Question

A distribution of values is normal with a mean of 60 and a standard deviation of 16. From this distribution, you are drawing samples of size 25. Find the interval containing the middle-most 76% of sample means.

Answer 1

Answer:

The interval containing the middle-most 76% of sample means is between 56.24 and 63.76.

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.

A distribution of values is normal with a mean of 60 and a standard deviation of 16.

This means that $\mu = 60, \sigma = 16$

Samples of size 25:

This means that $n = 25, s = \frac{16}{\sqrt{25}} = 3.2$

Find the interval containing the middle-most 76% of sample means.

Between the 50 - (76/2) = 12th percentile and the 50 + (76/2) = 88th percentile.

12th percentile:

X when Z has a p-value of 0.12, so X when Z = -1.175.

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

By the Central Limit Theorem

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

$-1.175 = \frac{X - 60}{3.2}$

$X - 60 = -1.175*3.2$

$X = 56.24$

88th percentile:

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

$1.175 = \frac{X - 60}{3.2}$

$X - 60 = 1.175*3.2$

$X = 63.76$

The interval containing the middle-most 76% of sample means is between 56.24 and 63.76.