Question

According to a survey, 10% of Americans are afraid to fly. Suppose 1,100 Americans are sampled. What is the probability percentage that 121 or more Americans in the survey are afraid to fly

Answer 1

0.137.

Explanation:

Given that

Sample size = 1100

Afraid to fly = 10 %

Mean = n x p

Mean = 1100 x 0.1 = 110

Variance = n p q

Variance = 1100 x 0.1 x 0.9 = 99

Standard deviation =
`√(n* p* q)`

S.D.=
`√(1100* 0.1* 0.9)`

S.D.=9.94

By using Normal distribution

`Z=(X-\mu)/(\sigma)`

`Z=(121-110)/(9.94)`

Z=1.16

`P(X>121)=P(Z>1.16)=1-0.8643\\P(X>121)=0.137\\`

Thus the answer will be 0.137.