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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
In this problem:
Item a:
This probability is P(X = 7), so:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 7) = C_{12,7}.(0.267)^{7}.(0.733)^{5} = 0.0162[/tex]
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:
[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]
Then
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{12,0}.(0.267)^{0}.(0.733)^{12} = 0.0241[/tex]
[tex]P(X = 1) = C_{12,1}.(0.267)^{1}.(0.733)^{11} = 0.1052[/tex]
[tex]P(X = 2) = C_{12,2}.(0.267)^{2}.(0.733)^{10} = 0.2107[/tex]
[tex]P(X = 3) = C_{12,3}.(0.267)^{3}.(0.733)^{9} = 0.2558[/tex]
Then
[tex]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[/tex]
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:
[tex]P(X < 6) = P(X < 4) + P(X = 5) = 0.5958 + P(X = 5)[/tex]
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 https://brainly.com/question/24923631
