Question

The mean height of women in a country (ages 20-29) is 64 4 inches A random sample of 50 women in this age group is selected What is the probability that the mean height for the sample is greater than 65 inches? Assume o = 2.91 The probability that the mean height for the sample is greater than 65 inches is​

Answer 1

Answer:

0.0721 = 7.21% probability that the mean height for the sample is greater than 65 inches.

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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 64.4 inches, standard deviation of 2.91

This means that $\mu = 64.4, \sigma = 2.91$

Sample of 50 women

This means that $n = 50, s = \frac{2.91}{\sqrt{50}}$

What is the probability that the mean height for the sample is greater than 65 inches?

This is 1 subtracted by the p-value of Z when X = 65. So

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

By the Central Limit Theorem

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

$Z = \frac{65 - 64.4}{\frac{2.91}{\sqrt{50}}}$

$Z = 1.46$

$Z = 1.46$ has a p-value of 0.9279

1 - 0.9279 = 0.0721

0.0721 = 7.21% probability that the mean height for the sample is greater than 65 inches.