NCERT Class 9 Maths Solutions: Statistics

Question:

Concept in a Minute:
This question involves understanding and representing data using a histogram, which is a graphical representation of the distribution of numerical data. It also requires identifying the modal class from a frequency distribution.


Part 1: Drawing a Histogram

A histogram is a bar graph where the bars represent the frequency of data within specific intervals (class intervals).

Steps to draw the histogram:
1. Identify the axes: The horizontal axis (x-axis) will represent the “Number of letters” (the class intervals), and the vertical axis (y-axis) will represent the “Number of surnames” (the frequencies).
2. Determine the scale:
* For the x-axis, the class intervals are given: 1-4, 4-6, 6-8, 8-12, 12-20. Notice that the class intervals are not of equal width. This is important for drawing the histogram correctly.
* For the y-axis, the frequencies range from 4 to 44. A suitable scale would be to mark intervals of 5 or 10 (e.g., 0, 5, 10, 15, 20, 25, 30, 35, 40, 45).
3. Draw the bars: For each class interval, draw a rectangle (bar) whose width corresponds to the class interval and whose height corresponds to the frequency of that interval.
* Important Note for Unequal Class Widths: When class intervals are not of equal width, we need to adjust the height of the bars to maintain the correct area representation. The adjusted height (or frequency density) is calculated as: Frequency Density = Frequency / Class Width.
* Class 1-4: Class Width = 4 – 1 = 3. Frequency = 6. Frequency Density = 6 / 3 = 2.
* Class 4-6: Class Width = 6 – 4 = 2. Frequency = 30. Frequency Density = 30 / 2 = 15.
* Class 6-8: Class Width = 8 – 6 = 2. Frequency = 44. Frequency Density = 44 / 2 = 22.
* Class 8-12: Class Width = 12 – 8 = 4. Frequency = 16. Frequency Density = 16 / 4 = 4.
* Class 12-20: Class Width = 20 – 12 = 8. Frequency = 4. Frequency Density = 4 / 8 = 0.5.

Now, when drawing the histogram, the height of each bar will be its frequency density, and the width will be its actual class width.

* Bar 1 (1-4): Width = 3 units, Height = 2.
* Bar 2 (4-6): Width = 2 units, Height = 15.
* Bar 3 (6-8): Width = 2 units, Height = 22.
* Bar 4 (8-12): Width = 4 units, Height = 4.
* Bar 5 (12-20): Width = 8 units, Height = 0.5.

The bars in a histogram are adjacent to each other.

Part 2: Writing the class interval in which the maximum number of surnames lie.

This is asking for the modal class. The modal class is the class interval that has the highest frequency.

From the table:
* 1-4: 6 surnames
* 4-6: 30 surnames
* 6-8: 44 surnames
* 8-12: 16 surnames
* 12-20: 4 surnames

The highest frequency is 44, which corresponds to the class interval 6-8.

Therefore, the class interval in which the maximum number of surnames lie is 6-8.

Question:

Concept in a Minute:
A histogram is a graphical representation of the distribution of numerical data. It is an estimate of the probability distribution of a continuous variable (quantitative variable). To draw a histogram, you need to represent the data using bars where the width of the bar represents the class interval and the height of the bar represents the frequency of that class interval. For variable class widths, the area of the bar should be proportional to the frequency. The height of the bar is calculated as frequency divided by the class width, which is known as the frequency density.


The question asks to draw a histogram to represent the given data. A histogram uses bars to show the frequency of data points within specific ranges or intervals (class intervals). The age groups are the class intervals, and the number of children represents the frequency for each age group. Since the class intervals are not of uniform width (e.g., 1-2 has a width of 1, while 3-5 has a width of 2, and 7-10 has a width of 3), we need to calculate the frequency density for each class interval. The frequency density is calculated by dividing the frequency by the class width. This ensures that the area of each bar is proportional to the frequency, which is essential for accurate representation in a histogram with varying class widths.

Steps to draw the histogram:
1. Identify the class intervals (Age in years) and their corresponding frequencies (Number of children).
2. Calculate the width of each class interval.
3. Calculate the frequency density for each class interval using the formula: Frequency Density = Frequency / Class Width.
4. Draw the x-axis and label it with the class intervals.
5. Draw the y-axis and label it as “Frequency Density”.
6. Draw rectangular bars for each class interval. The width of each bar will correspond to the class interval on the x-axis, and the height of each bar will correspond to its calculated frequency density on the y-axis. Ensure that adjacent bars touch each other.

