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Q3:
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 :
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 :
Given: The frequency distribution of monthly household expenditure of 200 families.
| Expenditure (in ₹) | Number of Families ($f_i$) |
|---|---|
| 1000 - 1500 | 24 |
| 1500 - 2000 | 40 |
| 2000 - 2500 | 33 |
| 2500 - 3000 | 28 |
| 3000 - 3500 | 30 |
| 3500 - 4000 | 22 |
| 4000 - 4500 | 16 |
| 4500 - 5000 | 7 |
To Find: The Modal monthly expenditure and the Mean monthly expenditure.
Part 1: Finding the Mode
Step 1: Identify the Modal Class. The modal class is the class interval with the highest frequency. Here, the highest frequency is $40$, which corresponds to the class interval $1500 - 2000$.
Step 2: Apply the Mode Formula. The formula for mode is: $Mode = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
Where:
$l$ (lower limit of modal class) = $1500$
$f_1$ (frequency of modal class) = $40$
$f_0$ (frequency of class preceding modal class) = $24$
$f_2$ (frequency of class succeeding modal class) = $33$
$h$ (class size) = $500$
Step 3: Calculation.
$Mode = 1500 + \left( \frac{40 - 24}{2(40) - 24 - 33} \right) \times 500$
$Mode = 1500 + \left( \frac{16}{80 - 57} \right) \times 500$
$Mode = 1500 + \left( \frac{16}{23} \right) \times 500$
$Mode = 1500 + \frac{8000}{23} \approx 1500 + 347.83 = 1847.83$
Part 2: Finding the Mean
Step 1: Calculate Class Marks ($x_i$). $x_i = \frac{\text{Upper Limit} + \text{Lower Limit}}{2}$. Let assumed mean $a = 2750$.
| Expenditure | $f_i$ | $x_i$ | $d_i = x_i - 2750$ | $u_i = \frac{d_i}{500}$ | $f_i u_i$ |
|---|---|---|---|---|---|
| 1000-1500 | 24 | 1250 | -1500 | -3 | -72 |
| 1500-2000 | 40 | 1750 | -1000 | -2 | -80 |
| 2000-2500 | 33 | 2250 | -500 | -1 | -33 |
| 2500-3000 | 28 | 2750 | 0 | 0 | 0 |
| 3000-3500 | 30 | 3250 | 500 | 1 | 30 |
| 3500-4000 | 22 | 3750 | 1000 | 2 | 44 |
| 4000-4500 | 16 | 4250 | 1500 | 3 | 48 |
| 4500-5000 | 7 | 4750 | 2000 | 4 | 28 |
| Total | 200 | - | - | - | -35 |
Step 2: Apply the Step-Deviation Formula. $\bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h$
$\bar{x} = 2750 + \left( \frac{-35}{200} \right) \times 500$
$\bar{x} = 2750 + (-0.175 \times 500)$
$\bar{x} = 2750 - 87.5 = 2662.5$
Final Answer: The modal monthly expenditure is ₹1847.83 and the mean monthly expenditure is ₹2662.50.
More Questions from Class 10 Mathematics Statistics EXERCISE 13.2
- Q1: The following table shows the ages of the patients admitted in a hospital during a year: Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
- Q2: The following data gives the information on the observed lifetimes (in hours) of 225 electrical components : Determine the modal lifetimes of the components.
- Q4: The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.
- Q5: The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches. Find the mode of the data.
- Q6: A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data:
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