Question

A social media survey found that 70​% 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. b. What is the probability that 102 or more parents from this sample are​ "friends" with their children on this online networking​ site?

Answer 1

Standard Error of proportion

Part b)

Using Normal approximation to Binomial

Mean = n * P = ( 140 * 0.7 ) = 98
Variance = n * P * Q = ( 140 * 0.7 * 0.3 ) = 29.4
Standard deviation = = 5.4222

P ( X >= 102 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 102 - 0.5 ) =P ( X > 101.5 )

P ( X > 101.5 ) = 1 - P ( X < 101.5 )
Standardizing the value

Z = ( 101.5 - 98 ) / 5.4222
Z = 0.65

P ( Z > 0.65 )
P ( X > 101.5 ) = 1 - P ( Z < 0.65 )
P ( X > 101.5 ) = 1 - 0.7422
P ( X > 101.5 ) = 0.2578