NCERT Solutions for Class 9 Maths Chapter 12: Statistics — Free PDF

Data: Information collected for a specific purpose. Data can be primary (collected first-hand) or secondary (obtained from existing sources).

Raw data: Unorganised data as originally collected.

Frequency: The number of times a particular observation occurs in the data set.

Ungrouped frequency distribution: A table listing each distinct observation alongside its frequency. Best used when the range of data is small.

Grouped frequency distribution: Data is organised into non-overlapping class intervals (e.g., $0$–$10$, $10$–$20$). Used when the range is large. The class intervals must be continuous (the upper limit of one class equals the lower limit of the next).

Class mark (midpoint): $\dfrac{\text{Upper limit} + \text{Lower limit}}{2}$. This value represents all observations falling in that class.

Class width (size): Upper boundary $-$ Lower boundary of a class interval.

A measure of central tendency is a single value that represents the centre of the data set.

• Mean: The arithmetic average. Uses every data point but is sensitive to outliers.
  - Median: The middle value when data is sorted. Robust to extreme values.
  - Mode: The most frequently occurring value. The only measure applicable to categorical data.

Each measure has different strengths, and choosing the right one depends on the nature of the data.

Bar Graph: Used for discrete or categorical data. Bars have equal gaps between them.

Histogram: Used for continuous grouped data. Bars touch each other (no gaps). The width of each bar equals the class width.

Frequency Polygon: Formed by joining the midpoints of the tops of histogram bars with straight lines. Must be closed at both ends with zero-frequency points.

The blood groups of 30 students are: A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O, A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O. Prepare a frequency distribution table.

Solution:

Verification: $9 + 6 + 3 + 12 = 30$ ✓

This is an ungrouped frequency distribution because each category is listed individually.

The distance (in km) of 40 engineers from their residence to their place of work were found. Construct a grouped frequency distribution table with class size 5.

Solution:
Classes: $0$–$5$, $5$–$10$, $10$–$15$, $15$–$20$, $20$–$25$, $25$–$30$, $30$–$35$.

Count the number of observations falling in each class interval. The class $0$–$5$ includes values $\geq 0$ and $< 5$.

In a grouped frequency distribution, the upper boundary of one class equals the lower boundary of the next class (continuous classes).

Construct a frequency distribution for the marks of 30 students with class intervals $0$–$10$, $10$–$20$, etc.:
6, 15, 27, 34, 41, 8, 22, 19, 37, 48, 11, 25, 33, 16, 44, 9, 29, 38, 21, 47, 12, 31, 40, 7, 23, 36, 18, 43, 14, 28.

Solution:

Always verify that the total frequency equals the number of data points.

Draw a histogram for the following distribution:

Solution:
Steps:
1. Mark class intervals on the x-axis.
2. Mark frequencies on the y-axis.
3. Draw rectangles with widths equal to class size and heights equal to frequencies.
4. There should be no gap between consecutive bars.

The tallest bar corresponds to the class $30$–$40$ with frequency $12$.

For the same data, draw a frequency polygon.

Solution:
Find the midpoints of each class: $5, 15, 25, 35, 45$.
Plot points: $(5, 5)$, $(15, 10)$, $(25, 8)$, $(35, 12)$, $(45, 6)$.
Add closure points at $(-5, 0)$ and $(55, 0)$ to close the polygon.
Join all points with straight lines.

The frequency polygon gives a clearer picture of the trend in the data compared to the histogram.

The following table gives the number of students in each age group:

Solution:
1. Draw the histogram with bars for each class.
2. Mark the midpoints: $7.5$, $12.5$, $17.5$, $22.5$, $27.5$.
3. Plot points at $(7.5, 40)$, $(12.5, 32)$, $(17.5, 48)$, $(22.5, 36)$, $(27.5, 24)$.
4. Add closure points $(2.5, 0)$ and $(32.5, 0)$.
5. Join all points to form the polygon.

Note: the frequency polygon can be drawn independently without the histogram by directly plotting midpoint-frequency pairs.

Find the mean of the first five natural numbers.

Solution:

If the mean of $6, 8, 5, x, 4$ is $7$, find $x$.

Solution:

Find the mean of the following frequency distribution:

Solution:
$\sum f_i x_i = 10(3) + 15(5) + 20(7) + 25(4) + 30(1) = 30 + 75 + 140 + 100 + 30 = 375$

$\sum f_i = 3 + 5 + 7 + 4 + 1 = 20$

Find the median of: 13, 16, 12, 14, 19, 12, 14, 13, 14.

