Chapter 7, Problem 80CE

Chlorine concentration in a municipal water supply is a uniformly distributed random variable that ranges between 0.74 ppm and 0.98 ppm. (a) What is the mean chlorine concentration? (b) The standard deviation? (c) What is the probability that the chlorine concentration will exceed 0.80 ppm on a given day? (d) Will be under 0.85 ppm? (e) Will be between 0.80 ppm and 0.90 ppm?

To determine

Find the mean of chlorine concentration.

Answer to Problem 80CE

The mean chlorine concentration is 0.86 ppm.

Explanation of Solution

Calculation:

It is given that the chlorine concentration follows Uniform distribution within the range of 0.74 ppm and 0.98 ppm.

Uniform distribution:

A random variable X said to follow a uniform distribution, $U\left(a,b\right)$, if the probability density function of X is,

$f\left(x\right)=\frac{1}{b-a}$ with domain of $a\le x\le b$, where a and b are the lower and upper limits, respectively.

The mean of the uniform distribution is,

$\mu =\frac{a+b}{2}$ .

Let the random variable X defines the chlorine concentration.

The uniform model is $U\left(0.74,0.98\right)$.

In the uniform model the lower limit is $a=2,500$ and the upper lower limit is $b=4,500$.

Hence, the mean of chlorine concentration is,

$\begin{array}{c}\mu =\frac{0.74+0.98}{2}\\ =\frac{1.72}{2}\\ =0.86\end{array}$.

Hence, the mean chlorine concentration is 0.86 ppm.

To determine

Find the standard deviation of chlorine concentration.

Answer to Problem 80CE

The standard deviation of chlorine concentration is 0.0692 ppm.

Explanation of Solution

Calculation:

The standard deviation of the uniform distribution is,

$\sigma =\sqrt{\frac{{\left(b-a\right)}^2}{12}}$.

Hence, the standard deviation of chlorine concentration is,

$\begin{array}{c}\sigma =\sqrt{\frac{{\left(0.98-0.74\right)}^2}{12}}\\ =\sqrt{\frac{{\left(0.24\right)}^2}{12}}\\ =\sqrt{\frac{0.0576}{12}}\\ =\sqrt{0.0048}\\ =0.0692\end{array}$.

Hence, the standard deviation of chlorine concentration is 0.0692 ppm.

To determine

Find the probability that the chlorine concentration will exceed 0.80 ppm on a given day.

Answer to Problem 80CE

The probability that the chlorine concentration will exceed 0.80 ppm on a given day is 0.75.

Explanation of Solution

Calculation:

The cumulative distribution function of the uniform distribution is,

$P\left(X\le x\right)=\frac{x-a}{b-a}$.

As the random variable X defines the chlorine concentration then the probability that chlorine concentration will exceed 0.80 ppm on a given day implies that $P\left(X>0.80\right)$.

The probability $P\left(X>0.80\right)$ can be rewritten as,

$P\left(X>0.80\right)=1-P\left(X\le 0.80\right)$.

Thus,

$\begin{array}{c}P\left(X<0.80\right)=\frac{0.80-0.74}{0.98-0.74}\\ =\frac{0.06}{0.24}\\ =0.25\end{array}$.

Hence,

$\begin{array}{c}P\left(X>0.80\right)=1-P\left(X\le 0.80\right)\\ =1-0.25\\ =0.75\end{array}$

Thus, the probability that the chlorine concentration will exceed 0.80 ppm on a given day is 0.75.

To determine

Find the probability that the chlorine concentration will be under 0.85 ppm.

Answer to Problem 80CE

The probability that the chlorine concentration will be under 0.85 ppm is 0.45833.

Explanation of Solution

Calculation:

The cumulative distribution function of the uniform distribution is,

$P\left(X\le x\right)=\frac{x-a}{b-a}$.

As the random variable X defines the chlorine concentration then the probability that the chlorine concentration will be under 0.85 ppm implies that $P\left(X<0.85\right)$.

Thus, the probability that the chlorine concentration will be under 0.85 ppm is,

$\begin{array}{c}P\left(X<0.85\right)=\frac{0.85-0.74}{0.98-0.74}\\ =\frac{0.11}{0.24}\\ =0.4583\end{array}$

Thus, the probability that the chlorine concentration will be under 0.85 ppm is 0.45833.

To determine

Find the probability that the chlorine concentration will be between 0.80 ppm and 0.90 ppm.

Answer to Problem 80CE

The probability that the chlorine concentration will be between 0.80 ppm and 0.90 ppm is 0.41.

Explanation of Solution

Calculation:

The cumulative distribution function of the uniform distribution is,

$P\left(X\le x\right)=\frac{x-a}{b-a}$.

As the random variable X defines the chlorine concentration then the probability that the chlorine concentration will be between 0.80 ppm and 0.90 ppm implies that $P\left(0.80<X<0.90\right)$.

The probability $P\left(0.80<X<0.90\right)$ can be rewritten as,

$P\left(0.80<X<0.90\right)=P\left(X<0.90\right)-P\left(X<0.80\right)$

Thus, the probability that the chlorine concentration will be between 0.80 ppm and 0.90 ppm is,

$\begin{array}{c}P\left(0.80<X<0.90\right)=P\left(X<0.90\right)-P\left(X<0.80\right)\\ =\left(\frac{0.90-0.74}{0.98-0.74}\right)-\left(\frac{0.80-0.74}{0.98-0.74}\right)\\ =\frac{0.16}{0.24}-\frac{0.06}{0.24}\\ =0.66-0.25\\ =\underset{\_}{0.41}.\end{array}$.

Thus, the probability that the chlorine concentration will be between 0.80 ppm and 0.90 ppm is 0.41.