NCERT MCQ Solutions for Class 9 Maths Chapter 12 Statistics

Bar graphs are pictorial representations where bars of uniform width are drawn with equal spacing on one axis and heights proportional to values on the other.
They help in easily visualising comparisons among individual data items and make interpretation simpler than using only tables.

Bar graphs are most suitable for comparing individual data items visually. Each bar represents a distinct value or category.
Since comparisons are easier to grasp with visual cues than raw numbers, bar graphs help show trends and differences effectively using equal-width bars.

In bar graphs, equal spacing is maintained between the bars on one axis to ensure visual clarity and uniformity.
This spacing helps in comparing the height of bars effectively.
The width of each bar is also kept the same to avoid misleading interpretations of the data.

In most bar graphs, the horizontal axis or x-axis represents the variable such as category names or time periods.
The values corresponding to these variables are shown on the vertical axis.
This arrangement helps read the graph easily and relate variables to their respective values.

The height of each bar in a bar graph represents the value of the variable.
Taller bars indicate higher values and shorter bars indicate lower values.
Equal bar widths ensure that only the height influences perception of value and comparisons remain accurate.

The class mark and class size are given as 25 and 10, respectively. The class mark is the midpoint of the class interval, and the class size is the difference between the upper and lower limits of the class interval.
Let the class interval be represented as:
Lower Limit = L, Upper Limit = U.
The class mark is calculated as:
Class Mark = (L + U) / 2.
Substitute the given class mark (25):
25 = (L + U) / 2.
Multiply through by 2:
L + U = 50. — (1)
The class size is the difference between the upper and lower limits:
Class Size = U – L.
Substitute the given class size (10):
10 = U – L. — (2)
Now solve the system of equations (1) and (2):
From (2): U = L + 10.
Substitute U = L + 10 into (1):
L + (L + 10) = 50,
2L + 10 = 50,
2L = 40,
L = 20.
Substitute L = 20 into U = L + 10:
U = 20 + 10 = 30.

The first ten prime numbers are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29.
To find the median, arrange the numbers in ascending order (already done above). Since there are 10 numbers (an even count), the median is the average of the two middle numbers. The two middle numbers are the 5th and 6th numbers in the list:
5th number = 11,
6th number = 13.
The median is calculated as:
Median = (11 + 13) / 2,
Median = 24 / 2,
Median = 12.

In histograms, there are no gaps between bars because it represents continuous grouped data.
Each bar touches the adjacent bar, unlike bar graphs where spacing exists between categories.
This visual continuity reflects that the data is part of a continuous interval without interruptions.

Histograms are used to represent continuous class interval data such as age, weight or time.
Unlike bar graphs which show discrete categories, histograms show frequencies over continuous intervals and the area of the bars is proportional to the frequency of the class interval.

The class mark of any class interval is the average of its upper and lower limits.
For the class 20 – 30, the class mark = (20 + 30) ÷ 2 = 25.
Class marks are used in frequency polygons to represent class intervals as single points on a graph.

A frequency polygon uses class marks as x-values and corresponding frequencies as y-values to plot points. These points are then joined with line segments to form the polygon.
Class marks are crucial because they represent the central values of each class interval.

To complete a frequency polygon properly, two additional class marks are added—one before the first and one after the last class interval.
These have zero frequencies and ensure the polygon begins and ends on the horizontal axis, giving the shape a closed form.

The first ten multiples of 5 are:
5, 10, 15, 20, 25, 30, 35, 40, 45, 50.
To calculate the mean, use the formula:
Mean = (Sum of all observations) / (Number of observations).
Calculate the sum of the first ten multiples of 5.
Sum = 5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45 + 50.
This is an arithmetic progression (AP) with:
– First term (a) = 5,
– Last term (l) = 50,
– Number of terms (n) = 10.
The sum of an AP is given by:
Sum = n/2 × (a + l).
Substitute the values:
Sum = 10/2 × (5 + 50),
Sum = 5 × 55,
Sum = 275.
Calculate the mean.
Mean = Sum / Number of observations,
Mean = 275 / 10,
Mean = 55.

The first ten multiples of 2 are:
2, 4, 6, 8, 10, 12, 14, 16, 18, 20.
To calculate the mean, use the formula:
Mean = (Sum of all observations) / (Number of observations).
Calculate the sum of the first ten multiples of 2.
Sum = 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20.
This is an arithmetic progression (AP) with:
– First term (a) = 2,
– Last term (l) = 20,
– Number of terms (n) = 10.
The sum of an AP is given by:
Sum = n/2 × (a + l).
Substitute the values:
Sum = 10/2 × (2 + 20),
Sum = 5 × 22,
Sum = 110.
Calculate the mean.
Mean = Sum / Number of observations,
Mean = 110 / 10,
Mean = 11.

The area enclosed by a frequency polygon represents the total frequency of the data set.
When plotted accurately with frequencies and class marks, the polygon approximates the distribution of the data and the area enclosed is proportional to the sum of all frequencies.

A histogram consists of adjacent rectangles representing continuous class intervals. These bars are drawn without gaps to show that the data is continuous.
The area of each rectangle corresponds to the frequency of the class interval it represents in the distribution.

When class intervals are unequal, bar heights must be adjusted proportionally to ensure accurate representation.
This means calculating the frequency per unit width and then scaling the height accordingly, so the area of each bar remains proportional to its frequency.

Class marks are not required for drawing a histogram. Instead, class intervals and frequencies are necessary. Histograms represent grouped frequency data with continuous intervals.
Class marks are essential for frequency polygons but are not used when plotting histogram bars.

Total frequency is the sum of frequencies of all class intervals.
Here, 3 + 7 + 5 = 15.
This total represents the number of observations in the data set and is important when computing measures like mean, median or when interpreting graphs.

