Statistics — Class 9 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 9 Mathematics chapter "Statistics" — 7 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 9 Mathematics chapter "Statistics" — 7 important questions with detailed answers for CBSE board exam prepa…
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NCERT Solutions for Class 9 Mathematics — Statistics. Step-by-step answers to all 7 textbook questions from this chapter, written for CBSE board preparation and free to use.
Statistics — All 7 Questions Solved
Q1. Define mean, median, and mode for a set of data. For the data set {12, 15, 12, 18, 20, 12, 25}, find the mean, median, and mode.
Step 1: Definitions:
Mean: The average of all values in the data set.
Median: The middle value when data is arranged in order.
Mode: The value that appears most frequently.
Step 2: Given data set: {12, 15, 12, 18, 20, 12, 25}
Number of values (n) = 7
Step 3: Calculate MEAN:
Mean = (Sum of all values) / (Number of values)
Sum = 12 + 15 + 12 + 18 + 20 + 12 + 25
Sum = 114
Mean = 114 / 7
Mean ≈ 16.29
Step 4: Calculate MEDIAN:
Arrange data in ascending order: {12, 12, 12, 15, 18, 20, 25}
Since n = 7 (odd number), median is the middle value at position (n+1)/2 = 4th position.
Median = 15
Step 5: Calculate MODE:
Frequency count:
12 appears 3 times
15 appears 1 time
18 appears 1 time
20 appears 1 time
25 appears 1 time
The value 12 appears most frequently (3 times).
Mode = 12
Step 6: Summary:
Mean = 114/7 ≈ 16.29
Median = 15
Mode = 12
Answer: Mean ≈ 16.29; Median = 15; Mode = 12.
Q2. For the frequency distribution: Class 0-10 (frequency 5), 10-20 (frequency 8), 20-30 (frequency 12), 30-40 (frequency 7), 40-50 (frequency 3). Find the mean of the grouped data.
Step 1: Create frequency distribution table:
Class | Midpoint (x) | Frequency (f) | f×x
0-10 | 5 | 5 | 25
10-20 | 15 | 8 | 120
20-30 | 25 | 12 | 300
30-40 | 35 | 7 | 245
40-50 | 45 | 3 | 135
Step 2: Calculate midpoint of each class:
Midpoint = (Lower limit + Upper limit) / 2
0-10: (0 + 10)/2 = 5
10-20: (10 + 20)/2 = 15
20-30: (20 + 30)/2 = 25
30-40: (30 + 40)/2 = 35
40-50: (40 + 50)/2 = 45
Step 3: Calculate f × x for each class:
5 × 5 = 25
15 × 8 = 120
25 × 12 = 300
35 × 7 = 245
45 × 3 = 135
Step 4: Calculate totals:
Total frequency (Σf) = 5 + 8 + 12 + 7 + 3 = 35
Total of f×x (Σf×x) = 25 + 120 + 300 + 245 + 135 = 825
Step 5: Calculate mean:
Mean = Σ(f×x) / Σf
Mean = 825 / 35
Mean = 23.57 (approximately)
Step 6: The mean lies in the 20-30 class, which makes sense as it has the highest frequency.
Answer: Mean of grouped data = 825/35 ≈ 23.57.
Q3. Construct a frequency distribution table and draw a bar graph for the following data: Marks 40, 45, 50, 55, 60, 50, 45, 60, 65, 70, 60, 55, 50, 65, 70, 70. Use class intervals 40-50, 50-60, 60-70, 70-80.
Step 1: Raw data:
40, 45, 50, 55, 60, 50, 45, 60, 65, 70, 60, 55, 50, 65, 70, 70
(Total: 16 values)
Step 2: Count frequencies for each class:
Class 40-50: 40, 45, 45, 50, 50, 50 → Frequency = 6
Class 50-60: 55, 55, 60, 60, 60 → Frequency = 5
Class 60-70: 65, 65, 70, 70, 70 → Frequency = 5
Class 70-80: None → Frequency = 0
Note: 70 is often considered in the 70-80 class (right-inclusive), but here included in 60-70.
Correction: Let's use left-inclusive intervals.
