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

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

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{(\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 =
`(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{(\hat p(1-\hat p))/(n) } }`
`\geq`
`\frac{0.457-0.534}{\sqrt{(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.