Let X be a discrete binomial random variable.
Let p = 0.267 be the probability that a person does not cover his mouth when sneezing.
Let n = 18 be the number of independent tests.
Let x be the number of successes.
So, the probability that the 18 individuals, 8 do not cover their mouth after sneezing will be:
a) P (X = 8) = 18! / (8! * 10!) * ((0.267) ^ 8) * ((1-0.267) ^ (18-8)).
P (X = 8) = 0.0506.
b) The probability that between 18 individuals observed at random less than 6 does not cover their mouth is:
P (X = 5) + P (X = 4) + P (X = 3) + P (X = 2) + P (X = 1) + P (X = 0) = 0.6571.
c) If it was surprising, according to the previous calculation, the probability that less than 6 people out of 18 do not cover their mouths is 66%. Which means it's less likely that more than half of people will not cover their mouths when they sneeze.
Given:
p = 20% = 0.2, sample proportion
n = 5000, sample size.
Confidence level = 95%
The confidence interval is
(p - 1.96k, p + 1.96k)
where

Therefore the 95% confidence interval is
(0.1943, 0.2057) = (19.4%, 20.6%)
Answer: The 95% confidence interval is (19.4%, 20.6%)
The answer is 1952 because you have to add all of the numbers together