Question

A population has mean 187 and standard deviation 32. If a random sample of 64 observations is selected at random from this population, what is the probability that the sample average will be less than 182

Answer 1

Answer:

11.51% probability that the sample average will be less than 182

Step-by-step explanation:

To solve this question, we have 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 $\mu$ 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.

Central Limit theorem:

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 $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 187, \sigma = 32, n = 64, s = \frac{32}{\sqrt{64}} = 4$

What is the probability that the sample average will be less than 182

This is the pvalue of Z when X = 182. So

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

By the Central Limit Theorem

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

$Z = \frac{182 - 187}{4}$

$Z = -1.2$

$Z = -1.2$ has a pvalue of 0.1151

11.51% probability that the sample average will be less than 182