Question

A social media survey found that 71%

of parents are​ "friends" with their children on a certain online networking site. A random sample of 140

parents was selected. Complete parts a through d below.

a. Calculate the standard error of the proportion.

sigma Subscript p

equals0.0383

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

b. What is the probability that

105

or more parents from this sample are​ "friends" with their children on this online networking​ site?

​P(105

or more parents from this sample are​ "friends" with their

​children)equals

nothing

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

Answer 1

Part a

We are given

Population proportion = p = 0.71

Sample size = n = 140

Standard error = sqrt(p*(1 – p)/n) = sqrt(0.71*(1 – 0.71)/140) = sqrt(0.71*0.29/140) = sqrt(0.001471)

Standard error = sqrt(0.001471) = 0.03835

Standard error = 0.0384

Part b

We are given

Population proportion = p = 0.71, so q = 1 – p = 1 – 0.71 = 0.29

Sample size = n = 140

We have to find P(X≥105)

Here, we have to use normal approximation to binomial distribution.

Mean = n*p = 140*0.71 = 99.4

Standard deviation = SD = sqrt(n*p*q) = sqrt(140*0.71*0.29) = 5.368985

We have to find P(X≥105), so we subtract the continuity correction 0.5 from 105. This means we have to find P(X≥104.5)

P(X≥104.5) = 1 – P(X<104.5)

Z = (X – mean) / SD

Z = (104.5 - 99.4) / 5.368985

Z = 0.9499

P(Z<0.9499) = 0.828919 (by using z-table/excel/Ti-83/84 calculator)

P(X<104.5) = 0.828919

P(X≥104.5) = 1 – P(X<104.5)

P(X≥104.5) = 1 – 0.828919

P(X≥104.5) = 0.171081

Required probability = 0.1711