0.75     (B)       0.25                 (C)       0.8                   (D)       0.2

8. NORMAL DISTRIBUTION:

Q1. (A) Suppose that Z is distributed according to the standard normal distribution.

1) the area under the curve to the left of  is:

(A) 0.0764       (B) 0.9236       (C) 0    (D) 0.8133

2) the area under the curve to the left of  z = 1.39 is:

(A) 0.7268       (B) 0.9177       (C) .2732         (D) 0.0832

3) the area under the curve to the right of is:

(A) 0. 7815      (B) 0.8133      (C) 0.1867      (D) 0.0154

4) the area under the curve between  and  is:

(A) 0.7576       (B) 0.8665      (C) 0.0154      (D) 0.2424

5) the value of  k such that  is:

(A) 0.8665        (B) -1.11        (C) 1.11          (D) 1.00

(B) Suppose that Z is distributed according to the standard normal distribution. Find:

1) P(Z< -3.9)

2) P(Z> 4.5)

1) P(Z< 3.7)

2) P(Z> -4.1)

Q2. The finished inside diameter of a piston ring is normally distributed with a mean of 12 centimeters and a standard deviation of 0.03 centimeter. Then,

1) the proportion of rings that will have inside diameter less than 12.05

centimeters is:

(A) 0.0475       (B) 0.9525      (C) 0.7257      (D) 0.8413

2) the proportion of rings that will have inside diameter exceeding 11.97

centimeters is:

(A) 0.0475       (B) 0.8413      (C) 0.1587      (D) 0.4514

3) the probability that a piston ring will have an inside diameter between 11.95

and 12.05 centimeters is:

(A) 0.905         (B) -0.905      (C) 0.4514      (D) 0.7257

Q3. The average life of a certain type of small motor is 10 years with a standard deviation of 2 years. Assume the live of the motor is normally distributed. The manufacturer replaces free all motors that fail while under guarantee. If he is willing to replace only 1.5% of the motors that fail, then he should give a guarantee of :

(A) 10.03 years            (B) 8 years       (C) 5.66 years (D) 3 years

Q4. A machine makes bolts (that are used in the construction of an electric transformer). It produces bolts with diameters (X) following a normal distribution with a mean of 0.060 inches and a standard deviation of 0.001 inches. Any bolt with diameter less than 0.058 inches or greater than 0.062 inches must be scrapped. Then

 (1) The proportion of bolts that must be scrapped is equal to (A) 0.0456 (B) 0.0228 (C) 0.9772 (D) 0.3333 (E) 0.1667 (2) If P(X>a)= 0.1949, then a equals to: (A) 0.0629 (B) 0.0659 (C) 0.0649 (D) 0.0669 (E) 0.0609

Q5. The diameters of ball bearings manufactured by an industrial process are normally distributed with a mean m = 3.0 cm and a standard deviation s = 0.005 cm. All ball bearings with diameters not within the specifications m ± d cm (d > 0) will be scrapped.

(1) Determine the value of d such that 90% of ball bearings manufactured by this process will

not be scrapped.

(2) If d = 0.005, what is the percentage of manufactured ball bearings that will be scraped?

Q6. The weight of a large number of fat persons is nicely modeled with a normal distribution with mean of 128 kg and a standard deviation of 9 kg.

 (1) The percentage of fat persons with weights at most 110 kg is (A) 0.09 % (B) 90.3 % (C) 99.82 % (D) 2.28 % (2) The percentage of fat persons with weights more than 149 kg is (A) 0.09 % (B) 0.99 % (C) 9.7 % (D) 99.82 % (3) The weight x above which 86% of those persons will be (A) 118.28 (B) 128.28 (C) 154.82 (D) 81.28 (4) The weight x below which 50% of those persons will be (A) 101.18 (B) 128 (C) 154.82 (D) 81

Q7. The random variable X, representing the lifespan of a certain electronic device, is normally distributed with a mean of 40 months and a standard deviation of 2 months. Find

1. P(X<38).                                                         (0.1587)
2. P(38<X<40).                                                   (0.3413)
3. P(X=38).                                                         (0.0000)
4. The value of x such that P(X<x)=0.7324.          (41.24)

Q8. If the random variable X has a normal distribution with the mean m and the variance s2, then P(X<m+2s) equals to

(A) 0.8772       (B) 0.4772       (C) 0.5772                   (D) 0.7772       (E) 0.9772

Q9. If the random variable X has a normal distribution with the mean m and the variance 1, and if P(X<3)=0.877, then m equals to

(A) 3.84           (B) 2.84           (C) 1.84           (D) 4.84           (E) 8.84

Q10. Suppose that the marks of the students in a certain course are distributed according to a normal distribution with the mean 70 and the variance 25. If it is known that 33% of the student failed the exam, then the passing mark x is

