Question

The mean weight of an adult is 6767 kilograms with a variance of 121121. If 164164 adults are randomly selected, what is the probability that the sample mean would be greater than 64.864.8 kilograms

Answer 1

99.48% probability that the sample mean would be greater than 64.8 kilograms.

Explanation:

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

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean
`\mu` and standard deviation(which is the square root of the variance)
`\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))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question:

`\mu = 67, \sigma = √(121) = 11, n = 164, s = (11)/(√(164)) = 0.86`

What is the probability that the sample mean would be greater than 64.8 kilograms?

This is 1 subtracted by the pvalue of Z when X = 64.8.

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

By the Central Limit Theorem

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

`Z = (64.8 - 67)/(0.86)`

`Z = -2.56`

`Z = -2.56` has a pvalue of 0.0052

1 - 0.0052 = 0.9948

99.48% probability that the sample mean would be greater than 64.8 kilograms.