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

Answer:

0.137.

Step-by-step 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 = $\sqrt{n\times p\times q}$

S.D.=$\sqrt{1100\times 0.1\times 0.9}$

S.D.=9.94

By using Normal distribution

$Z=\dfrac{X-\mu}{\sigma}$

$Z=\dfrac{121-110}{9.94}$

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.