(A) 67.8          (B) 60.8           (C) 57.8                       (D) 50.8           (E) 70.8

Q11. If the random variable X has a normal distribution with the mean 10 and the variance 36, then

1. The value of X above which an area of 0.2296 lie is

(A) 14.44         (B) 16.44         (C) 10.44                     (D) 18.44         (E) 11.44

2. The probability that the value of X is greater than 16 is

(A) 0.9587       (B) 0.1587       (C) 0.7587                   (D) 0.0587       (E) 0.5587

Q12. Suppose that the marks of the students in a certain course are distributed according to a normal distribution with the mean 65 and the variance 16. A student fails the exam if he obtains a mark less than 60. Then the percentage of students who fail the exam is

(A) 20.56%      (B) 90.56%      (C) 50.56%                  (D) 10.56%                  (E)40.56%

Q13. The average rainfall in a certain city for the month of March is 9.22 centimeters. Assuming a normal distribution with a standard deviation of 2.83 centimeters, then the probability that next March, this city will receive:

(1) less than 11.84 centimeters of rain is:

(A) 0.8238                    (B) 0.1762                   (C) 0.5             (D) 0.2018

(2) more than 5 centimeters but less than 7 centimeters of rain is:

(A) 0.8504                    (B) 0.1496                   (C) 0.6502       (D) 0.34221

(3) more than 13.8 centimeters  of rain is:

(A) 0.0526                    (B) 0.9474                   (C) 0.3101       (D) 0.4053

9. EXPONENTIAL DISTRIBUTION

Q1. If the random variable X has an exponential distribution with the mean 4, then:

1. P(X<8) equals to

(A) 0.2647       (B) 0.4647       (C) 0.8647                   (D) 0.6647       (E) 0.0647

2. The variance of X is

(A) 4                (B) 16              (C) 2                            (D) 1/4             (E) 1/2

Q2. Suppose that the failure time (in hours) of a certain electrical device is distributed with a probability density function given by:

1) the probability that a randomly selected device will fail within the first 50 hours is:

(A) 0.4995       (B) 0.7001       (C) 0.5105      (D) 0.2999

2) the probability that a randomly selected device will last more than 150 hours is:

(A) 0.8827       (B) 0.2788      (C) 0.1173      (D) 0.8827

3) the average failure time of the electrical device is:

(A) 1/70           (B) 70            (C) 140            (D) 35

4) the variance of the failure time of the electrical device is:

(A) 4900          (B) 1/49000    (C) 70             (D) 1225

[Hint: ]

Q3. The lifetime of a specific battery is a random variable X with probability density function given by:

 (1) The mean life time of the battery equals to (A) 200 (B) 1/200 (C) 100 (D) 1/100 (E) Non of these (2) P(X>100) = (A) 0.5 (B) 0.6065 (C) 0.3945 (D) 0.3679 (E) 0.6321 (3) P(X=200) = (A) 0.5 (B) 0.0 (C) 0.3945 (D) 0.3679 (E) 1.0

Q4. Suppose that the lifetime of a certain electrical device is given by T. The random variable T is modeled nicely by an exponential distribution with mean of 6 years. A random sample of four of these devices are installed in different systems. Assuming that these devices work independently, then:

 (1) the variance of the random variable  T  is (A) 136 (B) (36)2 (C) 6 (D) 36 (2) the probability that at most one of the devices in the sample will be functioning more than 6 years is (A) 0.4689 (B) 0.6321 (C) 0.5311 (D) 0.3679 (3) the probability that at least two of the devices in the sample will be functioning more than 6 years is (A) 0.4689 (B) 0.6321 (C) 0.5311 (D) 0.3679 (4) the expected number of devices in the sample which will be functioning more than  6 years is approximately equal to (A) 3.47 (B) 1.47 (C) 4.47 (D) 1.47

Q5. Assume the length (in minutes) of a particular type of a telephone conversation is a random variable with a probability density function of the form:

1.  is:

(a) 0.587    (b) -0.413        (c) 0.413          (d) 0.758

1. For this random variable,  will have an exact value equals:

(a) 0.250    (b) 0.750          (c) 0.950          (d) 0.3175

1. For this random variable,  will have a lower bound valued according to chebyshev’s theory equals:

(a) 0.750    (b) 0.250          (c) 0.950          (d) 0.3175

Q6. The length of time for one customer to be served at a bank is a random variable X that follows the exponential distribution with a mean of 4 minutes.

