3. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure :

Understand the Question We are given the monthly household expenditure data for families. Modal Monthly Expenditure (Mode): Represents the expenditure value that occurs most frequently. It is found by locating the modal class (class with the highest frequency) and applying the grouped mode formula. Mean Monthly Expenditure (Mean): Represents the average expenditure across all families. It is calculated using the Step-Deviation Method with an assumed mean and class size . Part (i) Find the modal monthly expenditure of the families. Step 1 · Calculate the Mode The highest frequency is , corresponding to the modal class . Lower limit of modal class () Frequency of modal class () Frequency of preceding class () Frequency of succeeding class () Class size () Answer > (i) Part (ii) Find the mean monthly expenditure of the families. Step 1 · Calculate the Mean using Step-Deviation Method Let Assumed Mean and Class Size . Using the step-deviation formula: Answer > (ii) Common Mistakes Frequencies Confusion: Swapping , , and . Note that is the modal class frequency, is the preceding class frequency, and is the succeeding class frequency. Sign Errors in Step-Deviation: Forgetting negative signs for deviations below the assumed mean (), which leads to an incorrect sum . Denominator in Mode Formula: Incorrectly computing . Always evaluate first before subtracting and .

Statistics | Exercise 13.2

Question 3

The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure :

Solution

Understand the Question

We are given the monthly household expenditure data for $200$ families.
Modal Monthly Expenditure (Mode): Represents the expenditure value that occurs most frequently. It is found by locating the modal class (class with the highest frequency) and applying the grouped mode formula.
Mean Monthly Expenditure (Mean): Represents the average expenditure across all families. It is calculated using the Step-Deviation Method with an assumed mean $a = 2750$ and class size $h = 500$ .

(i) Find the modal monthly expenditure of the families.

Step 1 · Calculate the Mode

Diagram 1

The highest frequency is $40$ , corresponding to the modal class $1500 - 2000$ .

Lower limit of modal class ( $l$ ) $= 1500$
Frequency of modal class ( $f_1$ ) $= 40$
Frequency of preceding class ( $f_0$ ) $= 24$
Frequency of succeeding class ( $f_2$ ) $= 33$
Class size ( $h$ ) $= 2000 - 1500 = 500$

$\text{Mode} = l + \left( \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$

\begin{aligned} \text{Mode} &= 1500 + \left( \dfrac{40 - 24}{2 \times 40 - 24 - 33} \right) \times 500 \\[0.6em] &= 1500 + \left( \dfrac{16}{80 - 24 - 33} \right) \times 500 \\[0.6em] &= 1500 + \left( \dfrac{16}{23} \right) \times 500 \\[0.6em] &= 1500 + \dfrac{8000}{23} \\[0.6em] &= 1500 + 347.8260\dots \\[0.6em] &\approx 1847.83 \end{aligned}

Answer

(i) $\text{₹}1847.83$

(ii) Find the mean monthly expenditure of the families.

Step 1 · Calculate the Mean using Step-Deviation Method

Let Assumed Mean $a = 2750$ and Class Size $h = 500$ .

\begin{array}{|c|c|c|c|c|c|} \hline \text{Expenditure (in ₹)} & \text{Number of families } (f_i) & \text{Class Mark } (x_i) & d_i = x_i - 2750 & u_i = \dfrac{d_i}{500} & f_i u_i \\ \hline 1000 - 1500 & 24 & 1250 & -1500 & -3 & -72 \\ \hline 1500 - 2000 & 40 & 1750 & -1000 & -2 & -80 \\ \hline 2000 - 2500 & 33 & 2250 & -500 & -1 & -33 \\ \hline 2500 - 3000 & 28 & 2750 & 0 & 0 & 0 \\ \hline 3000 - 3500 & 30 & 3250 & 500 & 1 & 30 \\ \hline 3500 - 4000 & 22 & 3750 & 1000 & 2 & 44 \\ \hline 4000 - 4500 & 16 & 4250 & 1500 & 3 & 48 \\ \hline 4500 - 5000 & 7 & 4750 & 2000 & 4 & 28 \\ \hline \text{Total} & \sum f_i = 200 & & & & \sum f_i u_i = -35 \\ \hline \end{array}

Using the step-deviation formula: $\text{Mean } (\bar{x}) = a + \left( \dfrac{\sum f_i u_i}{\sum f_i} \right) \times h$

\begin{aligned} \bar{x} &= 2750 + \left( \dfrac{-35}{200} \right) \times 500 \\[0.6em] &= 2750 + (-0.175) \times 500 \\[0.6em] &= 2750 - 87.5 \\[0.6em] &= 2662.50 \end{aligned}

Answer

(ii) $\text{₹}2662.50$

Common Mistakes

Frequencies Confusion: Swapping $f_0$ , $f_1$ , and $f_2$ . Note that $f_1 = 40$ is the modal class frequency, $f_0 = 24$ is the preceding class frequency, and $f_2 = 33$ is the succeeding class frequency.
Sign Errors in Step-Deviation: Forgetting negative signs for deviations below the assumed mean ( $u_i = -3, -2, -1$ ), which leads to an incorrect sum $\sum f_i u_i$ .
Denominator in Mode Formula: Incorrectly computing $2f_1 - f_0 - f_2$ . Always evaluate $2f_1 = 2(40) = 80$ first before subtracting $f_0$ and $f_2$ .

Question 3

More questions in Exercise 13.2