CBSE Class 10 Mathematics Statistics Worksheet Set B - Free Printable

Let's solve each question from the worksheet step by step.

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Section A (1 mark each)

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#### Q1. Find the sum of lower limits of median and modal class for the following distribution:

| Class | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 |
|-------------|------|------|-------|-------|-------|
| Frequency | 10 | 15 | 12 | 20 | 9 |

Step 1: Total frequency (N)
$$
N = 10 + 15 + 12 + 20 + 9 = 66
$$

Step 2: Find median class

Median position:
$$
\frac{N}{2} = \frac{66}{2} = 33
$$

Cumulative frequencies:

| Class | Frequency | Cumulative Freq |
|----------|-----------|------------------|
| 0-5 | 10 | 10 |
| 5-10 | 15 | 25 |
| 10-15 | 12 | 37 |
| 15-20 | 20 | 57 |
| 20-25 | 9 | 66 |

The cumulative frequency just greater than 33 is 37, which lies in class 10–15 → Median class = 10–15

Lower limit of median class = 10

Step 3: Find modal class

Modal class is the one with highest frequency → 15–20 has frequency 20 → Modal class = 15–20

Lower limit of modal class = 15

Sum of lower limits = 10 + 15 = 25

✔ Answer: 25

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#### Q2. Find the upper limit of the median class of the following frequency distribution:

| Class | 0-5 | 6-11 | 12-17 | 18-23 | 24-29 |
|-----------|------|------|-------|-------|-------|
| Frequency | 15 | 10 | 15 | 8 | 11 |

Total frequency N = 15 + 10 + 15 + 8 + 11 = 59

Median position:
$$
\frac{N}{2} = \frac{59}{2} = 29.5
$$

Cumulative frequencies:

| Class | Frequency | Cumulative Freq |
|-----------|-----------|------------------|
| 0-5 | 15 | 15 |
| 6-11 | 10 | 25 |
| 12-17 | 15 | 40 |
| 18-23 | 8 | 48 |
| 24-29 | 11 | 59 |

Cumulative frequency just exceeding 29.5 is 40 → Median class = 12–17

Upper limit of this class = 17

✔ Answer: 17.5? Wait — let’s check the answer given: (Ans: 17.5)

But our class is 12–17, so upper limit should be 17, not 17.5.

Wait! Is the class interval continuous?

Let’s check if classes are continuous:

- 0–5, 6–11 → gap between 5 and 6 → not continuous.

So we need to make it continuous.

We can adjust the boundaries:

- 0–5 → 0.5 to 5.5
- 6–11 → 5.5 to 11.5
- 12–17 → 11.5 to 17.5
- 18–23 → 17.5 to 23.5
- 24–29 → 23.5 to 29.5

So actual class intervals are:

| Class (adjusted) | 0.5–5.5 | 5.5–11.5 | 11.5–17.5 | 17.5–23.5 | 23.5–29.5 |
|------------------|---------|----------|-----------|-----------|-----------|
| Frequency | 15 | 10 | 15 | 8 | 11 |

Now cumulative freq:

- 0.5–5.5: 15
- 5.5–11.5: 25
- 11.5–17.5: 40 → exceeds 29.5

So median class is 11.5–17.5

Upper limit = 17.5

✔ Answer: 17.5

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#### Q3. Find the mean of the numbers 1, 2, 3, ... n

Mean = $\frac{\text{Sum}}{\text{Number of terms}}$

Sum of first $n$ natural numbers = $\frac{n(n+1)}{2}$

Number of terms = $n$

So,
$$
\text{Mean} = \frac{n(n+1)/2}{n} = \frac{n+1}{2}
$$

✔ Answer: $\boxed{\frac{n+1}{2}}$

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#### Q4. Which measure of central tendency is obtained from the abscissa of the point of intersection of the less than type and more than type cumulative frequency curves of a grouped data?

- The less than ogive and more than ogive intersect at the median.
- The x-coordinate (abscissa) of that point gives the median.

✔ Answer: Median

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#### Q5. For the following distribution, find the modal class.

| Marks | Below 10 | Below 20 | Below 30 | Below 40 | Below 50 | Below 60 |
|---------------|----------|----------|----------|----------|----------|----------|
| No. of students | 3 | 12 | 27 | 57 | 75 | 80 |

This is a cumulative frequency distribution (less than type).