Solution:
Arranging in ascending order: $12, 12, 13, 13, 14, 14, 14, 16, 19$
$n = 9$ (odd), so median $= \left(\dfrac{9+1}{2}\right)^{\text{th}} = 5^{\text{th}}$ observation $= 14$.

Find the median of: 3, 7, 2, 5, 8, 1, 4, 6.

Solution:
Arranging in ascending order: $1, 2, 3, 4, 5, 6, 7, 8$
$n = 8$ (even).
Median $= \dfrac{1}{2}\left[4^{\text{th}} + 5^{\text{th}}\right] = \dfrac{1}{2}[4 + 5] = 4.5$

Find the mode of: 2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4.

Solution:
Frequencies: $0$(1), $2$(3), $3$(3), $4$(3), $5$(2), $6$(1).
Three values ($2$, $3$, $4$) each appear $3$ times. This is a trimodal distribution with modes $2$, $3$, and $4$.

A data set can have no mode (all values occur equally), one mode (unimodal), two modes (bimodal), or more (multimodal).

The mean of the following distribution is $22$. Find the missing frequency $p$.

Solution:
$\sum f_i = 4 + p + 8 + 2 = 14 + p$

$\sum f_i x_i = 10(4) + 20p + 30(8) + 40(2) = 40 + 20p + 240 + 80 = 360 + 20p$

The daily wages (in Rs) of 10 workers: 120, 120, 130, 130, 130, 140, 160, 160, 200, 1000. Compute the mean, median, and mode. Which best represents the data?

Solution:
Mean $= \dfrac{2290}{10} = 229$

Median: $n = 10$, so median $= \dfrac{5^{\text{th}} + 6^{\text{th}}}{2} = \dfrac{130 + 140}{2} = 135$

Mode $= 130$ (appears $3$ times)

The mean ($229$) is pulled up by the outlier ($1000$). The median ($135$) best represents the typical wage since it is not affected by extreme values.

If $5$ is added to each observation, how do the mean, median, and mode change?

Solution:
All three increase by $5$. In general, if a constant $k$ is added to every observation:
- New mean $=$ old mean $+ k$
- New median $=$ old median $+ k$
- New mode $=$ old mode $+ k$

Similarly, if every observation is multiplied by $k$, then the mean, median, and mode are all multiplied by $k$.

Find the range of: $15, 22, 9, 47, 31, 18, 6, 40$.

Solution:
Range $=$ Maximum value $-$ Minimum value $= 47 - 6 = 41$.

The range gives a rough measure of spread but is heavily influenced by extreme values. It uses only two data points, so it does not tell the full story about how data is distributed.

The median is the middle value of the sorted data, not the value in the middle of the original list. Always arrange the observations in ascending (or descending) order before finding the median.

If the given classes have gaps (e.g., $1$–$10$, $11$–$20$, $21$–$30$), you must convert them to continuous classes ($0.5$–$10.5$, $10.5$–$20.5$, $20.5$–$30.5$) by subtracting $0.5$ from each lower limit and adding $0.5$ to each upper limit.

A frequency polygon must close at both ends by adding points with zero frequency — one class-width before the first class midpoint and one class-width after the last. Without these closure points, the polygon is incomplete.

For data with frequencies, the mean is $\dfrac{\sum f_i x_i}{\sum f_i}$, NOT $\dfrac{\sum x_i}{n}$. Each observation must be weighted by its frequency. Forgetting the frequency weights is a very common error.

A data set can have no mode (if all values occur equally often), one mode (unimodal), two modes (bimodal), or more (multimodal). Do not force a single mode when the data has multiple values with the same highest frequency.

The marks of 10 students are: 45, 60, 55, 72, 38, 55, 80, 55, 62, 48. Find the mean, median, and mode.

Answer:
Sorted: $38, 45, 48, 55, 55, 55, 60, 62, 72, 80$.
Mean $= \dfrac{570}{10} = 57$.
Median ($n = 10$, even): $\dfrac{55 + 55}{2} = 55$.
Mode $= 55$ (appears $3$ times).

If the mean of $5, 8, 11, p, 20$ is $12$, find $p$.

Answer: $\dfrac{5 + 8 + 11 + p + 20}{5} = 12 \implies 44 + p = 60 \implies p = 16$.

Find the class mark of the class interval $25$–$35$.

Answer: Class mark $= \dfrac{25 + 35}{2} = 30$.

A survey records the favourite colours of $50$ students. Should you use a histogram or a bar graph to represent this data? Explain.

Answer: Use a bar graph. Favourite colour is categorical (discrete) data, not continuous grouped data. Bars should have gaps between them. A histogram is used only for continuous grouped numerical data.