(1) The probability that a customer will be served in less than 2 minutes is:

(A) 0.9534                    (B) 0.2123                   (C) 0.6065                   (D) 0.3935

(2) The probability that a customer will be served in more than 4 minutes is:

(A)        0.6321             (B) 0.3679                   (C) 0.4905                   (D) 0.0012

(3)  The probability that a customer will be served in more than  2 but less than  5 minutes is:

(A) 0.6799                    (B) 0.32                       (C) 0.4018                   (D) 0.5523

(4) The variance of service time at this bank is

(A) 2                             (B) 4                            (C) 8                            (D) 16

10. SAMPLING DISTRIBUTIONS

Q.1 A machine is producing metal pieces that are cylindrical in shape. A sample was taken and the diameters were found to be 1.70, 2.11, 2.20, 2.31 and 2.28 centimeters. Then,

(a) The sample mean is:

(A) 2.22           (B) 2.32           (C) 2.90           (D) 2.20           (E) None of these

(b) The sample variance is:

(A) 0.597         (B) 0.285         (C) 0.356         (D) 0.534         (E) None of these

10.1. Single Mean:

Q1. A machine is producing metal pieces that are cylindrical in shape. A random sample of size 5 is taken and the diameters are 1.70, 2.11, 2.20, 2.31 and 2.28 centimeters. Then,

1) The sample mean is:

(A) 2.12           (B) 2.32          (C) 2.90          (D) 2.20           (E) 2.22

2) The sample variance is:

(A) 0.59757     (B) 0.28555    (C) 0.35633    (D) 0.06115     (E) 0.53400

Q2. The average life of a certain battery is 5 years, with a standard deviation of 1 year. Assume that the live of the battery approximately follows a normal distribution.

1) The sample mean  of a random sample of 5 batteries selected from this product has a mean

equal to:

(A) 0.2              (B) 5              (C) 3                (D) None of these

2) The variance  of the sample mean  of a random sample of 5 batteries

selected from this product is equal to:

(A) 0.2              (B) 5              (C) 3                (D) None of these

3) The probability that the average life of a random sample of size 16 of such batteries will be

between 4.5 and 5.4 years is:

(A) 0.1039        (B) 0.2135      (C) 0.7865      (D) 0.9224

4) The probability that the average life of a random sample of size 16 of such batteries will be

less than 5.5 years is:

(A) 0.9772        (B) 0.0228      (C) 0.9223      (D) None of these

5) The probability that the average life of a random sample of size 16 of such batteries will be

more than 4.75 years is:

(A) 0.8413       (B) 0.1587      (C) 0.9452      (D) None of these

6) If  where  represents the sample mean for a random sample of size

9 of such batteries, then the numerical value of  is:

(A) 4.653         (B) 6.5            (C) 5.347        (D) None of these

Q3. The random variable X, representing the lifespan of a certain light bulb, is distributed normally with a mean of 400 hours and a standard deviation of 10 hours.

1. What is the probability that a particular light bulb will last for more than 380 hours?
2. Light bulbs with lifespan less than 380 hours are rejected. Find the percentage of light bulbs that will be rejected.
3. If 9 light bulbs are selected randomly, find the probability that their average lifespan will be less than 405.

Q4. Suppose that you take a random sample of size n=64 from a distribution with mean m=55 and standard deviation s=10. Let  be the sample mean.

(a)    What is the approximated sampling distribution of ?

(b)   What is the mean of ?

(c)    What is the standard error (standard deviation) of ?

(d)   Find the probability that the sample mean  exceeds 52.

Q5. The amount of time that customers using ATM (Automatic Teller Machine) is a random variable with the mean 3.0 minutes and the standard deviation of 1.4 minutes. If a random sample of 49 customers is observed, then

 (1) the probability that their mean time will be at least 2.8 minutes is (A) 1.0 (B) 0.8413 (C) 0.3274 (D) 0.4468 (2) the probability that their mean time will be between 2.7 and 3.2 minutes is (A) 0.7745 (B) 0.2784 (C) 0.9973 (D) 0.0236

Q6. The average (mean) life of a certain type of batteries is 5 years, with a standard deviation of 1 year. Assuming the live of the battery follows approximately a normal distribution.

(a)    If a random sample of 5 batteries (to be selected from this production) has a mean of 3 years with a standard deviation of one year, then: the random variable  (the mean of all possible samples of size 5  batteries) has a mean  equal to:

(A) 0.2              (B) 5               (C) 3                (D) None of these

(b)   The variance of the random variable , , is equal to:

(A) 0.2              (B) 5              (C) 3                (D) None of these

(c)    The probability that the mean life of a random sample of size 16 of such batteries  will be between 4.5 and 5.4 years is:

(A) 0.1039        (B) 0.2135      (C) 0.7865      (D) None of these

(d)   The probability that the mean life of a random sample of size 16 of such batteries  will be less than 5.5 years is:

(A) 0.9772        (B) 0.0228      (C) 0.9223      (D) None of these

(e)    If  where  represents the sample mean for a random sample of size 9 to be selected  from the production lot, then the numerical value of  is:

(A) 4.653         (B) 6.5            (C) 5.347        (D) None of these

Q7. The average life of an industrial machine is 6 years, with a standard deviation of 1 year. Assume the life of such machines follows approximately a normal distribution. A random sample of 4 of such machines is selected. The sample mean life of the machines in the sample is .

