Question

According to Net Market Share, Microsoft's Internet Explorer browser has 53.4% of the global market. A random sample of 70 users was selected. What is the probability that 32 or more from this sample used Internet Explorer as their browser?

Answer 1

Answer:

Probability that 32 or more from this sample used Internet Explorer as their browser is 0.9015.

Step-by-step explanation:

We are given that according to Net Market Share, Microsoft's Internet Explorer browser has 53.4% of the global market.

A random sample of 70 users was selected.

Let $\hat p$ = sample proportion of users who used Internet Explorer as their browser.

The z score probability distribution for sample proportion is given by;

                            Z  =  $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$  ~ N(0,1)

where, p = population proportion of users who use internet explorer = 53.4%

           $\hat p$ = sample proportion = $\frac{32}{70}$ = 0.457

           n = sample of users = 70

Now, probability that 32 or more from this sample used Internet Explorer as their browser is given by = P( $\hat p$ $\geq$ 0.457)

      P( $\hat p$ $\geq$ 0.457) = P( $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ $\geq$ $\frac{0.457-0.534}{\sqrt{\frac{0.457(1-0.457)}{70} } }$ ) = P(Z $\geq$ -1.29)

                            = P(Z $\leq$ 1.29) = 0.9015

The above probability is calculated by looking at the value of x = 1.29 in the z table which has an area of 0.9015.