Answer:
The probability that 121 or more Americans in the survey are afraid to fly is 0.1335.
Step-by-step 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:
[tex]np=1100\times 0.10=110>10\\\\n(1-p)=1100\times (1-0.10)=990>10[/tex]
Thus, a Normal approximation to binomial can be applied.
So, [tex]X\sim N(np,\ np(1-p))[/tex]
Compute the probability that 121 or more Americans in the survey are afraid to fly as follows:
[tex]P(X\geq 121)=P(\frac{X-np}{\sqrt{np(1-p)}}>\frac{121-110}{\sqrt{99}})\\\\=P(Z>1.11)\\\\=1-P(Z<1.11)\\\\=1-0.86650\\\\=0.1335[/tex]
Thus, the probability that 121 or more Americans in the survey are afraid to fly is 0.1335.