(1) The sample mean has a mean  equals to:

(A) 5                            (B) 6                (C) 7                            (D) 8

(2) The sample mean has a variance  equals to:

(A) 1                            (B) 0.5             (C) 0.25                       (D) 0.75

(3)

(A) 0.4602                   (B) 0.8413       (C) 0.1587                   (D) 0.5398

(4) If , then the numerical value of  is:

(A) 0.8508                   (B) 1.04           (C) 6.52                       (D) 0.2

10.2. Two Means:

Q1. A random sample of size n1 = 36 is taken from a normal population with a mean m1 = 70 and a standard deviation s1 = 4. A second independent random sample of size n2 = 49 is taken from a normal population with a mean m2 = 85 and a standard deviation s2 = 5. Let  and  be the averages of the first and second samples, respectively.

a) Find E( ) and Var( ).

b) Find E( ) and Var( ).

c) Find P( 70 <  < 71).

d) Find P(   > -16).

Q2. A random sample of size 25 is taken from a normal population (first population) having a mean of 100 and a standard deviation of 6. A second random sample of size 36 is taken from a different normal population (second population) having a mean of 97 and a standard deviation of 5. Assume that these two samples are independent.

 (1) the probability that the sample mean of the first sample will exceed the sample mean of the second population by at least 6 is (A) 0.0013 (B) 0.9147 (C) 0.0202 (D) 0.9832 (2) the probability that the difference between the two sample means will be less than 2 is (A) 0.099 (B) 0.2483 (C) 0.8499 (D) 0.9499

10.3. Single Proportion:

Q1. Suppose that 20% of the students in a certain university smoke cigarettes. A random sample of 5 students is taken from this university. Let  be the proportion of smokers in the sample.

(1) Find , the mean .

(2) Find , the variance of .

(3) Find an approximate distribution of .

(4) Find P( >0.25).

Q2: Suppose that you take a random sample of size n=100 from a binomial population with parameter p=0.25 (proportion of successes). Let  be the sample proportion of successes, where X is the number of successes in the sample.

(a)    What is the approximated sampling distribution of ?

(b)   What is the mean of ?

(c)    What is the standard error (standard deviation) of ?

(d)   Find the probability that the sample proportion  is less than 0.2.

10.4. Two Proportions:

Q1. Suppose that 25% of the male students and 20% of the female students in a certain university smoke cigarettes. A random sample of 5 male students is taken. Another random sample of 10 female students is independently taken from this university. Let  and  be the proportions of smokers in the two samples, respectively.

(1) Find , the mean of .

(2) Find , the variance of .

(3) Find an approximate distribution of .

(4) Find P(0.10< <0.20).

10.5 t-distribution:

Q1. Using t-table with degrees of freedom df=14, find , .

Q2. From the table of t-distribution with degrees of freedom , the value of  equals to

 (A) 2.131 (B) 1.753 (C) 3.268 (D) 0

11. ESTIMATION AND CONFIDENCE INTERVALS:

11.1. Single Mean:

Q1. An electrical firm manufacturing light bulbs that have a length of life that is normally distributed with a standard deviation of 30 hours. A sample of 50 bulbs were selected randomly and found to have an average of 750 hours. Let m be the population mean of life lengths of all bulbs manufactured by this firm.

(1) Find a point estimate for m.

(2) Construct a 94% confidence interval for m.

Q2. Suppose that we are interested in making some statistical inferences about the mean, m, of a normal population with standard deviation s=2.0. Suppose that a random sample of size n=49 from this population gave a sample mean   =4.5.

