Statistics | Exercise 13.3

Question 6

100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.

Solution

Understand the Question

To analyse the given grouped frequency distribution, we need to calculate three central tendencies:
- Median: Found using the median class where the cumulative frequency just exceeds $\dfrac{N}{2} = 50$ , via $\text{Median} = l + \left( \dfrac{\frac{N}{2} - cf}{f} \right) \times h$ .
- Mean: Calculated using the step-deviation method, $\bar{x} = a + \left( \dfrac{\sum f_i u_i}{\sum f_i} \right) \times h$ .
- Mode: Determined from the modal class with the highest frequency, using $\text{Mode} = l + \left( \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$ .

Step 1 · Find the Median

Construct the cumulative frequency table:

\begin{array}{|c|c|c|} \hline \text{Number of letters} & \text{Frequency } (f_i) & \text{Cumulative frequency } (cf) \\ \hline 1 - 4 & 6 & 6 \\ \hline 4 - 7 & 30 & 36 \\ \hline 7 - 10 & 40 & 76 \\ \hline 10 - 13 & 16 & 92 \\ \hline 13 - 16 & 4 & 96 \\ \hline 16 - 19 & 4 & 100 \\ \hline \text{Total } (N) & 100 & \\ \hline \end{array}

Here, $N = 100 \implies \dfrac{N}{2} = \dfrac{100}{2} = 50$ .

The cumulative frequency just greater than $50$ is $76$ , which belongs to the class interval $7 - 10$ . Therefore, the median class is $7 - 10$ .

From the median class:

Lower limit ( $l$ ) $= 7$
Cumulative frequency preceding median class ( $cf$ ) $= 36$
Frequency of median class ( $f$ ) $= 40$
Class size ( $h$ ) $= 3$

Applying the median formula:

\begin{aligned} \text{Median} &= l + \left[ \dfrac{\frac{N}{2} - cf}{f} \right] \times h \\[0.6em] &= 7 + \left[ \dfrac{50 - 36}{40} \right] \times 3 \\[0.6em] &= 7 + \left[ \dfrac{14}{40} \right] \times 3 \\[0.6em] &= 7 + 0.35 \times 3 \\[0.6em] &= 7 + 1.05 = 8.05 \end{aligned}

Step 2 · Find the Mean

Using the step-deviation method with assumed mean $a = 11.5$ and class size $h = 3$ :

\begin{array}{|c|c|c|c|c|c|} \hline \text{Number of letters} & f_i & x_i & d_i = x_i - 11.5 & u_i = \dfrac{d_i}{3} & f_i u_i \\ \hline 1 - 4 & 6 & 2.5 & -9 & -3 & -18 \\ \hline 4 - 7 & 30 & 5.5 & -6 & -2 & -60 \\ \hline 7 - 10 & 40 & 8.5 & -3 & -1 & -40 \\ \hline 10 - 13 & 16 & 11.5 & 0 & 0 & 0 \\ \hline 13 - 16 & 4 & 14.5 & 3 & 1 & 4 \\ \hline 16 - 19 & 4 & 17.5 & 6 & 2 & 8 \\ \hline \text{Total} & \sum f_i = 100 & & & & \sum f_i u_i = -106 \\ \hline \end{array}

Applying the mean formula:

\begin{aligned} \text{Mean } (\bar{x}) &= a + \left( \dfrac{\sum f_i u_i}{\sum f_i} \right) \times h \\[0.6em] &= 11.5 + \left( \dfrac{-106}{100} \right) \times 3 \\[0.6em] &= 11.5 + (-1.06) \times 3 \\[0.6em] &= 11.5 - 3.18 = 8.32 \end{aligned}

Step 3 · Find the Mode

The maximum frequency is $40$ , corresponding to the class interval $7 - 10$ . Therefore, the modal class is $7 - 10$ .

From the modal class:

Lower limit ( $l$ ) $= 7$
Class size ( $h$ ) $= 3$
Frequency of modal class ( $f_1$ ) $= 40$
Frequency of preceding class ( $f_0$ ) $= 30$
Frequency of succeeding class ( $f_2$ ) $= 16$

Applying the mode formula:

\begin{aligned} \text{Mode} &= l + \left[ \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \right] \times h \\[0.6em] &= 7 + \left[ \dfrac{40 - 30}{2(40) - 30 - 16} \right] \times 3 \\[0.6em] &= 7 + \left[ \dfrac{10}{80 - 46} \right] \times 3 \\[0.6em] &= 7 + \left[ \dfrac{10}{34} \right] \times 3 \\[0.6em] &= 7 + \dfrac{30}{34} \\[0.6em] &\approx 7 + 0.88 = 7.88 \end{aligned}

Answer

$\text{Median} = 8.05$ , $\text{Mean} = 8.32$ , $\text{Mode} = 7.88$

Common Mistakes

Wrong $cf$ in Median Formula: Using the cumulative frequency of the median class ( $76$ ) instead of the preceding class ( $cf = 36$ ).
Frequency Identification in Mode: Confusing $f_0$ (class before modal class $= 30$ ) and $f_2$ (class after modal class $= 16$ ).
Sign Error in Step-Deviation: Forgetting the negative sign when summing $f_i u_i$ , which results in $-106$ , not $+106$ .

Question 6

More questions in Exercise 13.3