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​?

nothing

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

Answer 1

p =0.37 , n=200

a)std deviation =(p(1-p)/n)^1/2

                          =0.034

b)P(X>0.37) = P(z> (p-μ / σ ) = P(z> (0.37-0.37 / 0.034 ) = P(z>0) = 1 - P(z<0.00) = 1 - 0.500 = 0.5

c)P(0.32<X<0.42) =P((0.32-0.37)/0.034 <Z<(0.42-0.37)/0.034)=P(-1.47<Z<1.47)=0.9292-0.0708 =0.858

d) P(X<0.30)=P(Z<(0.30-0.37)/0.034) =P(Z<-2.06) =0.02