91Ó°ÊÓ

From the study of bottled water brands it is found that 25% of bottled water is just tap water packed in a bottle.

X be thenumber of these brands that use tap water.

a.

Let x be the number of bottled water brands that use tap water

Here, n = 5

Andp= 0.25

Hence, we can say that x follows a binomial distribution with the parameters n = 5 and p = 0.25

b.

Probability distribution for X is given by:

$P\left(X\right|x)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{q}^{n-x}$

Where and x=0,1,2...

X follows a binomial distribution with parameters npq

c.

Given n=5,p=0.25, x=2

Therefore,

$$\begin{gathered} P(x=2)=\left(\begin{array}{c}5\\ 2\end{array}\right){\left(0.25\right)}^2{(1-0.25)}^3 \\ =10×0.062×0.421 \\ =0.2750 \end{gathered}$$

d.

Given n=5,p=0.25, x=2

$$\begin{gathered} P(x≤2)=\underset{x=0}{\overset{1}{∑}}\left(\begin{array}{c}5\\ x\end{array}\right){(0.25)}^2{(1-0.25)}^3 \\ =0.63281 \end{gathered}$$

e.

Here n=65, p=0.25

Mean is given by npthat is:

$$\begin{gathered} n×p=65×0.25 \\ =16.25 \end{gathered}$$

Standard deviation is given by$\sqrt{\mathbf{n}\mathbf{p}\mathbf{q}}$that is:

$$\begin{gathered} SD=\sqrt{n×p×(1-p)} \\ =\sqrt{65×0.25×0.75} \\ =3.491 \\ P(x>20)=P\left(7>\frac{x-mean}{variance}\right) \\ =P\left(7>\frac{20-16.25}{3.491}\right) \\ =P(7.1.074) \\ P(x>20)=1-P(7<1.074) \\ =0.1414 \end{gathered}$$

Hence in a random sample of 65 bottled water brands, is it likely that 20 or more brands will contain tap water and the probability will be 0.1414.