 (1) The distribution of  is (A) N(0,1) (B) t(48) (C) N(m, 0.2857) (D) N(m, 2.0) (E) N(m, 0.3333) (2) A good point estimate of m is (A) 4.50 (B) 2.00 (C) 2.50 (D) 7.00 (E) 1.125 (3) The standard error of  is (A) 0.0816 (B) 2.0 (C) 0.0408 (D) 0.5714 (E) 0.2857 (4) A 95% confidence interval for m is (A) (3.44,5.56) (B) (3.34,5.66) (C) (3.54,5.46) (D) (3.94,5.06) (E) (3.04,5.96) (5) If the upper confidence limit of a confidence interval is 5.2, then the lower confidence limit is (A) 3.6 (B) 3.8 (C) 4.0 (D) 3.5 (E) 4.1 (6) The confidence level of the confidence interval (3.88, 5.12)  is (A) 90.74% (B) 95.74% (C) 97.74% (D) 94.74% (E) 92.74% (7) If we use  to estimate m, then we are 95% confident that our estimation error will not exceed (A) e=0.50 (B) E=0.59 (C) e=0.58 (D) e=0.56 (E) e=0.51 (8) If we want to be 95% confident that the estimation error will not exceed e=0.1 when we use  to estimate m, then the sample size n must be equal to (A) 1529 (B) 1531 (C) 1537 (D) 1534 (E) 1530

Q3. The following measurements were recorded for lifetime, in years, of certain type of machine: 3.4, 4.8, 3.6, 3.3, 5.6, 3.7, 4.4, 5.2, and 4.8. Assuming that the measurements represent a random sample from a normal population, then a 99% confidence interval for the mean life time of the machine is

(A)              (B)

(C)                  (D)

Q4. A researcher wants to estimate the mean lifespan of a certain light bulbs. Suppose that the distribution is normal with standard deviation of 5 hours.

1.      Determine the sample size needed on order that the researcher will be 90% confident that the error will not exceed 2 hours when he uses the sample mean as a point estimate for the true mean.

2.      Suppose that the researcher selected a random sample of 49 bulbs and found that the sample mean is 390 hours.

(i)                  Find a good point estimate for the true mean m.

(ii)                Find a 95% confidence interval for the true mean m.

Q5. The amount of time that customers using ATM (Automatic Teller Machine) is a random variable with a standard deviation of 1.4 minutes. If we wish to estimate the population mean m by the sample mean , and if we want to be 96% confident that the sample mean will be within 0.3 minutes of the population mean, then the sample size  needed is

 (A) 98 (B) 100 (C) 92 (D) 85

Q6: A random sample of size n=36 from a normal quantitative population produced a mean  and a variance .

(a)    Give a point estimate for the population mean m.

(b)   Find a 95% confidence interval for the population mean m.

Q7. A group of 10 college students were asked to report the number of hours that they spent doing their homework during the previous weekend and the following results were obtained:

7.25, 8.5, 5.0, 6.75, 8.0, 5.25, 10.5, 8.5, 6.75, 9.25

, }

It is assumed that this sample is a random sample from a normal distribution with unknown variance s2. Let m be the mean of the number of hours that the college student spend doing his/her homework during the weekend.

(a)    Find the sample mean and the sample variance.

(b)   Find a point estimate for m.

(c)    Construct a 80% confidence interval for m.

Q8. An electronics company wanted to estimate its monthly operating expenses in thousands riyals (m). It is assumed that the monthly operating expenses (in thousands riyals) are distributed according to a normal distribution with variance s2=0.584.

(I) Suppose that a random sample of 49 months produced a sample mean .

(a)    Find a point estimate for m.

(b)   Find the standard error of .

(c)    Find a 90% confidence interval for m.

(II) Suppose that they want to estimate m by . Find the sample size (n) required if they want

their estimate to be within 0.15 of the actual mean with probability equals to 0.95.

Q9. The tensile strength of a certain type of thread is approximately normally distributed with standard deviation of 6.8 kilograms. A sample of 20 pieces of the thread has an average tensile strength of 72.8 kilograms. Then,

(a) A point estimate of the population mean of the tensile strength (m) is:

(A) 72.8           (B) 20              (C) 6.8             (D) 46.24         (E) None of these

(b) Suppose that we want to estimate the population mean (m) by the sample mean . To be

95% confident that the error of our estimate of the mean of tensile strength will be less than

3.4 kilograms, the minimum sample size should be at least:

(A) 4                (B) 16              (C) 20              (D) 18              (E) None of these

(c) For a 98% confidence interval for the mean of tensile strength, we have the lower bound

equal to:

(A) 68.45         (B) 69.26         (C) 71.44         (D) 69.68         (E) None of these

(d) For a 98% confidence interval for the mean of tensile strength, we have the upper bound

equal to:

(A) 74.16         (B) 77.15         (C) 75.92         (D) 76.34         (E) None of these

11.2. Two Means:

Q1.(I) The tensile strength of type I thread is approximately normally distributed with standard deviation of 6.8 kilograms. A sample of 20 pieces of the thread has an average tensile strength of 72.8 kilograms. Then,

1) To be 95% confident that the error of estimating the mean of tensile strength by the sample

mean will be less than 3.4 kilograms, the minimum sample size should be:

(A) 4     (B) 16              (C) 20             (D) 18              (E) None of these

2) The lower limit of a 98% confidence interval for the mean of tensile strength is

