Question

On average, 28 percent of 18 to 34 year olds check their social media profiles before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a random variable X, which has a standard deviation of five percent. Find the probability that the percent of 18 to 34 year olds who check social media before getting out of bed in the morning is, at most, 32. Round your answer to four decimal places.

Answer 1

0.7881 = 78.81% probability that the percent of 18 to 34 year olds who check social media before getting out of bed in the morning is, at most, 32.

Explanation:

When the distribution is normal, we use 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.

In this question:

`\mu = 28, \sigma = 5`

Find the probability that the percent of 18 to 34 year olds who check social media before getting out of bed in the morning is, at most, 32.

This is the pvalue of Z when X = 32. So

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

`Z = (32 - 28)/(5)`

`Z = 0.8`

`Z = 0.8` has a pvalue of 0.7881

0.7881 = 78.81% probability that the percent of 18 to 34 year olds who check social media before getting out of bed in the morning is, at most, 32.