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 $\left(a\right)$ through $\left(c\right)$

Would you be surprised if, after observing $12$ individuals, fewer than half covered their mouth when sneezing? Why?

Answer 1

Solution:

It is given that,

Sample size, $n=12$

Probability of a randomly selected individual will not cover his or her mouth when sneezing is, $P=0.267$ and

Probability of a randomly selected individual will cover his or her mouth when sneezing is,

$$\begin{gathered} Q=1-0.267 \\ =0.73 \end{gathered}$$

The probability of a mass function of a binomial distribution is

$P(X=x)={}^{n}C_x{p}^x{\left(1-p\right)}^{n-x},\mathrm{for}x=0,1,2,...,n$.

Now, the probability that among $12$ randomly observed individuals, exactly $7$ do not cover their mouth when sneezing is,

$$\begin{gathered} P(X\le 6)={}^{12}C_0\times {\left(0.73\right)}^0{\left(1-0.267\right)}^{12-0}+{}^{12}C_1\times {\left(0.73\right)}^1{\left(1-0.267\right)}^{12-1}+{}^{12}C_2\times {\left(0.73\right)}^2{\left(1-0.267\right)}^{12-2} \\ +{}^{12}C_3\times {\left(0.73\right)}^3{\left(1-0.267\right)}^{12-3}+{}^{12}C_4\times {\left(0.73\right)}^4{\left(1-0.267\right)}^{12-4}+{}^{12}C_5\times {\left(0.73\right)}^5{\left(1-0.267\right)}^{12-5} \\ +{}^{12}C_6\times {\left(0.73\right)}^6{\left(1-0.267\right)}^{12-6} \\ =0.076 \end{gathered}$$

Hence, as the probability of observing $12$ individuals, fewer than half covered their mouth when sneezing is $0.076$.