Question

Assume that weights of adult females are normally distributed with a mean of 79 kg and a standard deviation of 22 kg. What percentage 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?

Answer 1

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 $\mu$ and standard deviation $\sigma$, a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$.

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

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

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = 79, \sigma = 22$.

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

This percentage is the pvalue of Z when $X = 83$ subtracted by the pvalue of Z when $X = 75$. So:

X = 83

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{83 - 79}{22}$

$Z = 0.18$

$Z = 0.18$ has a pvalue of 0.5714.

X = 75

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{75- 79}{22}$

$Z = -0.18$

$Z = -0.18$ 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 $n = 100$. So $s = \frac{22}{\sqrt{100}} = 2.2.$

X = 83

$Z = \frac{X - \mu}{s}$

$Z = \frac{83 - 79}{2.2}$

$Z = 1.8$

$Z = 1.8$ has a pvalue of 0.9641.

X = 75

$Z = \frac{X - \mu}{s}$

$Z = \frac{75-79}{2.2}$

$Z = -1.8$

$Z = -1.8$ 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.