Download Study Questions Ch5,6,7 File

Chapter 5
1) An automatic machine inserts mixed vegetables into a plastic bag. Past experience revealed
that some packages were underweight and some were overweight, but most of them had
satisfactory weight.
What is the probability of selecting three packages that are satisfactory?
Answer:
P(all 3 satisfactory) = (0.9) (0.9) (0.9) = 0.729
2) A study of interior designers' opinions with respect to the most desirable primary color for
executive offices showed that:
What is the probability that a designer does not prefer blue?
Answer: Total is 400. So, P(prefer Blue) = 37/400 = 0.0925
P( not prefer Blue) = 1 – P (prefer Blue) = 1 – 0.0925 = 0.9075.
3) In a survey of employee satisfaction, the following table summarizes the results in terms of
employee satisfaction and gender.
a) What is the probability that an employee is Female and Dissatisfied? Answer: 0.22
b) What is the probability that an employee is Male or Dissatisfied?
Answer: P(M or D) = P(M) + P(D) – P(M&D) = 0.60 + 0.55 – 0.33
= 0.82
c) What is the probability that an employee is Satisfied given that the employee is Male?
Answer: Total male= 27 +33 = 60 out of 100.
P(S |M) = 27/60 = 0.45
4) In a study of student preference of energy drinks, the researcher has identified a population of 10
students. To get a quick result, the researcher will select a sample of 2 students. How many different
samples are possible?
Answer: Using the combination formula ;
N
C n
10 x9 x8 x7 x6 x5 x 4 x3x 2 x1
 45
(2 x1)(8 x7 x6 x5 x 4 x3x 2 x1))
Chapter 6
1) The probabilities and the number of automobiles lined up at a Lakeside Olds at opening time (7:30
a.m.) for service are
a) On a typical day, how many automobiles should Lakeside Olds expect to be lined up at opening?
Answer:    XP(X ) = 1(0.05) + 2(0.3) + 3(0.4) + 4(0.25) = 2.85
3) Carlson Jewelers permits the return of their diamond wedding rings, provided the return occurs
within two weeks of the purchase date. Their records reveal that 10 percent of the diamond
wedding rings are returned. Five different customers buy five rings. What is the probability
that none will be returned?
Answer: Given that probability of success is 0.10, P(X = 0) = 0.590 from the binomial
probability distribution table or using the binomial formula as;
5!
P(X=0) = 0!(5−0)! 𝜋 𝑥 (1 − 𝜋)(𝑛−𝑥) = 0.59
4) Chances are 50-50 that a newborn baby will be a girl. For families with five children, what is
the probability that all the children are girls?
Answer: Given that probability of success is 0.50 P(X = 5) = 0.031 from the binomial
probability distribution table or using the binomial formula.
5) Which of the following is NOT a characteristic of a binomial probability distribution?
A. Each outcome is mutually exclusive
B. Each trial is independent
C. Probability of success remains constant from trial to trial
D. Each outcome results from two trials (Correct Answer)
Chapter 7
1) The time to fly between New York City and Chicago is uniformly distributed with a minimum
of 120 minutes and a maximum of 150 minutes.
a) What is the mean? Answer:

a  b 120  150

 135 minutes
2
2
(b  a) 2
(150  120) 2
b) What is the standard deviation? Answer:  

 8.66 minutes
12
12
c) The time to fly between New York City and Chicago is uniformly distributed with a
minimum of 120 minutes and a maximum of 150 minutes. What is the probability that a
flight is less than 135 minutes?
Answer: P( X  135) 
1
(135  120)  0.50
150  120
2) A new extended-life light bulb has an average service life of 750 hours, with a standard
deviation of 50 hours. If the service life of these light bulbs approximates a normal distribution,
about what percent of the distribution will be between 600 hours and 900 hours?
Answer: 900 – 750 = 150 So, 900 is 3 standard deviation above the mean (150/50 = 3)
600 – 750 = 150 So, 600 is 3 standard deviation below the mean (150/50 = 3)
Therefore, using the Empirical Rule, 99.7% of all values are between 600 and 900.
3) The employees of Cartwright Manufacturing are awarded efficiency ratings. The distribution
of the ratings approximates a normal distribution. The mean is 400, the standard deviation 50.
What is the probability that ratings will be between 400 and 482?
a) Answer: z 
X 


482  400
 1.64 for this z value, the prob. value = 0.4495
50
b) What is the probability that ratings will be above 482?
Answer: P(X  482) = 0.5 – 0.4495 = 0.0505
4) The American Auto Association reports the mean price per gallon of regular gasoline is $3.10
with a population standard deviation of $0.20. Assume a random sample of 16 gasoline stations
is selected and their mean cost for regular gasoline is computed.
a) What is the standard error of the mean in this experiment?
Answer:  X 

n

0.2
 0.05
16
b) What is the probability that the sample mean is between $2.98 and $3.12?
Answer: z 
z
X   2.98  3.10  0.12


 2.4
0.05
/ n
0.2 / 16
p = 0.4918
X   3.12  3.10 0.02


 0.4 p = 0.1554
0.05
/ n
0.2 / 16
P( 2.98  X  3.12) = 0.4918+0.1554 = 0.6472
Chapter 9
1) A random sample of 42 college graduates revealed that they worked an average of 5.5 years on
the job before being promoted. The population standard deviation was 1.1 years. Using the
0.99 degree of confidence, what is the confidence interval for the population mean?
  
Answer: 99% confidence interval will be found by X  z

 n
 1.1 
5.5  2.58
  5.5  0.4379 interval is between 5.06 and 5.94
 42 