STA STATISTICS MODULE 2 TEST(Units
4 & 5) - Test#2 Question and answers
correctly solved
Multiple-choice questions each have five possible answers (a, b, c, d, e), one
of which is correct. Assume that you guess the answers to three such
questions. - correct answer ✔P(CWW) = (1/5)(4/5)(4/5) = 0.128
what is the probability of getting exactly one correct answer when three
guesses are made?
P(CWW) / P(WCW)/ P(WWC) = 3 * 0.128 = 0.384
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. -
correct 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. - correct
answer ✔mean = np
For the binomial distribution, which formula finds the standard deviation? -
correct 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. - correct 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. - correct 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. - correct
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. - correct 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. - correct 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)
A pharmaceutical company receives large shipments of aspirin tablets. The
acceptance sampling plan is to randomly select and test 46 tablets, then
accept the whole batch if there is only one or none that doesn't meet the
required specifications. If one shipment of 6000 aspirin tablets actually has a
4% rate of defects, what is the probability that this whole shipment will be
accepted? Will almost all such shipments be accepted, or will many be
rejected? - correct answer ✔Use stat crunch
n = 46
p = 0.04
p (x <= 3) = 0.4460
The probability that this whole shipment will be accepted is
0.4460
4 & 5) - Test#2 Question and answers
correctly solved
Multiple-choice questions each have five possible answers (a, b, c, d, e), one
of which is correct. Assume that you guess the answers to three such
questions. - correct answer ✔P(CWW) = (1/5)(4/5)(4/5) = 0.128
what is the probability of getting exactly one correct answer when three
guesses are made?
P(CWW) / P(WCW)/ P(WWC) = 3 * 0.128 = 0.384
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. -
correct 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. - correct
answer ✔mean = np
For the binomial distribution, which formula finds the standard deviation? -
correct 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. - correct 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. - correct 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. - correct
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. - correct 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. - correct 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)
A pharmaceutical company receives large shipments of aspirin tablets. The
acceptance sampling plan is to randomly select and test 46 tablets, then
accept the whole batch if there is only one or none that doesn't meet the
required specifications. If one shipment of 6000 aspirin tablets actually has a
4% rate of defects, what is the probability that this whole shipment will be
accepted? Will almost all such shipments be accepted, or will many be
rejected? - correct answer ✔Use stat crunch
n = 46
p = 0.04
p (x <= 3) = 0.4460
The probability that this whole shipment will be accepted is
0.4460