Detailed steps for calculation:
Class Interval | Frequency | Class Width | Frequency Density (Frequency / Class Width)
————-|———–|————-|———————————————
1 – 2 | 5 | 1 | 5 / 1 = 5
2 – 3 | 3 | 1 | 3 / 1 = 3
3 – 5 | 6 | 2 | 6 / 2 = 3
5 – 7 | 12 | 2 | 12 / 2 = 6
7 – 10 | 9 | 3 | 9 / 3 = 3
10 – 15 | 10 | 5 | 10 / 5 = 2
15 – 17 | 4 | 2 | 4 / 2 = 2

After calculating these values, plot the histogram with age on the x-axis and frequency density on the y-axis.

Question:

Concept in a Minute:
Histograms are graphical representations of the distribution of numerical data. They are similar to bar graphs but are used for continuous data, where bars touch each other. The x-axis represents the class intervals, and the y-axis represents the frequency of each interval. To find the number of items in a range, we sum the frequencies of the relevant intervals.


Part 1: Representing the information with a histogram.
To draw a histogram, we will use the given ‘Life time (in hours)’ as the class intervals on the x-axis and the ‘Number of lamps’ as the frequencies on the y-axis.
The class intervals are: 300-400, 400-500, 500-600, 600-700, 700-800, 800-900, 900-1000.
The corresponding frequencies are: 14, 56, 60, 86, 74, 62, 48.
Since the class intervals are of equal width (100 hours), we can directly plot the frequencies as the heights of the rectangular bars. The bars will be adjacent to each other.
On the x-axis, mark the boundaries of the class intervals: 300, 400, 500, 600, 700, 800, 900, 1000.
On the y-axis, choose a suitable scale to represent the frequencies. For example, 1 unit on the y-axis can represent 10 lamps.
Then, draw rectangles for each class interval with heights corresponding to the number of lamps.
The first bar will be from 300 to 400 with a height of 14.
The second bar will be from 400 to 500 with a height of 56.
The third bar will be from 500 to 600 with a height of 60.
The fourth bar will be from 600 to 700 with a height of 86.
The fifth bar will be from 700 to 800 with a height of 74.
The sixth bar will be from 800 to 900 with a height of 62.
The seventh bar will be from 900 to 1000 with a height of 48.
Label the x-axis as “Life time (in hours)” and the y-axis as “Number of lamps”.

Part 2: How many lamps have a life time of more than 700 hours?
To find the number of lamps with a life time of more than 700 hours, we need to sum the frequencies of all class intervals that start from 700 hours or more. These intervals are:
700 – 800 hours
800 – 900 hours
900 – 1000 hours
The number of lamps in these intervals are 74, 62, and 48 respectively.
Total number of lamps with a life time of more than 700 hours = 74 + 62 + 48 = 184.

Question:

Concept in a Minute:
Bar graphs are used to represent data visually, where the length of each bar is proportional to the value it represents. This is useful for comparing different categories. To find the maximum value in a set of data, simply identify the largest number.


Part 1: Drawing a Bar Graph

1. Identify Axes: Decide which axis will represent the political parties (usually the horizontal x-axis) and which will represent the number of seats won (usually the vertical y-axis).

2. Label Axes: Label the x-axis as “Political Party” and the y-axis as “Seats Won”.

3. Choose a Scale: Determine a suitable scale for the y-axis. The highest number of seats won is 75. A scale where each unit represents, for example, 5 or 10 seats would be appropriate. Let’s choose a scale where each unit on the y-axis represents 10 seats. So the y-axis will be marked 0, 10, 20, 30, 40, 50, 60, 70, 80.

4. Draw Bars: For each political party, draw a rectangular bar whose height corresponds to the number of seats won by that party, according to the chosen scale.
* Party A: Draw a bar up to 75 on the y-axis.
* Party B: Draw a bar up to 55 on the y-axis.
* Party C: Draw a bar up to 37 on the y-axis.
* Party D: Draw a bar up to 29 on the y-axis.
* Party E: Draw a bar up to 10 on the y-axis.
* Party F: Draw a bar up to 37 on the y-axis.

5. Add Titles and Labels: Give the bar graph a title, such as “State Assembly Election Polling Results”. Ensure each bar is clearly labeled with the corresponding political party’s name (A, B, C, D, E, F) below it on the x-axis.

Part 2: Political Party with Maximum Seats

To find the political party that won the maximum number of seats, look at the “Seats Won” row in the given table and identify the largest number.

The seats won by each party are:
Party A: 75
Party B: 55
Party C: 37
Party D: 29
Party E: 10
Party F: 37

Comparing these numbers, the maximum number of seats won is 75. This number corresponds to Political Party A.

Answer:
Political Party A won the maximum number of seats.