NCERT Solutions for Class 10 Maths Chapter 13: Statistics — Free PDF

Direct method:

Assumed mean method:

Step deviation method:

The modal class is the class with the highest frequency. The mode formula is:

Important: The modal class is identified by the highest frequency, NOT by the highest cumulative frequency.

First compute cumulative frequencies, then identify the median class — the class whose cumulative frequency first exceeds $\dfrac{n}{2}$.

Empirical relationship: For moderately skewed distributions:

Cumulative frequency curve (Ogive): Plot cumulative frequency on the y-axis against upper class boundaries on the x-axis. The median is the x-coordinate where the ogive crosses $\dfrac{n}{2}$ on the y-axis.

Less-than ogive: Plot (upper boundary, cumulative frequency) — a rising curve.
More-than ogive: Plot (lower boundary, n minus cumulative frequency) — a falling curve.
The x-coordinate of their intersection gives the median.

Problem: Find the mean of the following distribution using the step deviation method.

Solution:
Let $a = 50$ (assumed mean), $h = 20$ (class width).

Answer: Mean $= 50$.

Problem: The following table gives the literacy rate (in %) of 35 cities. Find the mean literacy rate.

Solution (Direct method):

Answer: Mean literacy rate $\approx 69.43\%$.

Problem: Find the mean of the following distribution using the assumed mean method.

Solution:
Let $a = 47.5$ (midpoint of the most frequent class 40-55). Class width $h = 15$.

Answer: Mean $= 54.375$.

Problem: Find the mode of the following data.

Solution:
Modal class = 20-30 (highest frequency $= 36$).

$l = 20$, $f_1 = 36$, $f_0 = 16$, $f_2 = 34$, $h = 10$.

Answer: Mode $\approx 29.09$.

Problem: The following data gives the number of patients in a hospital during a particular week. Find the mode.

Solution:
Modal class = 30-40 (highest frequency $= 23$).

$l = 30$, $f_1 = 23$, $f_0 = 21$, $f_2 = 14$, $h = 10$.

Answer: Mode $\approx 31.82$ years. The most common age group visiting the hospital is around 31-32 years.

Problem: Find the median of the following data.

Solution:
Cumulative frequencies: $5, 13, 33, 48, 55$.

$n = 55$, so $\dfrac{n}{2} = 27.5$.

Median class: 20-30 (cumulative frequency $33$ first exceeds $27.5$).

$l = 20$, $cf = 13$ (cumulative frequency before median class), $f = 20$, $h = 10$.

Answer: Median $= 27.25$.

Problem: Find the median for the following data with total frequency 170.

Solution:
$n = 170$, $\dfrac{n}{2} = 85$.

Cumulative frequencies: $25, 60, 110, 150, 170$.

Median class: 40-60 (cumulative frequency $110$ is the first to exceed $85$).

$l = 40$, $cf = 60$, $f = 50$, $h = 20$.

Answer: Median $= 50$.

Problem: Find the median wage from the following distribution.

Solution:
Cumulative frequencies: $12, 26, 34, 40, 50$.

$n = 50$, $\dfrac{n}{2} = 25$.

Median class: 120-140 (cumulative frequency $26$ first exceeds $25$).

$l = 120$, $cf = 12$, $f = 14$, $h = 20$.

Answer: Median wage $\approx$ Rs $138.57$.

Steps:
1. Construct a cumulative frequency table using upper class boundaries.
2. Plot points: (upper boundary, cumulative frequency).
3. Join the points with a smooth freehand curve.
4. Start the curve from (lower boundary of first class, 0).

Finding the median from the ogive:
Locate $\dfrac{n}{2}$ on the y-axis, draw a horizontal line to the curve, then drop a perpendicular to the x-axis. The x-coordinate is the median.

Example: For the data:
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
|---|---|---|---|---|---|
| Frequency | 5 | 8 | 20 | 15 | 7 |

Cumulative frequency table:
| Less than | 10 | 20 | 30 | 40 | 50 |
|---|---|---|---|---|---|
| CF | 5 | 13 | 33 | 48 | 55 |

Plot: $(10, 5), (20, 13), (30, 33), (40, 48), (50, 55)$.

Locate $\dfrac{55}{2} = 27.5$ on y-axis. The corresponding x-value is approximately $27.25$, which matches our earlier calculation.

More-than ogive: Plot (lower boundary, $n$ minus cumulative frequency).

