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

Answer:

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

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = p, \sigma = \sqrt{\frac{p(1-p)}{n}}$

In this problem:

$\mu = 0.4, \sigma = \sqrt{\frac{0.4*0.6}{500}} = 0.0219$

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.44 - 0.4}{0.0219}$

$Z = 1.83$

$Z = 1.83$ 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%.