Use the Central Limit Theorem to find the mean and standard error of the mean of the indicated sampling distribution. The amounts of time employees of a telecommunications company have worked for the company are normally distributed with a mean of 5.10 years and a standard deviation of 2.00 years. Random samples of size 15 are drawn from the population and the mean of each sample is determined. Round the answers to the nearest hundredth. O A. 5.10 years, 0.52 years OB. 5.10 years, 0.13 years C. 1.32 years, 2.00 years O D. 1.32 years, 0.52 years

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Use the Central Limit Theorem to find the mean and standard error of the mean of the indicated sampling distribution.
The amounts of time employees of a telecommunications company have worked for the company are normally distributed with a mean of 5.10 years and a standard deviation of 2.00 years. Random samples of size 15
are drawn from the population and the mean of each sample is determined. Round the answers to the nearest hundredth.
A. 5.10 years, 0.52 years
OB. 5.10 years, 0.13 years
OC. 1.32 years, 2.00 years
D. 1.32 years, 0.52 years
Transcribed Image Text:Use the Central Limit Theorem to find the mean and standard error of the mean of the indicated sampling distribution. The amounts of time employees of a telecommunications company have worked for the company are normally distributed with a mean of 5.10 years and a standard deviation of 2.00 years. Random samples of size 15 are drawn from the population and the mean of each sample is determined. Round the answers to the nearest hundredth. A. 5.10 years, 0.52 years OB. 5.10 years, 0.13 years OC. 1.32 years, 2.00 years D. 1.32 years, 0.52 years
7 of 10 (0 com
Find the area under the standard normal curve to the right of z = 1.
O A. 0.5398
B. 0.8413
O C. 0.1587
O D. 0.1397
Click to select your answer.
Transcribed Image Text:7 of 10 (0 com Find the area under the standard normal curve to the right of z = 1. O A. 0.5398 B. 0.8413 O C. 0.1587 O D. 0.1397 Click to select your answer.
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From the provided information,

Mean (µ) = 5.10

Standard deviation (σ) = 2.00

X~N (5.10, 2.00)

Here X be a random variable which represents the amounts of time employees of a telecommunications company.

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