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 $\left(c\right)$.

$\left(b\right)$ 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.)

Answer 1

Explanation:

It is given that,

Sample size, $n=12$

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

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, fewer than $5$ do not cover their mouth when sneezing is,

$$\begin{gathered} P\left(X<5\right)=P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)+P\left(X=4\right) \\ ={}^{12}C_0\times {\left(0.267\right)}^0{\left(1-0.267\right)}^{12}+{}^{12}C_1\times {\left(0.267\right)}^1{\left(1-0.267\right)}^{12-1}+{}^{12}C_1\times {\left(0.267\right)}^1{\left(1-0.267\right)}^{12-1} \\ +{}^{12}C_2\times {\left(0.267\right)}^2{\left(1-0.267\right)}^{12-2}+{}^{12}C_3\times {\left(0.267\right)}^3{\left(1-0.267\right)}^{12-3}+{}^{12}C_4\times {\left(0.267\right)}^4{\left(1-0.267\right)}^{12-4} \\ =0.024+0.105+0.21+0.256+0.21 \\ =0.805 \end{gathered}$$

Hence, the probability that among $12$ randomly observed individuals, exactly $5$ do not cover their mouth when sneezing is $0.805$.