We need to convert to frequency distribution:

| Marks | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 | 50–60 |
|-----------|------|-------|-------|-------|-------|-------|
| C.F. | 3 | 12 | 27 | 57 | 75 | 80 |
| Freq. | 3 | 9 | 15 | 30 | 18 | 5 |

Frequency table:

| Class | Frequency |
|----------|-----------|
| 0–10 | 3 |
| 10–20 | 9 |
| 20–30 | 15 |
| 30–40 | 30 |
| 40–50 | 18 |
| 50–60 | 5 |

Highest frequency = 30 → class 30–40

✔ Answer: Modal class = 30–40

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Section B (2 marks each)

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#### Q6. The following is the distribution of weights (in kg) of 40 persons:

| Weight (kg) | 40-45 | 45-50 | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 | 75-80 |
|-------------|-------|-------|-------|-------|-------|-------|-------|-------|
| No. of persons | 4 | 6 | 13 | 5 | 6 | 5 | 2 | 1 |

Construct a cumulative frequency distribution (less than type) table.

Less than type: We find number of persons with weight less than upper limit of each class.

| Weight (kg) | No. of persons | Less than cumulative frequency |
|-------------|----------------|-------------------------------|
| 40–45 | 4 | 4 |
| 45–50 | 6 | 4 + 6 = 10 |
| 50–55 | 13 | 10 + 13 = 23 |
| 55–60 | 5 | 23 + 5 = 28 |
| 60–65 | 6 | 28 + 6 = 34 |
| 65–70 | 5 | 34 + 5 = 39 |
| 70–75 | 2 | 39 + 2 = 41 |
| 75–80 | 1 | 41 + 1 = 42 |

But total persons = 4+6+13+5+6+5+2+1 = 42 → correct.

So cumulative frequency table:

| Weight (kg) | Less than cumulative frequency |
|-------------|--------------------------------|
| Less than 45 | 4 |
| Less than 50 | 10 |
| Less than 55 | 23 |
| Less than 60 | 28 |
| Less than 65 | 34 |
| Less than 70 | 39 |
| Less than 75 | 41 |
| Less than 80 | 42 |

✔ Answer: Table constructed as above

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#### Q7. Calculate the modal agricultural holdings of the village.

Given:

| Area of land (ha) | 1-3 | 3-5 | 5-7 | 7-9 | 9-11 | 11-13 |
|--------------------|-----|-----|-----|-----|------|-------|
| No. of families | 20 | 45 | 80 | 55 | 40 | 12 |

We need to find mode.

Step 1: Identify modal class

Highest frequency = 80 → class 5–7

So modal class = 5–7

Use mode formula for grouped data:

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

Where:
- $l$ = lower limit of modal class = 5
- $f_1$ = frequency of modal class = 80
- $f_0$ = frequency of class before = 45
- $f_2$ = frequency of class after = 55
- $h$ = class width = 2

Plug in:

$$
\text{Mode} = 5 + \left( \frac{80 - 45}{2 \times 80 - 45 - 55} \right) \times 2
= 5 + \left( \frac{35}{160 - 100} \right) \times 2
= 5 + \left( \frac{35}{60} \right) \times 2
= 5 + \left( \frac{7}{12} \right) \times 2
= 5 + \frac{14}{12}
= 5 + 1.1667 = 6.1667
$$

So, modal agricultural holding ≈ 6.17 ha

✔ Answer: Mode ≈ 6.17 hectares

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#### Q8. Construct a cumulative frequency distribution (more than type) for the following distribution:

| Class | 12.5–17.5 | 17.5–22.5 | 22.5–27.5 | 27.5–32.5 | 32.5–37.5 |
|----------------|-----------|-----------|-----------|-----------|-----------|
| Frequency | 2 | 22 | 19 | 14 | 13 |

More than type cumulative frequency means: number of observations greater than or equal to the lower limit.

Start from the bottom:

| Class | Frequency | More than cumulative frequency |
|----------------|-----------|-------------------------------|
| 12.5–17.5 | 2 | 2 + 22 + 19 + 14 + 13 = 70 |
| 17.5–22.5 | 22 | 22 + 19 + 14 + 13 = 68 |
| 22.5–27.5 | 19 | 19 + 14 + 13 = 46 |
| 27.5–32.5 | 14 | 14 + 13 = 27 |
| 32.5–37.5 | 13 | 13 |

So:

| More than (lower limit) | Cumulative frequency |
|-------------------------|----------------------|
| More than 12.5 | 70 |
| More than 17.5 | 68 |
| More than 22.5 | 46 |
| More than 27.5 | 27 |
| More than 32.5 | 13 |

✔ Answer: Table constructed as above

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✔ Final Answers Summary:

1. 25
2. 17.5
3. $\frac{n+1}{2}$
4. Median
5. 30–40
6. Cumulative frequency (less than):
- <45: 4, <50: 10, <55: 23, <60: 28, <65: 34, <70: 39, <75: 41, <80: 42
7. Mode ≈ 6.17 ha
8. Cumulative frequency (more than):
- >12.5: 70, >17.5: 68, >22.5: 46, >27.5: 27, >32.5: 13

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