(A) 68.45         (B) 69.26        (C) 71.44        (D) 69.68         (E) None of these

3) The upper limit of a 98% confidence interval for the mean of tensile strength is

(A) 74.16         (B) 77.15         (C) 75.92        (D) 76.34         (E) None of these

Q1.(II). The tensile strength of type II thread is approximately normally distributed with standard deviation of 6.8 kilograms. A sample of 25 pieces of the thread has an average tensile strength of 64.4 kilograms. Then for the 98% confidence interval of the difference in tensile strength means between type I and type II , we have:

1) the lower bound equals to:

(A) 2.90           (B) 4.21          (C) 3.65          (D) 6.58           (E) None of these

2) the upper bound equals to:

(A) 13.90         (B) 13.15        (C) 12.59        (D) 10.22         (E) None of these

Q2. Two random samples were independently selected from two normal populations with equal variances. The results are summarized as follows.

 First Sample Second Sample sample size (n) sample mean ( ) sample variance (S2) 12 10.5 4 14 10.0 5

Let m1 and m2 be the true means of the first and second populations, respectively.

1. Find a point estimate for m1 - m2..
2. Find 95% confidence interval for m1 - m2.

Q3. A researcher was interested in comparing the mean score of female students, mf, with the mean score of male students, mm, in a certain test. Two independent samples gave the following results:

 Sample Observations mean Variance Scores of Females 89.2 81.6 79.6 80.0 82.8 82.63 15.05 Scores of Males 83.2 83.2 84.8 81.4 78.6 71.5 77.6 80.04 20.79

Assume the populations are normal with equal variances.

 (1) The pooled estimate of the variance  is (A) 17.994 (B) 17.794 (C) 18.094 (D) 18.294 (E) 18.494 (2) A point estimate of  mf - mm  is (A) 2.63 (B) -2.59 (C) 2.59 (D) 0.00 (E) 0.59 (3) The lower limit of a 90% confidence interval for mf - mm  is (A) -1.97 (B) -1.67 (C) 1.97 (D) 1.67 (E) -1.57 (4) The upper limit of a 90% confidence interval for mf - mm  is (A) 6.95 (B) 7.45 (C) -7.55 (D) 7.15 (E) 7.55

Q4. A study was conducted to compare to brands of tires A and B. 10 tires of brand A and 12 tires of brand B were selected randomly. The tires were run until they wear out. The results are:

Brand A:           = 37000 kilometers          SA = 5100

Brand B:            = 38000 kilometers          SB = 6000

Assuming the populations are normally distributed with equal variances,

(1) Find a point estimate for mAmB.

(2) Construct a 90% confidence interval for mAmB.

Q5. The following data show the number of defects of code of particular type of software program made in two different countries (assuming normal populations with unknown equal variances)

 Country observations mean standard dev. A 48 39 42 52 40 48 54 46.143 5.900 B 50 40 43 45 50 38 36 43.143 5.551

(a) A point estimate of  is

(A) 3.0             (B) -3.0          (C) 2.0             (D) -2.0           (E) None of these

(b) A 90% confidence interval for the difference between the two population means

is

(A)                 (B)

(C)               (D)

Q6. A study was made by a taxi company to decide whether the use of new tires (A) instead of the present tires (B) improves fuel economy. Six cars were equipped with tires (A) and driven over a prescribed test course. Without changing drivers and cares, test course was made with tires (B). The gasoline consumption, in kilometers per liter (km/L), was recorded as follows: (assume the populations are normal with equal unknown variances)

 Car 1 2 3 4 5 6 Type (A) 4.5 4.8 6.6 7 6.7 4.6 Type (B) 3.9 4.9 6.2 6.5 6.8 4.1

 (a) A 95% confidence interval for the true mean gasoline consumption for brand A is: (A) (B) (C) (D) (b) A 99% confidence interval for the difference between the true means consumption of type (A) and type (B) (  is: (A) (B) (C) (D)

Q7. A geologist collected 20 different ore samples, all of the same weight, and randomly divided them into two groups. The titanium contents of the samples, found using two different methods, are listed in the table:

 Method (A) Method (B) 1.1   1.3   1.3   1.5   1.4 1.3   1.0   1.3   1.1   1.2 1.1   1.6   1.3   1.2   1.5 1.2   1.7   1.3   1.4   1.5

(a) Find a point estimate of  is

(b) Find a 90% confidence interval for the difference between the two population means

. (Assume two normal populations with equal variances).

11.3. Single Proportion:

Q1. A random sample of 200 students from a certain school showed that 15 students smoke. Let p be the proportion of smokers in the school.