The first ten multiples of 3 are:
3, 6, 9, 12, 15, 18, 21, 24, 27, 30.
To find the median, arrange the numbers in ascending order (already done above). Since there are 10 numbers (an even count), the median is the average of the two middle numbers. The two middle numbers are the 5th and 6th numbers in the list:
5th number = 15,
6th number = 18.
The median is calculated as:
Median = (15 + 18) / 2,
Median = 33 / 2,
Median = 16.5.

To calculate the average of the numbers, use the formula:
Average = (Sum of all observations) / (Number of observations).
Calculate the sum of the numbers.
The numbers are: 10, 8, 9, 7, 8.
Sum = 10 + 8 + 9 + 7 + 8 = 42.
Count the number of observations.
There are 5 numbers in the list.
Calculate the average.
Average = Sum / Number of observations,
Average = 42 / 5,
Average = 8.4.

Histograms are best for grouped data with continuous intervals such as ages or marks.
The graphical format allows users to understand the shape and distribution of the data quickly.
It is not suitable for scattered or purely categorical data that lacks continuity.

If class intervals are not continuous, they must be adjusted by modifying boundaries to avoid overlapping or gaps.
This ensures that the histogram accurately reflects a continuous distribution and that the bars correctly represent adjacent intervals without breaks between them.

A frequency polygon can be drawn without a histogram if class marks and frequencies are known.
By plotting points using class marks on the x-axis and frequencies on the y-axis and connecting them with lines, a frequency polygon is formed showing distribution trends.

In frequency polygons, each plotted point corresponds to a class mark and its respective frequency.
The vertical position (height) of the point represents the frequency for that class.
Connecting these points reveals the distribution pattern and helps identify trends like peaks and symmetry.

Frequency polygons are ideal for comparing two or more sets of data on the same graph.
By plotting multiple polygons together using common class marks, we can visually compare how different groups perform or distribute across the same interval range.

The first 10 natural numbers are:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
To calculate the mean, use the formula:
Mean = (Sum of all observations) / (Number of observations).
Calculate the sum of the first 10 natural numbers.
This is an arithmetic progression (AP) with:
– First term (a) = 1,
– Last term (l) = 10,
– Number of terms (n) = 10.
The sum of an AP is given by:
Sum = n/2 × (a + l).
Substitute the values:
Sum = 10/2 × (1 + 10),
Sum = 5 × 11,
Sum = 55.
Calculate the mean.
Mean = Sum / Number of observations,
Mean = 55 / 10,
Mean = 5.5.

We are given the following information:
The width of each class in the frequency distribution is 5.
The lower class limit of the lowest class is 10.
There are 5 continuous classes.
Determine the class intervals.
Each class has a width of 5, so the class intervals can be written as:
1. First class: 10 – 15 (lower limit = 10, upper limit = 10 + 5 = 15),
2. Second class: 15 – 20,
3. Third class: 20 – 25,
4. Fourth class: 25 – 30,
5. Fifth class: 30 – 35.
Identify the upper class limit of the highest class.
From the above intervals, the highest class is the fifth class, which has an upper class limit of 35.

Class mark represents the central value of a class interval and is calculated as (lower limit + upper limit) ÷ 2.
For the class 140 – 150, the class mark is (140 + 150) ÷ 2 = 145.
It is used as the x-value in frequency polygons.

A frequency polygon is formed by joining the midpoints of the tops of the rectangles in a histogram. These midpoints correspond to class marks.
When connected, they form a polygon that approximates the data’s distribution pattern and is often smoother to interpret.

The class mark of 0 – 10 is (0 + 10) ÷ 2 = 5.
To complete the frequency polygon, a point before the first class is added with zero frequency.
This means starting from (5, 0), assuming the next adjacent class mark on the x-axis.

In histograms, the area of each bar (height × width) is proportional to the frequency of that class interval.
If widths vary, then height must be adjusted to maintain accurate frequency representation.
This makes histograms useful for both equal and unequal class widths.

A bar graph can be misleading if the bar widths are unequal because it distorts the visual impression of value.
Bar graphs require uniform width and equal spacing between bars to ensure fair comparisons.
Any deviation in width can cause misinterpretation of the data.

We are given the following information:
The mean of 5 numbers is 30.
When one number is excluded, the mean of the remaining 4 numbers becomes 28.
Calculate the total sum of the 5 numbers.
The formula for the mean is:
Mean = (Sum of all observations) / (Number of observations).
For the 5 numbers:
Mean = 30,
Number of observations = 5.
Substitute into the formula:
30 = (Sum of all 5 numbers) / 5.
Multiply through by 5:
Sum of all 5 numbers = 30 × 5 = 150.
Calculate the total sum of the remaining 4 numbers.
When one number is excluded, the mean of the remaining 4 numbers is 28. Using the same formula:
Mean = (Sum of remaining 4 numbers) / (Number of observations).
For the 4 numbers:
Mean = 28,
Number of observations = 4.
Substitute into the formula:
28 = (Sum of remaining 4 numbers) / 4.
Multiply through by 4:
Sum of remaining 4 numbers = 28 × 4 = 112.
Find the excluded number.
The excluded number is the difference between the total sum of the 5 numbers and the sum of the remaining 4 numbers:
Excluded number = (Sum of all 5 numbers) – (Sum of remaining 4 numbers).
Substitute the values:
Excluded number = 150 – 112 = 38.

Arrange the data in ascending order:
22, 34, 39, 45, 54, 54, 56, 68, 78, 84.
Count the number of observations.
There are 10 numbers (even count).
Find the median.
For an even number of observations, the median is the average of the two middle numbers. The two middle numbers are the 5th and 6th numbers:
5th number = 54,
6th number = 54.
Median = (54 + 54) / 2 = 54.