Class 40-50: 40, 45, 45 → Frequency = 3
Class 50-60: 50, 50, 50, 55, 55 → Frequency = 5
Class 60-70: 60, 60, 60, 65, 65 → Frequency = 5
Class 70-80: 70, 70, 70 → Frequency = 3
Total: 3 + 5 + 5 + 3 = 16 ✓
Step 3: Frequency Distribution Table:
Class | Frequency | Relative Frequency
40-50 | 3 | 3/16 = 0.1875
50-60 | 5 | 5/16 = 0.3125
60-70 | 5 | 5/16 = 0.3125
70-80 | 3 | 3/16 = 0.1875
Step 4: Bar Graph Description:
X-axis: Class intervals (40-50, 50-60, 60-70, 70-80)
Y-axis: Frequency (scale 0-5)
Bar heights:
Class 40-50: bar height = 3
Class 50-60: bar height = 5
Class 60-70: bar height = 5
Class 70-80: bar height = 3
Step 5: Observations:
- Maximum frequency in classes 50-60 and 60-70 (5 each)
- Symmetric distribution around classes 50-70
- The data is concentrated in the middle ranges (50-70)
Answer: Frequency distribution table created above. Bar graph shows bars of heights 3, 5, 5, 3 for classes 40-50, 50-60, 60-70, 70-80 respectively.
Q4. For the data {2, 4, 6, 8, 10}, calculate the mean, median, and verify that for this symmetric data set these values are equal.
Step 1: Given data set (already in order):
{2, 4, 6, 8, 10}
n = 5 (odd number)
Step 2: Calculate MEAN:
Mean = (Sum of all values) / n
Sum = 2 + 4 + 6 + 8 + 10 = 30
Mean = 30 / 5 = 6
Step 3: Calculate MEDIAN:
For odd n, median is the value at position (n+1)/2 = 3rd position
Arranged data: {2, 4, 6, 8, 10}
Median = 6 (the 3rd value)
Step 4: Compare:
Mean = 6
Median = 6
Mean = Median ✓
Step 5: Explanation:
This is a symmetric (arithmetic progression) data set.
The differences between consecutive terms are equal: 4-2=2, 6-4=2, 8-6=2, 10-8=2
In symmetric distributions, the mean and median are equal.
The middle value (6) is also the average of all values.
Step 6: Mode:
No value repeats in this data set.
Mode = No mode (or all values are equally frequent)
Answer: Mean = 6, Median = 6. For this symmetric data set, mean equals median.
Q5. The data shows test scores of 10 students: 70, 75, 80, 85, 80, 90, 85, 95, 80, 75. Calculate mean, median, mode, and range. Also determine how many students scored above the mean.
Step 1: Given data:
70, 75, 80, 85, 80, 90, 85, 95, 80, 75
n = 10
Step 2: Calculate MEAN:
Sum = 70 + 75 + 80 + 85 + 80 + 90 + 85 + 95 + 80 + 75
Sum = 815
Mean = 815 / 10 = 81.5
Step 3: Calculate MEDIAN:
Arrange in ascending order: {70, 75, 75, 80, 80, 80, 85, 85, 90, 95}
n = 10 (even), so median = average of 5th and 6th values
5th value = 80
6th value = 80
Median = (80 + 80) / 2 = 80
Step 4: Calculate MODE:
Frequency count:
70: 1 time
75: 2 times
80: 3 times
85: 2 times
90: 1 time
95: 1 time
Mode = 80 (appears 3 times, most frequent)
Step 5: Calculate RANGE:
Range = Maximum value - Minimum value
Range = 95 - 70 = 25
Step 6: Count students scoring ABOVE the mean:
Mean = 81.5
Scores above 81.5: 85, 90, 85, 95
Number of students above mean = 4
Step 7: Summary:
Mean = 81.5
Median = 80
Mode = 80
Range = 25
Students above mean = 4 out of 10 = 40%
Answer: Mean = 81.5; Median = 80; Mode = 80; Range = 25; 4 students scored above the mean.
Q6. Create a grouped frequency distribution for the following age data (in years): 22, 25, 23, 29, 31, 26, 24, 28, 32, 25, 27, 30, 23, 26, 29 using class intervals 20-25, 25-30, 30-35. Calculate the mean.
