Chapter 4, Problem 89CE

The self-reported number of hours worked per week by 204 top executives is given below. (a) Estimate the mean, standard deviation, and coefficient of variation using an Excel worksheet to organize your calculations. (b) Do the unequal class sizes hamper your calculations? Why do you suppose that was done?

To determine

Estimate the mean, standard deviation and coefficient of variation from the grouped data and frequencies.

Answer to Problem 89CE

The mean, standard deviation and coefficient of variation from the grouped data and frequencies are:

Measures	Values
Mean	60.196
Standard deviation	8.536
Coefficient of variation	14.2%

Explanation of Solution

Calculation:

The given information is about the distribution of number of hours worked per week by 204 top executives.

The mean for the grouped data is:

$\overline{x}=\frac{{\displaystyle \sum _{j=1}^k{f}_j{m}_j}}{n}$

The standard deviation for the grouped data is:

$s=\sqrt{{\displaystyle \sum _{j=1}^k\frac{f_j\left(m_j-\overline{x}\right)}{n-1}}}$

Where,

$f_j$ is the frequency for j^th interval,

$m_j$ is the mid-point for j^th interval,

n is the total sample size $\left({\displaystyle \sum _{j=1}^k{f}_j}\right)$, when j lies between 1 and k.

If $x_l$ and $x_u$ are the lower and upper limits of the j^th interval, then $m_j=\frac{x_l+x_u}{2}$.

Coefficient of variation:

The coefficient of variation (CV) is a measure of relative variability which is the ratio of the standard deviation to the mean. If the coefficient of variation is high, then level of dispersion around the mean is high. The CV is often expressed as a percentage.

$CV=100\times \frac{s}{\overline{x}}$,

Where, s be the sample standard deviation and $\overline{x}$ be the sample mean.

The table below gives the mean for the grouped data:

From $x_l$	To $x_u$	Frequency $f_j$	Mid-point $m_j=\frac{x_l+x_u}{2}$	$f_j\times m_j$	${\left(m_j-\overline{x}\right)}^2$	$f_j{\left(m_j-\overline{x}\right)}^2$
40	50	12	$\frac{40+50}{2}=45$	540	$\begin{array}{l}{\left(45-60.196\right)}^2\\ =230.92\end{array}$	2771.04
50	60	116	$\frac{50+60}{2}=55$	6,380	$\begin{array}{l}{\left(55-60.196\right)}^2\\ =26.998\end{array}$	3131.768
60	80	74	$\frac{60+80}{2}=70$	5,180	$\begin{array}{l}{\left(70-60.196\right)}^2\\ =96.12\end{array}$	7112.88
80	100	2	$\frac{80+100}{2}=90$	180	$\begin{array}{l}{\left(90-60.196\right)}^2\\ =888.28\end{array}$	1776.56
		${\displaystyle \sum f_j}=204$		$\begin{array}{l}{\displaystyle \sum _{j=1}^4{f}_j{m}_j}\\ =12,280\end{array}$		$\begin{array}{l}{\displaystyle \sum _{j=1}^4{f}_j{\left(m_j-\overline{x}\right)}^2}\\ =14,792.248\end{array}$

The mean for the grouped data is:

Substitute the values ${\displaystyle \sum f_j}=204$ and ${\displaystyle \sum _{j=1}^4{f}_j{m}_j}=12,280$ to get the mean,

$\begin{array}{c}\overline{x}=\frac{12,280}{204}\\ =60.196\end{array}$

Thus, the mean from the grouped data and frequencies is 60.196.

The standard deviation for the grouped data is:

Substitute the values ${\displaystyle \sum f_j}=204$ and ${\displaystyle \sum _{j=1}^4{f}_j{\left(m_j-\overline{x}\right)}^2}=14,792.248$ to get the standard deviation,

$\begin{array}{c}s=\sqrt{\frac{14,792.248}{204-1}}\\ =\sqrt{\frac{14,792.248}{203}}\\ =\sqrt{72.868}\\ =8.536\end{array}$

Thus, the standard deviation from the grouped data and frequencies is 8.536.

The coefficient of variation is:

Substitute $\overline{x}=60.196$,$s=8.536$,

$\begin{array}{c}CV=100\times \frac{s}{\overline{x}}\\ =100\times \frac{8.536}{60.196}\\ =100\times 0.1418\\ \approx 14.18\%\end{array}$

Thus, the CV is 14.18%.

To determine

State whether the unequal class size hamper the calculations.

Answer to Problem 89CE

No, the unequal class sizes hamper the calculations.

Explanation of Solution

The unequal sample sizes in the observations disturb the calculations. The unequal class sizes means that none of the class size will hold the value zero.