Question

According to a study done by a university​ student, 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.
​(a) What is the probability that among 18 randomly observed individuals exactly 6 do not cover their mouth when​ sneezing?
​(b) What is the probability that among 18 randomly observed individuals fewer than 3 do not cover their mouth when​ sneezing?
​(c) Would you be surprised​ if, after observing 18 ​individuals, fewer than half covered their mouth when​ sneezing? Why?

Answer 1

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:

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

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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:

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

$P(X = 6) = C_{18,6}.(0.267)^{6}.(0.733)^{12} = 0.1618$

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:

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

$P(X = 0) = C_{18,0}.(0.267)^{0}.(0.733)^{18} = 0.0037$

$P(X = 1) = C_{18,1}.(0.267)^{1}.(0.733)^{17} = 0.0245$

$P(X = 2) = C_{18,2}.(0.267)^{2}.(0.733)^{16} = 0.0758$

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.

• $\sqrt{V(X)} = \sqrt{18(0.267)(0.733)} = 1.88$

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 brainly.com/question/24863377

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