Question

The distribution of wait times for customers at a certain department of motor vehicles in a large city is skewed to the right with mean 23 minutes and standard deviation 11 minutes. A random sample of 50 customer wait times will be selected. Let x¯W represent the sample mean wait time, in minutes. Which of the following is the best interpretation of P(x¯W>25)≈0.10 ? For a random sample of 50 customer wait times, the probability that the total wait time will be greater than 25 minutes is approximately 0.10.
A For a randomly selected customer from the population, the probability that the total customer wait time will be greater than 25 minutes is approximately 0.10.

B For a randomly selected customer from the population, the probability that the sample mean customer wait time will be greater than 25 minutes is approximately 0.10.

C For a random sample of 50 customer wait times, the probability that the sample mean customer wait time will be greater than 23 minutes is approximately 0.10.

D For a random sample of 50 customer wait times, the probability that the sample mean customer wait time will be greater than 25 minutes is approximately 0.10.

A sports magazine reports that the mean number of hot dogs sold by hot dog vendors at a certain sporting event is equal to 150. A random sample of 50 hot dog vendors was selected, and the mean number of hot dogs sold by the vendors at the sporting event was 140. For samples of size 50, which of the following is true about the sampling distribution of the sample mean number of hot dogs sold by hot dog vendors at the sporting event? For all random samples of 50 sporting events, the sample mean will be 150 hot dogs.

A For all random samples of 50 hot dog vendors, the sample mean will be 140 hot dogs.

B The mean of the sampling distribution of the sample mean is 150 hot dogs.

C The mean of the sampling distribution of the sample mean is 140 hot dogs.

D All random samples of 50 hot dog vendors will have a sample mean within 10 hot dogs of the population mean.

A certain company produces fidget spinners with ball bearings made of either plastic or metal. Under standard testing conditions, fidget spinners from this company with plastic bearings spin for an average of 2.7 minutes, while those from this company with metal bearings spin for an average of 4.2 minutes. A random sample of three fidget spinners with plastic bearings is selected from company stock, and each is spun one time under the same standard conditions; let x¯1 represent the average spinning time for these three spinners. A random sample of seven fidget spinners with metal bearings is selected from company stock, and each is likewise spun one time under standard conditions; let x¯2 represent the average spinning time for these seven spinners. What is the mean μ(x¯1−x¯2) of the sampling distribution of the difference in sample means x¯1−x¯2 ? 3(2.7)−7(4.2)=−21.3

A 3−7=−4

B 2.7−4.2=−1.5

C 2.73−4.27=0.3

D 4.2−2.7=1.5

Consider two populations of coins, one of pennies and one of quarters. A random sample of 25 pennies was selected, and the mean age of the sample was 32 years. A random sample of 35 quarters was taken, and the mean age of the sample was 19 years.

For the sampling distribution of the difference in sample means, have the conditions for normality been met?

Yes, the conditions for normality have been met because the distributions of age for the two populations are approximately normal.

A

Yes, the conditions for normality have been met because the sample sizes taken from both populations are large enough.

B

No, the conditions for normality have not been met because neither sample size is large enough and no information is given about the distributions of the populations.

C

No, the conditions for normality have not been met because the sample size for the pennies is not large enough and no information is given about the distributions of the populations.

D

No, the conditions for normality have not been met because the sample size for the quarters is not large enough and no information is given about the distributions of the populations.

Answer 1

Answer:

1) D

2) D

3) B

4) A

Step-by-step explanation:

1) D,  P(50 > 25) ≈ 0.10 ⇒ the random sample of 50 customer wait times, the probability that the sample mean customer wait time will be greater than 25 minutes is approximately 0.10

2) D, (150 - 140) hot dogs = 10 hot dogs

3) B, 2.7- 4.7  = -1.5

4) A, (i) 25 + 25/4 = 25 + 6.25 = 31.25 ≈ 32 years

(ii) 35 + 35/4 = 35 + 8.75 = 43.75 ≈ 44 years  

Answer 2

Answer:

Step-by-step explanation:

1.

"D. For a random sample of 50 customer wait times, the probability that the sample mean customer wait time will be greater than 25 minutes is approximately 0.10" is correct.

2.

"B. The mean of the sampling distribution of the sample mean is 150 hot dogs" is correct.

(since the mean of all sample means is equal to the population mean which is 150, here).

3.

"B. 2.7−4.2=−1.5" is correct.

(Option D. is incorrect because we want μ(x¯1−x¯2) but not μ(x¯2−x¯1); Option A. is incorrect because 3 and 7 are sample sizes and we are not interested in the difference between sample sizes. Option C. is incorrect because there are no such values as 2.73 and 4.27, here).

4.

"C. No, the conditions for normality have not been met because the sample size for the pennies is not large enough and no information is given about the distributions of the populations" is correct.

(since 25 pennies < 30 is considered small sample and 35 quarters > 30 is considered large sample).