Question

A population has a mean of 180 and a standard deviation of 24. A sample of 100 observations will be taken. The probability that the mean from that sample will be between 183 and 186 is _____.

Answer 1

Answer:

The probability that the mean from that sample will be between 183 and 186 is 0.0994 = 9.94%.

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 estabilishes 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.

A population has a mean of 180 and a standard deviation of 24.

This means that $\mu = 180, \sigma = 24$

A sample of 100 observations will be taken.

This means that $n = 100, s = \frac{24}{\sqrt{100}} = 2.4$

The probability that the mean from that sample will be between 183 and 186 is:

This is the pvalue of Z when X = 186 subtracted by the pvalue of Z when X = 183. So

X = 186

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

By the Central Limit Theorem

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

$Z = \frac{186 - 180}{2.4}$

$Z = 2.5$

$Z = 2.5$ has a pvalue of 0.9938

X = 183

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

$Z = \frac{183 - 180}{2.4}$

$Z = 1.25$

$Z = 1.25$ has a pvalue of 0.8944

0.9938 - 0.8944 = 0.0994

The probability that the mean from that sample will be between 183 and 186 is 0.0994 = 9.94%.