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
and
.
If 1100 Americans are sampled, then the sample size is
.
The mean of the distribution is
.
This means 
The standard deviation is 
We substitute the values to get:

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: 
We now convert to Z-scores to get:

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:

We convert to z-scores:

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:

We convert to z-scores:

From the normal distribution, P(z>\:-2.16)=0.0154
As a percentage, we get 1.54%