Chapter 3, Problem 84CE

a.

To determine

Find the mean cost.

Answer to Problem 84CE

The mean cost is $141.20.

Explanation of Solution

Calculation:

Arithmetic mean of grouped data:

The mean of the data that is given in frequency distribution (grouped data) is calculated as,

$\overline{x}=\frac{{\displaystyle \sum fM}}{n}$

In the formula, $\overline{x}$ denotes the sample mean, M denotes the midpoint for each class, f denotes the frequency for each class, n denotes the total frequency.

Midpoint:

The midpoint is the average of the lower class limits of two consecutive classes. The formula is,

$\text{Midpoint}=\frac{\text{Lower}\text{limit}\text{of}\text{class}+\text{Lower}\text{limit}\text{of}\text{consecutive}\text{class}}{2}$

The frequency distribution table for mean is,

Electricity Cost	Frequency f	M	$fM$
$80 up to $100	3	90	270
100 up to 120	8	110	880
120 up to 140	12	130	1,560
140 up to 160	16	150	2,400
160 up to 180	7	170	1,190
180 up to 200	4	190	760
	$n=50$		${\displaystyle \sum fM}=7,060$

Substitute $n=50$ and ${\displaystyle \sum fM}=7,060$ in arithmetic mean formula.

$\begin{array}{c}\overline{x}=\frac{7,060}{50}\\ =141.20\end{array}$

Hence, the mean cost is $141.20.

b.

To determine

Find the standard deviation.

Answer to Problem 84CE

The standard deviation is $26.24.

Explanation of Solution

Calculation:

Standard deviation of grouped data:

The standard deviation of the data that is given in frequency distribution (grouped data) is calculated as,

$s=\sqrt{\frac{{\displaystyle \sum f{\left(M-\overline{x}\right)}^2}}{n-1}}$

In the formula, s denotes the sample standard deviation, $\overline{x}$ denotes the sample mean of grouped data, M denotes the midpoint for each class, f denotes the frequency for each class, n denotes the total frequency.

From part (a) the value of mean is 141.20. The frequency distribution table for standard deviation is,

Electricity Cost	Frequency f	M	$\left(M-\overline{x}\right)$	${\left(M-\overline{x}\right)}^2$	$f{\left(M-\overline{x}\right)}^2$
$80 up to $100	3	90	–51.2	2,621.44	7,864.32
100 up to 120	8	110	–31.2	973.44	7,787.52
120 up to 140	12	130	–11.2	125.44	1,505.28
140 up to 160	16	150	8.8	77.44	1,239.04
160 up to 180	7	170	28.8	829.44	5,806.08
180 up to 200	4	190	48.8	2,381.44	9,525.76
	$n=50$				${\displaystyle \sum f{\left(M-\overline{x}\right)}^2}=33,728$

Substitute $n=50$ and ${\displaystyle \sum f{\left(M-\overline{x}\right)}^2}=33,728$, in standard deviation formula.

$\begin{array}{c}s=\sqrt{\frac{33,728}{50-1}}\\ =\sqrt{\frac{33,728}{49}}\\ =\sqrt{688.3265}\\ =26.24\end{array}$

Hence, the standard deviation is $26.24.

c.

To determine

Find the limits to estimate the proportion of costs within two standard deviations of the mean.

Answer to Problem 84CE

The limits to estimate the proportion of costs within two standard deviations of the mean are $88.72 and $193.68.

Explanation of Solution

Calculation:

Empirical Rule:

Let $\overline{x}$ be the mean and s is the standard deviation of the symmetrical, bell-shaped distribution then,

• About 68% of the observations lie within plus and minus one standard deviation of the mean. That is, $\overline{x}\pm 1s$.
• About 95% of the observations lie within plus and minus two standard deviation of the mean. That is, $\overline{x}\pm 2s$.
• About 99.7% of the observations lie within plus and minus three standard deviation of the mean. That is, $\overline{x}\pm 3s$.

The values are,

$\begin{array}{c}\overline{x}-2s=141.20-2\left(26.24\right)\\ =141.20-52.48\\ =88.72\end{array}$

$\begin{array}{c}\overline{x}+2s=141.20+2\left(26.24\right)\\ =141.20+52.48\\ =193.68\end{array}$

Hence, the limits to estimate the proportion of costs within two standard deviations of the mean are $88.72 and $193.68.