For the same data:
| More than | 0 | 10 | 20 | 30 | 40 |
|---|---|---|---|---|---|
| CF | 55 | 50 | 42 | 22 | 7 |

Plot: $(0, 55), (10, 50), (20, 42), (30, 22), (40, 7)$.

Intersection method: Draw both ogives on the same graph. Their point of intersection gives the median on the x-axis. This is a commonly tested 4-5 mark question in board exams.

The less-than ogive is a rising curve and the more-than ogive is a falling curve. They always intersect at exactly one point, and the x-coordinate of that intersection is the median.

Problem: The mean of the following distribution is 50. Find the missing frequencies $f_1$ and $f_2$ if the total frequency is 120.

Solution:
$17 + f_1 + 32 + f_2 + 19 = 120$
$f_1 + f_2 = 52$ ...(i)

Using the step deviation method with $a = 50$, $h = 20$:
$\sum f_i u_i = 17(-2) + f_1(-1) + 32(0) + f_2(1) + 19(2) = -34 - f_1 + f_2 + 38 = 4 - f_1 + f_2$

Mean $= 50 + 20 \times \dfrac{4 - f_1 + f_2}{120} = 50$

So $4 - f_1 + f_2 = 0$, giving $f_2 - f_1 = -4$, i.e., $f_1 - f_2 = 4$ ...(ii)

From (i) and (ii): $f_1 = 28$ and $f_2 = 24$.

Problem: For a distribution, the mean is 35 and the mode is 32. Find the median using the empirical relationship.

Solution:
$3 \times \text{Median} = \text{Mode} + 2 \times \text{Mean}$
$3 \times \text{Median} = 32 + 2 \times 35 = 32 + 70 = 102$
$\text{Median} = \dfrac{102}{3} = 34$

Answer: Median $= 34$.

Note: This relationship is approximate and works best for moderately skewed distributions.

Problem: Find the mean of the following data using all three methods and verify they give the same answer.

Direct method:
$\sum f_i x_i = 4(15) + 8(25) + 10(35) + 12(45) + 6(55) = 60 + 200 + 350 + 540 + 330 = 1480$
$\bar{x} = \dfrac{1480}{40} = 37$

Assumed mean method ($a = 35$):
$\sum f_i d_i = 4(-20) + 8(-10) + 10(0) + 12(10) + 6(20) = -80 - 80 + 0 + 120 + 120 = 80$
$\bar{x} = 35 + \dfrac{80}{40} = 35 + 2 = 37$ ✓

Step deviation method ($a = 35$, $h = 10$):
$\sum f_i u_i = 4(-2) + 8(-1) + 10(0) + 12(1) + 6(2) = -8 - 8 + 0 + 12 + 12 = 8$
$\bar{x} = 35 + 10 \times \dfrac{8}{40} = 35 + 2 = 37$ ✓

All three methods give the same answer: **Mean $= 37$**.

Question: Find the mean of the following distribution.

Answer:
Let $a = 25$, $h = 10$.
$\sum f_i u_i = 7(-2) + 12(-1) + 18(0) + 10(1) + 3(2) = -14 - 12 + 0 + 10 + 6 = -10$
$\bar{x} = 25 + 10 \times \dfrac{-10}{50} = 25 - 2 = 23$.

**Mean $= 23$.**

Question: Find the mode.

Answer:
Modal class = 30-40. $l = 30$, $f_1 = 12$, $f_0 = 8$, $f_2 = 6$, $h = 10$.
$\text{Mode} = 30 + \dfrac{12 - 8}{24 - 8 - 6} \times 10 = 30 + \dfrac{4}{10} \times 10 = 30 + 4 = 34$.

**Mode $= 34$.**

Question: Find the median.

Answer:
$n = 80$, $\dfrac{n}{2} = 40$.
Cumulative frequencies: $10, 25, 50, 70, 80$.
Median class: 40-60 ($cf = 50$ first exceeds $40$).
$l = 40$, $cf = 25$, $f = 25$, $h = 20$.
$\text{Median} = 40 + \dfrac{40 - 25}{25} \times 20 = 40 + \dfrac{15}{25} \times 20 = 40 + 12 = 52$.

**Median $= 52$.**

Question: For a distribution, Mode $= 65$ and Mean $= 61.6$. Find the Median.

Answer:
$3 \times \text{Median} = 65 + 2 \times 61.6 = 65 + 123.2 = 188.2$
$\text{Median} = \dfrac{188.2}{3} = 62.73$.

**Median $\approx 62.73$.**