**Answer:**

14.28% of individual adult females have weights between 75 kg and 83 kg.

92.82% of the sample means are between 75 kg and 83 kg.

**Step-by-step explanation:**

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 .

**Problems of normally distributed samples can be 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.

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

Assume that weights of adult females are normally distributed with a mean of 79 kg and a standard deviation of 22 kg. This means that .

**What percentage of individual adult females have weights between 75 kg and 83 kg?**

This percentage is the pvalue of Z when subtracted by the pvalue of Z when . So:

**X = 83**

has a pvalue of 0.5714.

**X = 75**

has a pvalue of 0.4286.

This means that 0.5714-0.4286 = 0.1428 = 14.28% of individual adult females have weights between 75 kg and 83 kg.

**If samples of 100 adult females are randomly selected and the mean weight is computed for each sample, what percentage of the sample means are between 75 kg and 83 kg?**

Now we use the Central Limit THeorem, when . So

**X = 83**

has a pvalue of 0.9641.

**X = 75**

has a pvalue of 0.0359.

This means that 0.9641-0.0359 = 0.9282 = 92.82% of the sample means are between 75 kg and 83 kg.