For random samples of size n = 25 selected from a normal distribution with a mean of µ = 50 and a standard deviation of LaTeX: \sigmaσ= 20, find the range of sample means that defines the middle 95% of the distribution of sample means.

The range of sample means that defines the middle 95% of the distribution of sample means is between 42.16 and 57.84.

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 normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

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 normally distributed random variable X, with mean and standard deviation , the sample means with size n can be approximated to a normal distribution with mean and standard deviation

In this problem, we have that:

Find the range of sample means that defines the middle 95% of the distribution of sample means.

This is from the 50 - 95/2 = 2.5th percentile to the 50 + 95/2 = 97.5th percentile.

2.5th percentile

value of X when Z has a pvalue of 0.025. So X when Z = -1.96.

By the Central Limit Theorem

97.5th percentile

value of X when Z has a pvalue of 0.975. So X when Z = 1.96.

The range of sample means that defines the middle 95% of the distribution of sample means is between 42.16 and 57.84.

1/5^2 = 7.56/45