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Ivenika [448]
3 years ago
13

An investment website can tell what devices are used to access the site. The site managers wonder whether they should enhance th

e 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.
Mathematics
1 answer:
AveGali [126]3 years ago
3 0

Answer:

0.8418

Step-by-step 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

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