Step 1: Given data (ages):
22, 25, 23, 29, 31, 26, 24, 28, 32, 25, 27, 30, 23, 26, 29
(Total: 15 values)
Step 2: Create frequency distribution with class 20-25 (includes 20, excludes 25):
Class 20-25: 22, 23, 23, 24 → Frequency = 4
Class 25-30: 25, 25, 26, 26, 27, 28, 29, 29 → Frequency = 8
Class 30-35: 30, 31, 32 → Frequency = 3
Total frequency = 4 + 8 + 3 = 15 ✓
Step 3: Frequency Distribution Table:
Class | Midpoint (x) | Frequency (f) | f×x
20-25 | 22.5 | 4 | 90
25-30 | 27.5 | 8 | 220
30-35 | 32.5 | 3 | 97.5
Note: Midpoint = (20+25)/2 = 22.5, (25+30)/2 = 27.5, (30+35)/2 = 32.5
Step 4: Calculate f × x:
22.5 × 4 = 90
27.5 × 8 = 220
32.5 × 3 = 97.5
Step 5: Calculate totals:
Σf = 4 + 8 + 3 = 15
Σ(f × x) = 90 + 220 + 97.5 = 407.5
Step 6: Calculate mean:
Mean = Σ(f × x) / Σf
Mean = 407.5 / 15
Mean ≈ 27.17 years
Step 7: Verification:
The mean falls in the class 25-30, which has the highest frequency (8). This makes sense.
Answer: Frequency distribution created. Mean age = 407.5/15 ≈ 27.17 years.
Q7. For a data set with 11 values, how do you find the median? Given data {3, 7, 5, 9, 1, 8, 4, 6, 2, 10, 11}, find the median and verify by arranging in order.
Step 1: Method for finding median with odd number of values:
For n values (where n is odd), median is the value at position (n+1)/2 when arranged in ascending order.
Step 2: Given data:
{3, 7, 5, 9, 1, 8, 4, 6, 2, 10, 11}
n = 11 (odd)
Step 3: Find position of median:
Position = (n + 1) / 2 = (11 + 1) / 2 = 12 / 2 = 6th position
Step 4: Arrange data in ascending order:
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
Position: 1st, 2nd, 3rd, 4th, 5th, 6th, 7th, 8th, 9th, 10th, 11th
Step 5: Identify median:
Median = 6th value = 6
Step 6: Verification:
Values below median: {1, 2, 3, 4, 5} → 5 values
Median: {6}
Values above median: {7, 8, 9, 10, 11} → 5 values
The median divides the data into two equal halves. ✓
Step 7: Properties:
For any data set with odd n, the median always divides the data into two equal parts.
Number of values below median = Number of values above median = (n-1)/2 = 5
Answer: Median = 6. The 6th value when arranged in ascending order {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}.
Statistics — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. What is the term for facts or figures collected for a definite purpose?
- A. Information
- B. Statistics
- C. Data
- D. Analysis
Answer: C. Data — Data refers to the raw facts and figures collected for a specific purpose, which then can be processed to yield information or statistics.
MCQ 2. Data collected by an investigator himself for a definite purpose is known as:
- A. Secondary data
- B. Primary data
- C. Raw data
- D. Grouped data
Answer: B. Primary data — Primary data is original data collected directly by the investigator for the first time for a specific inquiry.
MCQ 3. Which of the following is an example of secondary data?
- A. Heights of students measured by a researcher in their class.
- B. Survey results collected by a student for a project.
- C. Data on population from a government census report.
- D. Records of daily temperature noted down directly from a thermometer.
Answer: C. Data on population from a government census report. — Secondary data is data that has been collected by someone else and is available for use. A government census report is an existing source.
MCQ 4. If the class marks of a frequency distribution are 10, 20, 30, 40, then the class size is:
- A. 5
- B. 10
- C. 15
- D. 20
Answer: B. 10 — The class size is the difference between two consecutive class marks. Here, 20 - 10 = 10, or 30 - 20 = 10.
MCQ 5. What is the range of the data: 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20?
- A. 10
- B. 15
- C. 26
- D. 32
Answer: C. 26 — The range is the difference between the maximum and minimum values in the data set. Here, Max = 32, Min = 6, so Range = 32 - 6 = 26.
MCQ 6. In a grouped frequency distribution, if the lower limit of a class is 10 and its upper limit is 20, then the class mark is:
- A. 5
- B. 10
- C. 15
- D. 20
Answer: C. 15 — The class mark is the midpoint of the class interval, calculated as (Lower Limit + Upper Limit) / 2. So, (10 + 20) / 2 = 15.
MCQ 7. Which type of graph is suitable for representing grouped frequency distributions with continuous classes?
- A. Bar graph
- B. Pie chart
- C. Histogram
- D. Line graph
Answer: C. Histogram — A histogram is a graphical representation used for continuous grouped frequency distributions, where the bars are adjacent to each other.
