STAT 2023 MODULE 4 TEST SOLUTIONS 2026
PRACTICE QUESTIONS AND SOLUTIONS
GUARANTEED PASS
◉ Based on a survey, assume that 32% of consumers are
comfortable having drones deliver their purchases. Suppose that we
want to find the probability that when six consumers are randomly
selected, exactly two of them are comfortable with delivery by
drones. Identify the values of n, x, p, and q. Answer: The value of n is
6
The value of x is 2
The value of p is 0.32
The value of q is 0.68
(p + q = 1)
◉ The expression for calculating the mean of a binomial
distribution. Answer: mean = np
◉ For the binomial distribution, which formula finds the standard
deviation? Answer: square root of npq
◉ 40.3% of consumers believe that cash will be obsolete in the next
20 years. Assume that 6 consumers are randomly selected. Find the
,probability that fewer than 3 of the selected consumers believe that
cash will be obsolete in the next 20 years. Answer: Use stat crunch
n=6
p = 0.403
p (x < 3) = 0.53809939
◉ Assume that random guesses are made for 4 multiple-choice
questions on a test with 2 choices for each question, so that there
are n=4 trials, each with probability of success (correct) given by
p=0.50. Find the probability of no correct answers. Answer: n = 4, x
= 4, p = 0.5
The probability of no correct answers is
0.0620.062.
◉ Assume that when human resource managers are randomly
selected,
57% say job applicants should follow up within two weeks. If 5
human resource managers are randomly selected, find the
probability that exactly
3 of them say job applicants should follow up within two weeks.
Answer: Use stat crunch
n=5
p = 0.57
p (x = 3) = 0.34242186
, ◉ Assume that when human resource managers are randomly
selected, 58% say job applicants should follow up within two weeks.
If 5 human resource managers are randomly selected, find the
probability that at least 3 of them say job applicants should follow
up within two weeks. Answer: Use stat crunch
n=5
p = 0.58
p (x >= 3) = 0.64745966
◉ Assume that hybridization experiments are conducted with peas
having the property that for offspring, there is a 0.75 probability that
a pea has green pods. Assume that the offspring peas are randomly
selected in groups of
30. Answer: Complete parts (a) through (c) below.
a. The value of the mean is μ = 22.5 peas. (30*0.75 = 22.5)
The value of the standard deviation is σ = 2.4 peas.
(square root of (30)(0.75)(0.25))=2.4
b. Use the range rule of thumb to find the values separating results
that are significantly low or significantly high.
Values of 17.7 peas or fewer are significantly low.
(22.5-2(2.4))=17.7) = (mean - 2* standard deviation)
Values of 27.3 peas or greater are significantly high.
(22.5+2(2.4))=27.7) = (mean + 2* standard deviation)
PRACTICE QUESTIONS AND SOLUTIONS
GUARANTEED PASS
◉ Based on a survey, assume that 32% of consumers are
comfortable having drones deliver their purchases. Suppose that we
want to find the probability that when six consumers are randomly
selected, exactly two of them are comfortable with delivery by
drones. Identify the values of n, x, p, and q. Answer: The value of n is
6
The value of x is 2
The value of p is 0.32
The value of q is 0.68
(p + q = 1)
◉ The expression for calculating the mean of a binomial
distribution. Answer: mean = np
◉ For the binomial distribution, which formula finds the standard
deviation? Answer: square root of npq
◉ 40.3% of consumers believe that cash will be obsolete in the next
20 years. Assume that 6 consumers are randomly selected. Find the
,probability that fewer than 3 of the selected consumers believe that
cash will be obsolete in the next 20 years. Answer: Use stat crunch
n=6
p = 0.403
p (x < 3) = 0.53809939
◉ Assume that random guesses are made for 4 multiple-choice
questions on a test with 2 choices for each question, so that there
are n=4 trials, each with probability of success (correct) given by
p=0.50. Find the probability of no correct answers. Answer: n = 4, x
= 4, p = 0.5
The probability of no correct answers is
0.0620.062.
◉ Assume that when human resource managers are randomly
selected,
57% say job applicants should follow up within two weeks. If 5
human resource managers are randomly selected, find the
probability that exactly
3 of them say job applicants should follow up within two weeks.
Answer: Use stat crunch
n=5
p = 0.57
p (x = 3) = 0.34242186
, ◉ Assume that when human resource managers are randomly
selected, 58% say job applicants should follow up within two weeks.
If 5 human resource managers are randomly selected, find the
probability that at least 3 of them say job applicants should follow
up within two weeks. Answer: Use stat crunch
n=5
p = 0.58
p (x >= 3) = 0.64745966
◉ Assume that hybridization experiments are conducted with peas
having the property that for offspring, there is a 0.75 probability that
a pea has green pods. Assume that the offspring peas are randomly
selected in groups of
30. Answer: Complete parts (a) through (c) below.
a. The value of the mean is μ = 22.5 peas. (30*0.75 = 22.5)
The value of the standard deviation is σ = 2.4 peas.
(square root of (30)(0.75)(0.25))=2.4
b. Use the range rule of thumb to find the values separating results
that are significantly low or significantly high.
Values of 17.7 peas or fewer are significantly low.
(22.5-2(2.4))=17.7) = (mean - 2* standard deviation)
Values of 27.3 peas or greater are significantly high.
(22.5+2(2.4))=27.7) = (mean + 2* standard deviation)
