Question

It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minutes per day that people spend standing or walking. Among mildly obese people, the mean number of minutes of daily activity (standing or walking) is approximately Normally distributed with 367 minutes and standard deviation 64 minutes. The mean number of minutes of daily activity for lean people is approximately Normally distributed with 525 minutes and standard deviation 110 minutes. A researcher records the minutes of activity for an SRS of 8 mildly obese people and an SRS of 8 lean people. (a) What is the probability that the mean number of minutes of daily activity of the 8 mildly obese people exceeds 400 minutes

Answer 1

0.0721 = 7.21% probability that the mean number of minutes of daily activity of the 8 mildly obese people exceeds 400 minutes

Explanation:

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

Normal Probability Distribution:

Problems of normal distributions can be 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.

Among mildly obese people, the mean number of minutes of daily activity (standing or walking) is approximately Normally distributed with 367 minutes and standard deviation 64 minutes.

This means that
`\mu = 367, \sigma = 64`

A researcher records the minutes of activity for an SRS of 8 mildly obese people.

This means that
`n = 8, s = (64)/(√(8)) = 22.63`

(a) What is the probability that the mean number of minutes of daily activity of the 8 mildly obese people exceeds 400 minutes?

This is 1 subtracted by the pvalue of Z when X = 400. So

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

By the Central Limit Theorem

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

`Z = (400 - 367)/(22.63)`

`Z = 1.46`

`Z = 1.46` has a pvalue of 0.9279

1 - 0.9279 = 0.0721

0.0721 = 7.21% probability that the mean number of minutes of daily activity of the 8 mildly obese people exceeds 400 minutes