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", so they want to estimate the proportion of users who access the site that way​ (even if they also use their computers​ sometimes). They draw a random sample of

200200

investors from their customers. Suppose that the true proportion of smart phone users is

3737​%.

​a) What would the standard deviation of the sampling distribution of the proportion of the smart phone users​ be?

. 034.034

​(Round to three decimal places as​ needed.)​b) What is the probability that the sample proportion of smart phone users is greater than

0.370.37​?

. 5.5

​(Round to three decimal places as​ needed.)​c) What is the probability that the sample proportion is between

0.320.32

and

0.420.42​?

. 858.858

​(Round to three decimal places as​ needed.)​d) What is the probability that the sample proportion is less than

0.300.30​?

. 02.02

​(Round to three decimal places as​ needed.)​e) What is the probability that the sample proportion is greater than

0.440.44​?

nothing

​(Round to three decimal places as​ needed.)

Answer 1

Solution:
No. of sample = 200
Population proportion = 0.37
Solution(a)
Standard deviation = sqrt(p*(1-p)/n) = sqrt(0.37*0.63/200) = 0.034
Solution(b)
P(p>0.37)
Z = (0.37-0.37)/0.034 = 0.0
From Z table we find p-values
P(p>0.37) = 1- 0.5 = 0.5
Solution(c)
P(0.32<p<0.42) = P(p<0.42) - P(p<0.32)
Z = (0.32-0.37)/0.034 = -1.46
Z = (0.42-0.37)/0.034 = 1.46
P(0.32<p<0.42) = P(p<0.42) - P(p<0.32) = 0.92785 - 0.0 7215 = 0.856
Solution(d)
P(p<0.30)
Z = (0.30-0.37)/0.034 = -2.05
From Z table we found p-values
P(p<0.30) = 0.020
Solution(e)
P(p>0.44)
Z = (0.44-0.37)/0.034 = 2.05
From Z table we found p-value
P(p>0.44) = 1 - 0.97982 = 0.02