1. Find a point Estimate for p.
2. Find 95% confidence interval for p.

Q2. A researcher was interested in making some statistical inferences about the proportion of females (p) among the students of a certain university. A random sample of 500 students showed that 150 students are female.

 (1) A good point estimate for p is (A) 0.31 (B) 0.30 (C) 0.29 (D) 0.25 (E) 0.27 (2) The lower limit of a 90% confidence interval for p is (A) 0.2363 (B) 0.2463 (C) 0.2963 (D) 0.2063 (E) 0.2663 (3) The upper limit of a 90% confidence interval for p is (A) 0.3337 (B) 0.3137 (C) 0.3637 (D) 0.2937 (E) 0.3537

Q3. In a random sample of 500 homes in a certain city, it is found that 114 are heated by oil. Let p be the proportion of homes in this city that are heated by oil.

1) Find a point estimate for p.

2) Construct a 98% confidence interval for p.

Q4. In a study involved 1200 car drivers, it was found that 50 car drivers do not use seat belt.

(1) A point estimate for the proportion of  car drivers that do not use seat built is:

(A) 50  (B) 0.0417      (C) 0.9583      (D) 1150          (E) None of these

(2) The lower limit of a 95% confidence interval of the proportion of car drivers not using seat

built is

(A) 0.0322       (B) 0.0416       (C) 0 .0304      (D) –0.3500     (E) None of these

(3) The upper limit of a 95% confidence interval of the proportion of car drivers not using seat

built is

(A) 0.0417       (B) 0.0530       (C) 0.0512       (D) 0.4333       (E) None of these

Q5. A study was conducted to make some inferences about the proportion of female employees (p) in a certain hospital. A random sample gave the following data:

 Sample size 250 Number of females 120

(a)    Calculate a point estimate ( ) for the proportion of female employees (p).

(b)   Construct a 90% confidence interval for p.

Q6. In a certain city, the traffic police was interested in knowing the proportion of car drivers who do not use seat built. In a study involved 1200 car drivers it was found that 50 car drivers do not use seat belt.

(a) A point estimate for the proportion of car drivers who do not use seat built is:

(A) 50              (B) 0.0417      (C) 0.9583      (D) 1150          (E) None of these

(b) A 95% confidence interval of the proportion of car drivers who do not use

seat built has the lower bound equal to:

(A) 0.0322       (B) 0.0416       (C) 0 .0304      (D) –0.3500     (E) None of these

(c) A 95% confidence interval of the proportion of car drivers who do not use

seat built has the upper bound equal to:

(A) 0.0417       (B) 0.0530       (C) 0.0512       (D) 0.4333       (E) None of these

11.4. Two Proportions:

Q1. A survey of 500 students from a college of science shows that 275 students own computers. In another independent survey of 400 students from a college of engineering shows that 240 students own computers.

 (a) a 99% confidence interval for the true proportion of college of science's student who own computers is (A) (B) (C) (D) (29) a 95% confidence interval for the difference between the proportions of students owning computers in the two colleges is (A) (B) (C) (D)

Q2. A food company distributes "smiley cow" brand of milk. A random sample of 200 consumers in the city (A) showed that 80 consumers prefer the "smiley cow" brand of milk. Another independent random sample of 300 consumers in the city (B) showed that 90 consumers prefer "smiley cow" brand of milk. Define:

pA = the true proportion of consumers in the city (A) preferring "smiley cow" brand.

pB = the true proportion of consumers in the city (B) preferring "smiley cow" brand.

 (a) A 96% confidence interval for the true proportion of consumers preferring brand (A) is: (A) (B) (C) (D) (b) A 99% confidence interval for the difference between proportions of consumers preferring brand (A) and (B) is: (A) (B) (C) (D)

Q3. A random sample of 100 students from school "A" showed that 15 students smoke. Another independent random sample of 200 students from school "B" showed that 20 students smoke. Let p1 be the proportion of smokers in school "A" and p2 is the proportion of smokers in school "B".

(1) Find a point Estimate for p1-p2.

(2) Find 95% confidence interval for p1-p2.

12. HYPOTHESES TESTING:

12.1. Single Mean:

Q1. Suppose that we are interested in making some statistical inferences about the mean, m, of a normal population with standard deviation s=2.0. Suppose that a random sample of size n=49 from this population gave a sample mean .

