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

Answer 1

The probability that 121 or more Americans in the survey are afraid to fly is 0.1335.

Explanation:

Let X denote the number of Americans afraid to fly.

The proportion of Americans afraid to fly is, p = 0.10/

A random sample of n = 1100 Americans are selected.

The random variable X follows a Binomial distribution with parameters n = 1100 and p = 0.10.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

1. np ≥ 10

2. n(1 - p) ≥ 10

Check the conditions as follows:

`np=1100* 0.10=110>10\\\\n(1-p)=1100* (1-0.10)=990>10`

Thus, a Normal approximation to binomial can be applied.

So,
`X\sim N(np,\ np(1-p))`

Compute the probability that 121 or more Americans in the survey are afraid to fly as follows:

`P(X\geq 121)=P((X-np)/(√(np(1-p)))>(121-110)/(√(99)))\\\\=P(Z>1.11)\\\\=1-P(Z<1.11)\\\\=1-0.86650\\\\=0.1335`

Thus, the probability that 121 or more Americans in the survey are afraid to fly is 0.1335.