Question

The mean weight of an adult is 69 kilograms with a variance of 121. If 31 adults are randomly selected, what is the probability that the sample mean would be greater than 70.5 kilograms? Round your answer to four decimal places.

Answer 1

0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Also, important to remember that the standard deviation is the square root of the variance.

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 random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sample means with size n of at least 30 can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`

In this problem, we have that:

`\mu = 69, \sigma = √(121) = 11, n = 31, s = (11)/(√(31)) = 1.97565`

What is the probability that the sample mean would be greater than 70.5 kilograms?

This is 1 subtracted by the pvalue of Z when X = 70.5. So

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

By the Central limit theorem

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

`Z = (70.5 - 69)/(1.97565)`

`Z = 0.76`

`Z = 0.76` has a pvalue of 0.7764

1 - 0.7764 = 0.2236

0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.