Question

According to a report from a business intelligence company, smartphone owners are using an average of 22 apps per month. Assume that number of apps used per month by smartphone owners is normally distributed and that the standard deviation is 4. Complete parts (a) through (d) below. If you select a random sample of 36 smartphone owners, what is the probability that the sample mean is between 21 and 22?

Answer 1

0.4332 = 43.32% probability that the sample mean is between 21 and 22.

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 z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

According to a report from a business intelligence company, smartphone owners are using an average of 22 apps per month.

This means that
`\mu = 22`

Standard deviation is 4:

This means that
`\sigma = 4`

Sample of 36:

This means that
`n = 36, s = \frac{4}{sqrt{36}}`

What is the probability that the sample mean is between 21 and 22?

This is the p-value of Z when X = 22 subtracted by the p-value of Z when X = 21.

X = 22

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

By the Central Limit Theorem

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

`Z = \frac{22 - 22}{\frac{4}{sqrt{36}}}`

`Z = 0`

`Z = 0` has a p-value of 0.5.

X = 21

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

`Z = \frac{21 - 22}{\frac{4}{sqrt{36}}}`

`Z = -1.5`

`Z = -1.5` has a p-value of 0.0668.

0.5 - 0.0668 = 0.4332

0.4332 = 43.32% probability that the sample mean is between 21 and 22.