Question

According to a survey, 10% of Americans are afraid to fly. Suppose 1100 Americans are sampled. a) What is the probability that 121 or more americans in the survey are afraid to fly? Write your answer as a percentage rounded to two decimal places. b) What is the probability that 165 or more americans in the survey are afraid to fly? Write your answer as a percentage rounded to two decimal places. c) What is the probability that 8% or less of the americans surveyed answered they were afraid to fly? Write your answer as a percentage rounded to two decimal places.

Answer 1

Answer:

a) 14.46%

b) 0.00%

c) 1.54%

Step-by-step explanation:

According to the survey, 10% of Americans are afraid to fly.

This means $p=0.10$ and $q=1-0.10=0.90$.

If 1100 Americans are sampled, then the sample size is $n=1100$ .

The mean of the distribution is $\mu=np$.

This means $\mu=1100\times 0.10=110$

The standard deviation is $\sigma=\sqrt{npq}$

We substitute the values to get:

$\sigma=\sqrt{1100\times 0.1\times 0.9}=9.95$

a) We want to find the probability that 121 or more Americans in the survey are afraid to fly.

We first apply the continuity correction factor to get: $P(X\ge121)=P(X\:>\:121-0.5)\\P(X\ge121)=P(X\:>\:120.5)$

We now convert to Z-scores to get:

$P(X\:>\:120.5)=P(z\:>\:\frac{120.5-110}{9.95})=1.06$

From the standard normal distribution table P(z>1.06)=0.1446

As a percentage, the probability is 14.46%

b) We want to find the probability that 165 or more Americans in the survey are afraid to fly.

We apply the CCF to get:

$P(X\ge165)=P(X\:>\:165-0.5)\\P(X\ge165)=P(X\:>\:164.5)$

We convert to z-scores:

$P(X\:>\:164.5)=P(z\:>\:\frac{164.5-110}{9.95})=5.48$

From the normal distribution, P(z>164.5)=0

c)  First, 8% of 1100 is 88.

We want to find the probability that 88 or less Americans in the survey are afraid to fly.

We apply the CCF to get:

$P(X\le88)=P(X\:$

We convert to z-scores:

$P(X\:$

From the normal distribution, P(z>\:-2.16)=0.0154

As a percentage, we get 1.54%