**Answer:**

**a) **By the Central Limit Theorem, the sampling distribution of ¯x will be bell-shaped(which is the same as normally distributed) with mean and standard error

**b) **13.59% probability that the mean overall physical activity level of the sample is between 300 and 310 cpm

**c) **0% probability that the mean overall physical activity level of the sample is greater than 360 cpm

**Step-by-step explanation:**

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

**Normal probability distribution:**

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

**Central Limit theorem:**

The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation , a large sample size can be approximated to a normal distribution with mean and standard deviation of the sample, which is also called standard error

**In this problem, we have that:**

**(a) Describe the sampling distribution of ¯x making sure to comment on all relevant features including its mean, standard error, and shape.**

By the Central Limit Theorem, the sampling distribution of ¯x will be bell-shaped(which is the same as normally distributed) with mean and standard error

**(b) What is the probability that the mean overall physical activity level of the sample is between 300 and 310 cpm?**

This is the pvalue of Z when X = 310 subtracted by the pvalue of Z when X = 300. So

**X = 310**

By the Central Limit Theorem

has a pvalue of 0.1587

**X = 300**

has a pvalue of 0.0228

0.1587 - 0.0228 = 0.1359

13.59% probability that the mean overall physical activity level of the sample is between 300 and 310 cpm

**(c) What is the probability that the mean overall physical activity level of the sample is greater than 360 cpm?**

This is 1 subtracted by the pvalue of Z when X = 360. So

has a pvalue of 1

1-1 = 0

0% probability that the mean overall physical activity level of the sample is greater than 360 cpm