Question

According to a study done by Nick Wilson of Otago University Wellington, the probability a randomly selected individual will not cover his or her mouth when sneezing is 0.267. Suppose you sit on a bench in a mall and observe people's habits as
they sneeze. Complete parts (a) through (c)
(a) What is the probability that among 12 randomly observed individuals, exactly 7 do not cover their mouth when sneezing?
Using the binomial distribution, the probability is
(Round to four decimal places as needed.)
(b) What is the probability that among 12 randomly observed individuals, fewer than 5 do not cover their mouth when sneezing?
Using the binomial distribution, the probability is 1
(Round to four decimal places as needed.)
(c) Would you be surprised if, after observing 12 individuals, fewer than half covered their mouth when sneezing? Why?
No, I would not be surprising, because using the binomial distribution, the probability is . which is v0.05
(Round to four decimal places as needed.)

Answer 1

Using the binomial distribution, it is found that:

a) There is a 0.0162 = 1.62% probability that among 12 randomly observed individuals, exactly 7 do not cover their mouth when sneezing.

b) There is a 0.5958 = 59.58% probability that among 12 randomly observed individuals, fewer than 5 do not cover their mouth when sneezing.

c) The probability is higher than 5%, so it would not be unusual.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, with p probability.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$  

$C_{n,x} = \frac{n!}{x!(n-x)!}$  

In this problem:

• Sample of 12, thus $n = 12$
• 0.267 probability that a person covers their mouth, thus $p = 0.27$.

Item a:

This probability is P(X = 7), so:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 7) = C_{12,7}.(0.267)^{7}.(0.733)^{5} = 0.0162$

There is a 0.0162 = 1.62% probability that among 12 randomly observed individuals, exactly 7 do not cover their mouth when sneezing.

Item b:

The probability is P(X < 4), which is:

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

Then

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{12,0}.(0.267)^{0}.(0.733)^{12} = 0.0241$

$P(X = 1) = C_{12,1}.(0.267)^{1}.(0.733)^{11} = 0.1052$

$P(X = 2) = C_{12,2}.(0.267)^{2}.(0.733)^{10} = 0.2107$

$P(X = 3) = C_{12,3}.(0.267)^{3}.(0.733)^{9} = 0.2558$

Then

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0241 + 0.1052 + 0.2107 + 0.2558 = 0.5958$

There is a 0.5958 = 59.58% probability that among 12 randomly observed individuals, fewer than 5 do not cover their mouth when sneezing.

Item c:

This probability is:

$P(X < 6) = P(X < 4) + P(X = 5) = 0.5958 + P(X = 5)$

Negative probabilities do no exist, thus, this probability is guaranteed to be above 0.05, which means that it would not be unusual if fewer than half covered their mouth when sneezing.

A similar problem is given at brainly.com/question/24923631