COMM 215 ch 6 Exam Questions and Answers Concordia
University
1. A department store will place a sale item in a special display for a one-day sale. Previous
experience suggests that 38 percent of all customers who pass such a special display will purchase
the item. If 1,619 customers will pass the display on the day of the sale, and if a one-item-per-
customer limit is placed on the sale item, how many units of the sale item should the store stock in
order to have at most a 1 percent chance of running short of the item on the day of the sale?
Assume here that customers make independent purchase decisions. (Round your answer to
nearest whole number.)
Number of units 661
μ =(1,619)(.38)=615, σ=381.4364−−−−−−−√=19.5304μ =(1,619)(.38)=615,
σ=381.4364=19.5304
P(x ≥ st) = .99
z=st − μσz=st − μσ
2. 33=st − 61519.53042.33=st − 61519.5304
st = 660.5
661 units
2. In order to gain additional information about respondents, some marketing researchers have used
ultraviolet ink to precode questionnaires that promise confidentiality to respondents. Of 170
randomly selected marketing researchers who participated in an actual survey, 152 said that they
disapprove of this practice. Suppose that, before the survey was taken, a marketing manager
claimed that at least 52 percent of all marketing researchers would disapprove of the practice.
(a) Assuming that the manager's claim is correct, calculate the probability that 152 or fewer of 170
randomly selected marketing researchers would disapprove of the practice. Use the normal
approximation to the binomial. (Round z value to 2 decimal places. Round your answer to 5
decimal places.)
P(x ≤ 152) 1.00000
(b) Based on your result of part a, do you believe the marketing manager's claim?
(Click to select)yes No
μ = (170) (.52) = 88.40, σ = 42.4320−−−−−−√ = 6.5140P(x≤152) = P(z
≤152.5−88.406.5140) = 1.00000μ = (170) (.52) = 88.40, σ = 42.4320 = 6.5140P(x≤152) = P(z
≤152.5−88.406.5140) = 1.00000
3. Find z when the area to the left of z is .05.
1.645
−1.645
1.00
−1.96
, From z table, z = −1.645.
4. Consider a normal population with a mean of 10 and a variance of 4. Find P(X < 6).
.8413
.0228
.1587
.9772
5. Given that X is a normal random variable, the probability that a given value of X is below its mean is
.
less than 0.5
1
equal to 0.5
greater than 0.5
6. An aptitude test has a mean score of 80 and a standard deviation of 5. The population of scores is normally
distributed. What raw score corresponds to the 70th percentile?
83.5
82.6
76.5
77.4
X = μ + zσ = 80 + (.52)(5) = 82.6
7. If the random variable X has a mean of µ and a standard deviation σ, then the mean and standard
deviation, respectively, of (X − µ)/σ are
1 and 0.
μ and σ.
x¯x¯ and s.
0 and 1.
8. An apple juice producer buys all his apples from a conglomerate of apple growers in one northwestern
state. The amount of juice obtained from each of these apples is approximately normally distributed with a
mean of 2.25 ounces and a standard deviation of 0.15 ounce. What is the probability that a randomly selected
apple will contain between 2.00 and 3.00 ounces?
.9525
.4525
.0475
.9554
9. Consider a normal population with a mean of 10 and a variance of 4. Find P(X ≥ 10).
0.00
0.50
University
1. A department store will place a sale item in a special display for a one-day sale. Previous
experience suggests that 38 percent of all customers who pass such a special display will purchase
the item. If 1,619 customers will pass the display on the day of the sale, and if a one-item-per-
customer limit is placed on the sale item, how many units of the sale item should the store stock in
order to have at most a 1 percent chance of running short of the item on the day of the sale?
Assume here that customers make independent purchase decisions. (Round your answer to
nearest whole number.)
Number of units 661
μ =(1,619)(.38)=615, σ=381.4364−−−−−−−√=19.5304μ =(1,619)(.38)=615,
σ=381.4364=19.5304
P(x ≥ st) = .99
z=st − μσz=st − μσ
2. 33=st − 61519.53042.33=st − 61519.5304
st = 660.5
661 units
2. In order to gain additional information about respondents, some marketing researchers have used
ultraviolet ink to precode questionnaires that promise confidentiality to respondents. Of 170
randomly selected marketing researchers who participated in an actual survey, 152 said that they
disapprove of this practice. Suppose that, before the survey was taken, a marketing manager
claimed that at least 52 percent of all marketing researchers would disapprove of the practice.
(a) Assuming that the manager's claim is correct, calculate the probability that 152 or fewer of 170
randomly selected marketing researchers would disapprove of the practice. Use the normal
approximation to the binomial. (Round z value to 2 decimal places. Round your answer to 5
decimal places.)
P(x ≤ 152) 1.00000
(b) Based on your result of part a, do you believe the marketing manager's claim?
(Click to select)yes No
μ = (170) (.52) = 88.40, σ = 42.4320−−−−−−√ = 6.5140P(x≤152) = P(z
≤152.5−88.406.5140) = 1.00000μ = (170) (.52) = 88.40, σ = 42.4320 = 6.5140P(x≤152) = P(z
≤152.5−88.406.5140) = 1.00000
3. Find z when the area to the left of z is .05.
1.645
−1.645
1.00
−1.96
, From z table, z = −1.645.
4. Consider a normal population with a mean of 10 and a variance of 4. Find P(X < 6).
.8413
.0228
.1587
.9772
5. Given that X is a normal random variable, the probability that a given value of X is below its mean is
.
less than 0.5
1
equal to 0.5
greater than 0.5
6. An aptitude test has a mean score of 80 and a standard deviation of 5. The population of scores is normally
distributed. What raw score corresponds to the 70th percentile?
83.5
82.6
76.5
77.4
X = μ + zσ = 80 + (.52)(5) = 82.6
7. If the random variable X has a mean of µ and a standard deviation σ, then the mean and standard
deviation, respectively, of (X − µ)/σ are
1 and 0.
μ and σ.
x¯x¯ and s.
0 and 1.
8. An apple juice producer buys all his apples from a conglomerate of apple growers in one northwestern
state. The amount of juice obtained from each of these apples is approximately normally distributed with a
mean of 2.25 ounces and a standard deviation of 0.15 ounce. What is the probability that a randomly selected
apple will contain between 2.00 and 3.00 ounces?
.9525
.4525
.0475
.9554
9. Consider a normal population with a mean of 10 and a variance of 4. Find P(X ≥ 10).
0.00
0.50
