Question

Suppose that a population of people has an average weight of 160 lbs, and standard deviation of 50 lbs, and that weight is normally distributed. A researcher samples 100 people, and measures their weight. Find the probability that the researcher observes an average weight of the 100 people to be between 150 and 170. [Round your answer to four decimal places]

Answer 1

0.9544 = 95.44% probability that the researcher observes an average weight of the 100 people to be between 150 and 170.

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))`.

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

In this question, we have that:

`\mu = 160, \sigma = 50, n = 100, s = (50)/(√(100)) = 5`

Find the probability that the researcher observes an average weight of the 100 people to be between 150 and 170.

This is the pvalue of Z when X = 170 subtracted by the pvalue of Z when X = 150. So

X = 170

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

By the Central Limit Theorem

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

`Z = (170 - 160)/(5)`

`Z = 2`

`Z = 2` has a pvalue of 0.9772

X = 150

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

`Z = (150 - 160)/(5)`

`Z = -2`

`Z = -2` has a pvalue of 0.0228

0.9772 - 0.0228 = 0.9544

0.9544 = 95.44% probability that the researcher observes an average weight of the 100 people to be between 150 and 170.