Question

According to statcounter, Google Chrome browser controls 62.8% of the market share worldwide. A random sample of 70 users was selected. What is the probability that 35 or more from this sample used Google Chrome as their browser

Answer 1

Answer:

The probability that 35 or more from this sample used Google Chrome as their browser is 0.9838.

Step-by-step explanation:

We are given that according to Statcounter, the Google Chrome browser controls 62.8% of the market share worldwide.

A random sample of 70 users was selected.

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

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

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

where, $\hat p$ = sample proportion = $\frac{35}{70}$ = 0.50

            p = population proportion = 62.8%

            n = sample of users = 70

Now, the probability that 35 or more from this sample used Google Chrome as their browser is given by = P( $\hat p$ $\geq$ 0.50)

       P( $\hat p$ $\geq$ 0.50) = P( $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ $\geq$ $\frac{0.50-0.628}{\sqrt{\frac{0.50(1-0.50)}{70} } }$ ) = P(Z $\geq$ -2.14)

                           = P(Z $\leq$ 2.14)  = 0.9838

The above probability is calculated by looking at the value of x = 2.14 in the z table which has an area of 0.9838.