Question

An investment website can tell what devices are used to access the site. The site managers wonder whether they should enhance the facilities for trading via "smart phones" therefore they want to estimate the proportion of users who access the site that way. They draw a random sample of 200 investors from their customers. Suppose that the true proportion of smart phone users in 36%. What is the probability that the sample proportion is between 0.30 and 0.40? Hint: You'll need to find one z-score for 0.30 and another for 0.40. Then find the area between the two z-scores.

Answer 1

0.8418

Explanation:

We have p = 0.36

Sample size = 200

We have to find the standard error of proportion

= √p(1-p)/n

= √0.36(1-0.36)/200

= 0.034

P(0.30<p^0.40)

= 0.30-0.36/0.034 = -1.76,

0.40-0.36/0.034 = 1.18

P(-1.76<z<1.18)

When we go to the standard normal distribution table

P(z<1.18) = 0.8810

P(z<-1.76) = 0.0392

0.8810 - 0.0392

= 0.8418

This is the probability of the sample proportion being between 0.30 and 0.40