Statistics | Exercise 13.3

Question 1

The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Solution

Understand the Question

We are given the grouped frequency distribution of monthly electricity consumption for $68$ consumers.
We need to find three measures of central tendency:
- Mean ( $\bar{x}$ ) using the step-deviation method: $\bar{x} = a + \left(\dfrac{\sum f_i u_i}{\sum f_i}\right) \times h$
- Mode using the formula: $\text{Mode} = l + \left(\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$
- Median using the formula: $\text{Median} = l + \left(\dfrac{\frac{n}{2} - cf}{f}\right) \times h$
Finally, we compare the calculated values of mean, median, and mode.

Step 1 · Calculate the Mean

Let assumed mean $a = 135$ and class size $h = 20$ .

\begin{array}{|c|c|c|c|c|c|} \hline \text{Monthly consumption (in units)} & \text{Number of consumers } (f_i) & \text{Class mark } (x_i) & d_i = x_i - 135 & u_i = \dfrac{d_i}{20} & f_i u_i \\ \hline 65 - 85 & 4 & 75 & -60 & -3 & -12 \\ \hline 85 - 105 & 5 & 95 & -40 & -2 & -10 \\ \hline 105 - 125 & 13 & 115 & -20 & -1 & -13 \\ \hline 125 - 145 & 20 & 135 & 0 & 0 & 0 \\ \hline 145 - 165 & 14 & 155 & 20 & 1 & 14 \\ \hline 165 - 185 & 8 & 175 & 40 & 2 & 16 \\ \hline 185 - 205 & 4 & 195 & 60 & 3 & 12 \\ \hline \mathbf{Total} & \sum f_i = 68 & & & & \sum f_i u_i = 7 \\ \hline \end{array}

Using the step-deviation formula for mean:

\begin{aligned} \bar{x} &= a + \left(\dfrac{\sum f_i u_i}{\sum f_i}\right) \times h \\[0.6em] &= 135 + \left(\dfrac{7}{68}\right) \times 20 \\[0.6em] &= 135 + \dfrac{140}{68} \\[0.6em] &= 135 + 2.0588\dots \\[0.6em] &= 137.058 \text{ units} \end{aligned}

Step 2 · Calculate the Mode

The maximum class frequency is $20$ , which lies in the class $125 - 145$ .

Therefore, the modal class is $125 - 145$ .

Lower limit of modal class ( $l$ ) $= 125$
Class size ( $h$ ) $= 20$
Frequency of modal class ( $f_1$ ) $= 20$
Frequency of class preceding modal class ( $f_0$ ) $= 13$
Frequency of class succeeding modal class ( $f_2$ ) $= 14$

\begin{aligned} \text{Mode} &= l + \left[\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\right] \times h \\[0.6em] &= 125 + \left[\dfrac{20 - 13}{2(20) - 13 - 14}\right] \times 20 \\[0.6em] &= 125 + \left[\dfrac{7}{40 - 27}\right] \times 20 \\[0.6em] &= 125 + \left[\dfrac{7}{13}\right] \times 20 \\[0.6em] &= 125 + \dfrac{140}{13} \\[0.6em] &= 125 + 10.769\dots \\[0.6em] &= 135.76 \text{ units} \end{aligned}

Step 3 · Calculate the Median

Construct the cumulative frequency table: Diagram 2

\begin{array}{|c|c|c|} \hline \text{Monthly consumption (in units)} & \text{Number of consumers } (f_i) & \text{Cumulative frequency } (cf) \\ \hline 65 - 85 & 4 & 4 \\ \hline 85 - 105 & 5 & 9 \\ \hline 105 - 125 & 13 & 22 \\ \hline 125 - 145 & 20 & 42 \\ \hline 145 - 165 & 14 & 56 \\ \hline 165 - 185 & 8 & 64 \\ \hline 185 - 205 & 4 & 68 \\ \hline \end{array}

Here, $n = 68 \implies \dfrac{n}{2} = \dfrac{68}{2} = 34$ .

The cumulative frequency just greater than $34$ is $42$ , which belongs to the class $125 - 145$ .

Median class $= 125 - 145$
Lower limit ( $l$ ) $= 125$
Cumulative frequency of preceding class ( $cf$ ) $= 22$
Frequency of median class ( $f$ ) $= 20$
Class size ( $h$ ) $= 20$

\begin{aligned} \text{Median} &= l + \left[\dfrac{\frac{n}{2} - cf}{f}\right] \times h \\[0.6em] &= 125 + \left[\dfrac{34 - 22}{20}\right] \times 20 \\[0.6em] &= 125 + \left[\dfrac{12}{20}\right] \times 20 \\[0.6em] &= 125 + 12 \\[0.6em] &= 137 \text{ units} \end{aligned}

Step 4 · Compare the Measures

Comparing the three measures of central tendency:

$\text{Mean} \approx 137.058 \text{ units}$
$\text{Median} = 137 \text{ units}$
$\text{Mode} \approx 135.76 \text{ units}$

All three measures of central tendency are approximately equal.

Answer

$\text{Median} = 137 \text{ units}$ , $\text{Mean} = 137.058 \text{ units}$ , $\text{Mode} = 135.76 \text{ units}$ . The three measures are approximately equal.

Common Mistakes

Cumulative Frequency Error: Using the cumulative frequency of the median class ( $42$ ) instead of the preceding class ( $cf = 22$ ) in the median formula.
Formula Confusion in Mode: Incorrectly taking $2f_1 - f_0 - f_2$ as $2f_1 - (f_0 - f_2)$ or mixing up $f_0$ (preceding) and $f_2$ (succeeding).
Step-Deviation Class Size: Forgetting to multiply by the class size $h$ after computing the quotient in the mean or median formulas.

Question 1

More questions in Exercise 13.3