 (1) If we want to test Ho:m=5.0 against H1:m¹5.0, then the test statistic equals to (A) Z= -1.75 (B) Z=1.75 (C) T= -1.70 (D) T= 1.70 (E) Z= -1.65 (2) If we want to test Ho: m=5.0 against H1: m>5.0 at a=0.05, then the Rejection Region of Ho  is (A) (1.96,¥) (B) (2.325,¥) (C) (-¥,-1.645) (D) (-¥,-1.96 ) (E) (1.645, ¥) (3) If we want to test Ho: m=5.0 against H1: m>5.0 at a=0.05, then we (A) Accept Ho (B) Reject Ho (C) (D) (E)

Q2. An electrical firm manufactures light bulbs that have a length of life that is normally distributed with a standard deviation of 30 hours. A sample of 50 bulbs were selected randomly and found to have an average of 750 hours. Let m be the population mean of life lengths of all bulbs manufactured by this firm. Test Ho: m = 740 against H1: m < 740? Use a 0.05 level of significance.

Q3. An electrical firm manufactures light bulbs that have a length of life that is normally distributed. A sample of 20 bulbs were selected randomly and found to have an average of 655 hours and a standard deviation of 27 hours. Let m be the population mean of life lengths of all bulbs manufactured by this firm. Test Ho: m = 660 against H1: m¹ 660? Use a 0.02 level of significance.

Q4: A random sample of size n=36 from a normal quantitative population produced a mean  and a variance . Test Ho: m = 15 against Ha: m ¹ 15, use a=0.05.

Q5. A group of 10 college students were asked to report the number of hours that they spent doing their homework during the previous weekend and the following results were obtained:

7.25, 8.5, 5.0, 6.75, 8.0, 5.25, 10.5, 8.5, 6.75, 9.25

, }

It is assumed that this sample is a random sample from a normal distribution with unknown variance s2. Let m be the mean of the number of hours that the college student spend doing his/her homework during the weekend. Test Ho: m = 7.5 against Ha: m > 7.5, use a=0.2.

Q6. An electronics company wanted to estimate its monthly operating expenses in thousands riyals (m). It is assumed that the monthly operating expenses (in thousands riyals) are distributed according to a normal distribution with variance s2=0.584. Suppose that a random sample of 49 months produced a sample mean .  Test Ho: m=5.5 against and Ha: m¹5.5. Use a=0.01.

Q7. The tensile strength of a certain type of thread is approximately normally distributed with standard deviation of 6.8 kilograms. The manufacturer claims that the mean of the tensile strength of this type of thread equals to 70.0 kilogram. Do you agree with this claim if a sample of 20 pieces of the thread had an average tensile strength of 72.8 kilograms? Use a=0.05.

12.2. Two Means:

Q1. Two random samples were independently selected from two normal populations with equal variances. The results are summarized as follows.

 First Sample Second Sample sample size (n) sample mean ( ) sample variance (S2) 12 10.5 4 14 10.0 5

Let m1 and m2 be the true means of the first and second populations, respectively. Test Ho: m1= m2  against  H1: m1 ¹ m2 . (use a=0.05)

Q2. A researcher was interested in comparing the mean score of female students, mf, with the mean score of male students, mm, in a certain test. Two independent samples gave the following results:

 Sample Observations mean variance Scores of Females 89.2 81.6 79.6 80.0 82.8 82.63 15.05 Scores of Males 83.2 83.2 84.8 81.4 78.6 71.5 77.6 80.04 20.79

Assume that the populations are normal with equal variances.

 (1) The pooled estimate of the variance  is (A) 17.994 (B) 17.794 (C) 18.094 (D) 18.294 (E) 18.494 (2) If we want to test Ho: mf = mm against H1: mf ¹  mm then the test statistic equals to (A) Z= 1.129 (B) T= -1.029 (C) T=1.029 (D) T=1.329 (E) T= -1.329 (3) If we want to test Ho: mf = mm  against H1: mf ¹  mm at a=0.1, then the Acceptance Region of Ho is (A) (-¥,1.812) (B) (-1.812,1.812) (C) (-1.372,¥) (D) (-1.372,1.372) (E) (-1.812,¥) (4) If we want to test Ho: mf = mm against H1: mf ¹ mm at a=0.1, then we (A) Reject Ho (B) Accept Ho (C) (D) (E)

Q3. A study was conducted to compare to brands of tires A and B. 10 tires of brand A and 12 tires of brand B were selected randomly. The tires were run until they wear out. The results are:

Brand A:           = 37000 kilometers          SA = 5100

Brand B:            = 38000 kilometers          SB = 6000

Assuming the populations are normally distributed with equal variances. Test Ho: mA = mB against H1: mA < mB. Use a 0.1 level of significance.

Q4. A study was made by a taxi company to decide whether the use of new tires (A) instead of the present tires (B) improves fuel economy. Six cars were equipped with tires (A) and driven over a prescribed test course. Without changing drivers and cares, test course was made with tires (B). The gasoline consumption, in kilometers per liter (km/L), was recorded as follows: (assume the populations are normal with equal unknown variances)

 Car 1 2 3 4 5 6 Type (A) 4.5 4.8 6.6