Answer :
Using the binomial distribution, it is found that:
a) There is a 0.1618 = 16.18% probability that among 18 randomly observed individuals exactly 6 do not cover their mouth when sneezing.
b) There is a 0.104 = 10.4% probability that among 18 randomly observed individuals fewer than 3 do not cover their mouth when sneezing.
c) 9 is more than 2.5 standard deviations below the mean, hence it would not be surprising if fewer than half covered their mouth when sneezing.
What is the binomial distribution formula?
The formula is:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
The values of the parameters are given as follows:
n = 18, p = 0.267.
Item a:
The probability is P(X = 6), hence:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 6) = C_{18,6}.(0.267)^{6}.(0.733)^{12} = 0.1618[/tex]
There is a 0.1618 = 16.18% probability that among 18 randomly observed individuals exactly 6 do not cover their mouth when sneezing.
Item b:
The probability is:
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2).
Then:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{18,0}.(0.267)^{0}.(0.733)^{18} = 0.0037[/tex]
[tex]P(X = 1) = C_{18,1}.(0.267)^{1}.(0.733)^{17} = 0.0245[/tex]
[tex]P(X = 2) = C_{18,2}.(0.267)^{2}.(0.733)^{16} = 0.0758[/tex]
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0037 + 0.0245 + 0.0758 = 0.104.
There is a 0.104 = 10.4% probability that among 18 randomly observed individuals fewer than 3 do not cover their mouth when sneezing.
item c:
We have to look at the mean and the standard deviation, given, respectively, by:
- E(X) = np = 18 x 0.267 = 4.81.
- [tex]\sqrt{V(X)} = \sqrt{18(0.267)(0.733)} = 1.88[/tex]
9 is more than 2.5 standard deviations below the mean, hence it would not be surprising if fewer than half covered their mouth when sneezing.
More can be learned about the binomial distribution at https://brainly.com/question/24863377
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