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

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

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