Question

A random sample of 36 observations has been drawn from a normal distribution with mean 50 and standard deviation 12. Find the probability that the sample mean is in the interval 47≤X¯≤53. (Express the result to four significant digits.) Enter your answer in accordance to the question statementEnter your answer in accordance to the question statement

Answer 1

P(47≤X¯≤53) = 0.8664

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
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (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 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 = (\sigma)/(√(n))`.

In this problem, we have that:

`\mu = 50, \sigma = 12, n = 36, s = (12)/(√(36)) = 2`

Find the probability that the sample mean is in the interval 47≤X¯≤53.

This is the pvalue of Z when X = 53 subtracted by the pvalue of Z when X = 47. So

X = 53

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (53 - 50)/(2)`

`Z = 1.5`

`Z = 1.5` has a pvalue of 0.9332

X = 47

`Z = (X - \mu)/(s)`

`Z = (47 - 50)/(2)`

`Z = -1.5`

`Z = -1.5` has a pvalue of 0.0668

0.9332 - 0.0668 = 0.8664

P(47≤X¯≤53) = 0.8664