Soft-drink dispenser. The manager of a local soft-drink bottling company believes that when a new beverage dispensing machine is set to dispense 7 ounces, it in fact dispenses an amount at random anywhere between 6.5and 7.5 ounces inclusive. Suppose has a uniform probability

distribution.

a.Is the amount dispensed by the beverage machine a discreteor a continuous random variable? Explain.

b. Graph the frequency function for$\mathit{X}$ , the amount of beverage the manager believes is dispensed by the new machine when it is set to dispense 7 ounces.

c. Find the mean and standard deviation for the distribution graphed in part b, and locate the mean and theinterval $\mathbf{\mu }\mathbf{\pm }\mathbf{2}\mathbf{\sigma }$on the graph.

d. Find $\mathbf{P}(\mathbf{x}\ge \mathbf{7})$.

e. Find$\mathbf{P}(\mathbf{x}\mathbf{<}\mathbf{6})$ .

f. Find$\mathbf{P}(\mathbf{6}\mathbf{.}\mathbf{5}\le \mathbf{x}\le \mathbf{7}\mathbf{.}\mathbf{25})$ .

g. What is the probability that each of the next six bottles filled by the new machine will contain more than7.25 ounces of beverage? Assume that the amount of beverage dispensed in one bottle is independent of the amount dispensed in another bottle.

X denotes the amount of soft drink.

X follows a uniform distribution.

Since the amount of soft drinks can be measured, so the amount dispensed by the beverage machine is a continuous random variable.

When dispense is 7 ounces, the graph of the frequency function X is as follows

Let,

$\begin{array}{l}c=6.5\\ d=7.5\end{array}$

Since X follows a uniform distribution,the mean is

$\begin{array}{c}\mu =\frac{c+d}{2}\\ =\frac{6.5+7.5}{2}\\ =7\end{array}$

Hence, the mean is $\mu =7$.

The standard deviation is

$\begin{array}{c}\sigma =\frac{d-c}{\sqrt{12}}\\ =\frac{7.5-6.5}{\sqrt{12}}\\ =0.2886\end{array}$

Hence, the standard deviation is$\sigma =0.2886$.

The interval $\mu \pm 2\sigma$is

$\begin{array}{c}\mu \pm 2\sigma =7\pm (2\times 0.2886)\\ =7\pm 0.5772\\ =(6.4228,7.5772)\end{array}$

Hence, the interval $\mu \pm 2\sigma$ is$(6.4228,7.5772)$.

The mean and the interval $\mu \pm 2\sigma$are located below the graph,

The p.d.f of X is

$\begin{array}{c}f\left(x\right)=\frac{1}{d-c};c\le x\le d\\ =\frac{1}{7.5-6.5};6.5\le x\le 7.5\\ =1\end{array}$

$\begin{array}{c}P(x\ge 7)=1-P(x<7)\\ =1-\underset{6.5}{\overset{7}{\int }}f\left(x\right)dx\\ =1-\underset{6.5}{\overset{7}{\int }}1dx\\ =1-(7-6.5)\\ =1-0.5\\ =0.5\end{array}$

Since,

$\begin{array}{c}f\left(x\right)=\frac{1}{7.5-6.5};6.5\le x\le 7.5\\ =1\end{array}$

So$P(x<6)=0$,

$\begin{array}{c}P(6.5\le x\le 7.25)=\underset{6.5}{\overset{7.25}{\int }}f\left(x\right)dx\\ =\underset{6.5}{\overset{7.25}{\int }}1dx\\ =(7.25-6.5)\\ =\text{0.75}\end{array}$

Hence, $P(6.5\le x\le 7.25)=\text{0.75}$.

The probability that six bottles will contain more than 7.25 ounces of beverage is .

$P(x=6)$

$\begin{array}{c}P(x>7.25)=\underset{7.25}{\overset{7.5}{\int }}f\left(x\right)dx\\ =\underset{7.25}{\overset{7.5}{\int }}1dx\\ =(7.5-7.25)\\ =0.25\end{array}$

Since,$x>7.25$ is a binomial vitiate $B(0.25,6)$.

So,

$\begin{array}{c}P(x=6)=\left(\begin{array}{c}6\\ 6\end{array}\right){\left(0.25\right)}^6{(1-0.25)}^{6-6}\\ ={\left(0.25\right)}^6\\ =0.00024\end{array}$

Hence, the probability that six bottles will contain more than 7.25 ounces of beverage is 0.00024.