Question

A recent report indicates that 2-year-old children from well-educated suburban families watched an average of μ = 60 minutes of television each day. Assuming that the distribution of television-watching times is normal with a standard deviation of σ = 25 minutes, find each of the following proportions.

a. What proportion of 2-year-old children watch more than 2 hours of television each day?
b. What proportion of 2-year-old children watch less than 30 minutes a day?

Answer 1

a) 0.82% of 2-year-old children watch more than 2 hours of television each day.

b) 11.51% of 2-year-old children watch less than 30 minutes a day.

Explanation:

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.

In this problem, we have that:

`\mu = 60, \sigma = 25`

a. What proportion of 2-year-old children watch more than 2 hours of television each day?

This is 1 subtracted by the pvalue of Z when X = 2*60 = 120 minutes.

So

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

`Z = (120 - 60)/(25)`

`Z = 2.4`

`Z = 2.4` has a pvalue of 0.9918

1 - 0.9918 = 0.0082

0.82% of 2-year-old children watch more than 2 hours of television each day.

b. What proportion of 2-year-old children watch less than 30 minutes a day?

This is the pvalue of Z when X = 30. So

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

`Z = (30 - 60)/(25)`

`Z = -1.2`

`Z = -1.2` has a pvalue of 0.1151.

11.51% of 2-year-old children watch less than 30 minutes a day.