Question

According to recent surveys, 60% of households have personal computers. If a random sample of 182 households is selected, what is the probability that more than 68 but fewer than 91 have a personal computer? Round z-value calculations to 2 decimal places and final answer to at least 4 decimal places.

Answer 1

To answer this question we are going to perform a normal approximation to the binomial distribution.

We know that if X is binomial random variable we can approximate it to a normal distribution with:

$\begin{gathered} \mu=np \\ \text{and} \\ \sigma=\sqrt[]{npq} \end{gathered}$

In this case we have that:

$\begin{gathered} p=0.6 \\ q=1-p=1-0.6=0.4 \\ n=182 \end{gathered}$

Then the approximate normal distribution will have mean and standard deviation:

$\begin{gathered} \mu=(0.6)(182)=109.2 \\ \sigma=\sqrt[]{(0.6)(0.4)(182)}=6.61 \end{gathered}$

Now that we have the normal approximate distribution we need to calculate for this normal distribution:

$P(68Using the z-score defined by:[tex]z=\frac{x-\mu}{\sigma}$

we have that:

[tex]\begin{gathered} P(68Using a normal distribution table:[tex]\begin{gathered} P(68Therefore the probability we are looking for is 0.00298