Question

The mean height of women in a country​ (ages 20-29) is 64.3 inches. A random sample of 75 women in this age group is selected. What is the probability that the mean height for the sample is greater than 65 ​inches? Assume sigma=2.81.

Answer 1

Answer:

z(65) = (65-64.2)/[2.81/sqrt(60)] = 0.8/(0.3279)

Step-by-step explanation:

Answer 2

Using the normal probability distribution and the central limit theorem, it is found that there is a 0.0154 = 1.54% probability that the mean height for the sample is greater than 65 ​inches.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

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

• It 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, which is the percentile of X.
• By the Central Limit Theorem, for samples of size n, the standard deviation is $s = \frac{\sigma}{\sqrt{n}}$

In this problem:

• Mean of 64.3 inches, thus $\mu = 64.3$
• Standard deviation of 2.81 inches, thus $\sigma = 2.81$
• Sample of 75, thus $n = 75$.

The probability that the mean height for the sample is greater than 65 ​inches is 1 subtracted by the p-value of Z when X = 65, thus:

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

By the Central Limit Theorem

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

$Z = \frac{65 - 64.3}{\frac{2.81}{\sqrt{75}}}$

$Z = 2.16$

$Z = 2.16$ has a p-value of 0.9846.

1 - 0.9846 = 0.0154

0.0154 = 1.54% probability that the mean height for the sample is greater than 65 ​inches.

A similar problem is given at https://brainly.com/question/24663213