Question

8. The distribution for the time it takes a student to complete the fall class registration has mean of 94 minutes and standard deviation of 10 minutes. For a random sample of 80 students, determine the mean and standard deviation (standard error) of the sample mean. What can you say about the sampling distribution of the sample mean and why

Answer 1

Answer:

Mean = 94

Standard deviation = 1.12

The sampling distribution of the sample mean is going to be normally distributed, beause the size of the samples are 80, which is larger than 30.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$, the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation, which is also called standard error $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 94, \sigma = 10$

By the Central Limit Theorem

The sampling distribution of the sample mean is going to be normally distributed, beause the size of the samples are 80, which is larger than 30.

Mean = 94

Standard deviation:

$s = \frac{10}{\sqrt{80}} = 1.12$