Question

Suppose 40% of the population of pre-teens have a TV in their bedroom. If a random sample of 500 pre-teens is drawn from the population, then the probability that 44% or fewer of the pre-teens have a TV in their bedroom is

Answer 1

The probability that 44% or fewer of the pre-teens have a TV in their bedroom is 96.64%.

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.

For a proportion p in a sample of size n, we have that
`\mu = p, \sigma = \sqrt{(p(1-p))/(n)}`

In this problem:

`\mu = 0.4, \sigma = \sqrt{(0.4*0.6)/(500)} = 0.0219`

The probability that 44% or fewer of the pre-teens have a TV in their bedroom is

This is the pvalue of Z when X = 0.44. So

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

`Z = (0.44 - 0.4)/(0.0219)`

`Z = 1.83`

`Z = 1.83` has a pvalue of 0.9664

The probability that 44% or fewer of the pre-teens have a TV in their bedroom is 96.64%.