MCQ 8. The mean of 5 observations is 10. If each observation is increased by 3, what is the new mean?
- A. 10
- B. 13
- C. 15
- D. 30
Answer: B. 13 — If each observation in a data set is increased by a constant, the mean also increases by the same constant. So, the new mean is 10 + 3 = 13.
MCQ 9. What is the mode of the data: 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18?
- A. 14
- B. 18
- C. 22
- D. 28
Answer: A. 14 — The mode is the observation that occurs most frequently in the data set. In this data, 14 appears 4 times, which is more than any other number.
MCQ 10. To draw a frequency polygon, we plot the frequencies against the:
- A. Upper limits of the classes
- B. Lower limits of the classes
- C. Class marks of the classes
- D. Width of the classes
Answer: C. Class marks of the classes — A frequency polygon is drawn by plotting the frequencies against the class marks of the respective classes and joining these points with line segments.
MCQ 11. Which of the following is NOT a measure of central tendency?
- A. Mean
- B. Median
- C. Mode
- D. Range
Answer: D. Range — Mean, Median, and Mode are measures of central tendency, representing the typical value of a data set. Range is a measure of dispersion, not central tendency.
MCQ 12. The median of the data: 40, 50, 30, 20, 60 is:
- A. 20
- B. 30
- C. 40
- D. 50
Answer: C. 40 — First, arrange the data in ascending order: 20, 30, 40, 50, 60. Since there are 5 observations (odd), the median is the middle value, which is the 3rd term, 40.
MCQ 13. If the mean of x, x+2, x+4, x+6, x+8 is 15, then the value of x is:
- A. 11
- B. 13
- C. 15
- D. 17
Answer: A. 11 — The sum of observations is (x + x+2 + x+4 + x+6 + x+8) = 5x + 20. Mean = (5x + 20) / 5 = x + 4. Given mean is 15, so x + 4 = 15, which means x = 11.
MCQ 14. A student collected data on the number of siblings each of their 30 classmates has. This is an example of:
- A. Secondary data
- B. Grouped data
- C. Primary data
- D. Qualitative data
Answer: C. Primary data — Since the student personally collected the data for their specific purpose, it is considered primary data.
MCQ 15. For constructing a histogram, the class intervals must be:
- A. Overlapping
- B. Discrete
- C. Continuous
- D. Unequal
Answer: C. Continuous — Histograms are used for continuous frequency distributions, meaning there should be no gaps between the classes. If the data is discrete, it needs to be made continuous by adjusting class boundaries.
MCQ 16. The class limits of a frequency distribution are 10-20, 20-30, 30-40. The observation 20 belongs to which class?
- A. 10-20
- B. 20-30
- C. Both 10-20 and 20-30
- D. None of these
Answer: B. 20-30 — In continuous class intervals, the upper limit of a class is not included in that class, but in the next class. So, 20 is included in the class 20-30.
MCQ 17. If the observations 6, 8, 9, x+2, 15, 16, 18 are arranged in ascending order and their median is 14, then x is:
- A. 12
- B. 13
- C. 14
- D. 15
Answer: B. 13 — There are 7 observations (odd). The median is the (7+1)/2 = 4th observation, which is x+2. Given median is 14, so x+2 = 14, which means x = 12.
MCQ 18. A bar graph is drawn by taking:
- A. Frequency on X-axis and categories on Y-axis.
- B. Categories on X-axis and frequency on Y-axis.
- C. Class marks on X-axis and frequency on Y-axis.
- D. Upper limits on X-axis and frequency on Y-axis.
Answer: B. Categories on X-axis and frequency on Y-axis. — In a bar graph, categories (or discrete variables) are typically represented on the X-axis, and their corresponding frequencies (or values) are represented on the Y-axis.
MCQ 19. If the mean of 6, 8, 5, 7, x and 4 is 7, then the value of x is:
- A. 9
- B. 10
- C. 11
- D. 12
Answer: D. 12 — The sum of observations is 6+8+5+7+x+4 = 30+x. There are 6 observations. Mean = (30+x)/6. Given mean is 7, so (30+x)/6 = 7, which implies 30+x = 42, so x = 12.
MCQ 20. The data that has been published in a 'Statistical Abstract of India' is an example of:
- A. Primary data
- B. Secondary data
- C. Raw data
- D. Organized data
Answer: B. Secondary data — Data published in a 'Statistical Abstract of India' is collected by various government agencies and compiled. When someone uses this existing data, it is secondary data for them.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.