Chapter 14: Statistics (NCERT Solutions)
Exercise 14.1
Five examples of data from daily life:
1. Marks scored by students in a class test (e.g., 45, 67, 78, 89, 95).
2. Temperature of a city recorded daily for a week (e.g., 22°C, 24°C,
20°C, ...).
3. Number of cars passing a signal every hour (traffic data).
4. Height (in cm) of all students in a class.
5. Electricity bills of households in a colony for a month.
Primary Data (collected personally by investigator):
1. Marks scored (collected by the teacher).
3. Number of cars (counted by an investigator at the signal).
4. Heights of students (measured personally).
Secondary Data (already collected and published by someone else):
2. Temperature (from newspaper / weather department records).
5. Electricity bills (from electricity board records).
Exercise 14.2
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.
Represent this data in the form of a frequency distribution table. Which is the most common blood group and which is the rarest?
Frequency Distribution Table:
| Blood Group | Number of Students (Frequency) |
|---|---|
| A | 9 |
| B | 6 |
| O | 12 |
| AB | 3 |
| Total | 30 |
Most common: O (frequency 12). Rarest: AB (frequency 3).
5, 3, 10, 20, 25, 11, 13, 7, 12, 31, 19, 10, 12, 17, 18, 11, 32, 17, 16, 2, 7, 9, 7, 8, 3, 5, 12, 15, 18, 3, 12, 14, 2, 9, 6, 15, 15, 7, 6, 12.
Construct a grouped frequency distribution table with class size 5. Discuss what is the problem with this data.
| Class Interval (km) | Frequency |
|---|---|
| 0 - 5 | 5 |
| 5 - 10 | 11 |
| 10 - 15 | 11 |
| 15 - 20 | 9 |
| 20 - 25 | 1 |
| 25 - 30 | 1 |
| 30 - 35 | 2 |
| Total | 40 |
Problem: The data is very spread out (2 to 32 km). Most engineers (31 out of 40) live within 20 km, while only 4 live more than 20 km away. The distribution is not uniform, which makes it harder to represent proportionally.
Exercise 14.3
| Causes | Female Fatality Rate (%) |
|---|---|
| Reproductive health conditions | 31.8 |
| Neuropsychiatric conditions | 25.4 |
| Injuries | 12.4 |
| Cardiovascular conditions | 4.3 |
| Respiratory conditions | 4.1 |
| Other causes | 22.0 |
A bar graph should be drawn with the causes on the X-axis and the female fatality rate (%) on the Y-axis. Each cause is represented by a bar with height equal to its percentage. The bars should be of equal width and equally spaced.
Section: Scheduled Caste (940), Scheduled Tribe (970), OBC (920), General (890), Urban (910), Rural (930).
Plot a bar graph with sections on X-axis and number of girls per 1000 boys on Y-axis.
Conclusions:
1. The number of girls per thousand boys is highest in Scheduled Tribes
(970).
2. The General category has the lowest ratio (890), indicating
social bias in general sections.
3. All sections have fewer girls than boys (ratio below 1000), indicating a gender imbalance
throughout society.
Exercise 14.4
2, 3, 4, 5, 0, 1, 3, 3, 4, 3. Find the mean, median and mode of these scores.
Mean:
Sum = 2 + 3 + 4 + 5 + 0 + 1 + 3 + 3 + 4 + 3 = 28.
Mean = 28 / 10 = 2.8.
Median:
Arrange in order: 0, 1, 2, 3, 3, 3, 3, 4, 4, 5.
n = 10 (even). Median = average of 5th and 6th terms = (3 + 3)/2 = 3.
Mode: The value that appears most often = 3 (appears 4 times).
41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60. Find mean, median and mode of this data.
Mean:
Sum = 41+39+48+52+46+62+54+40+96+52+98+40+42+52+60 = 822.
Mean = 822 / 15 = 54.8.
Median:
Arrange: 39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98.
n = 15 (odd). Median = [(15+1)/2]th = 8th value = 52.
Mode: 52 appears 3 times. Mode = 52.
Salary (Rs): 3000, 4000, 5000, 6000, 7000, 8000, 9000, 10000; Workers: 16, 12, 10, 8, 6, 4, 3, 1.
| Salary (x) | Workers (f) | fx |
|---|---|---|
| 3000 | 16 | 48000 |
| 4000 | 12 | 48000 |
| 5000 | 10 | 50000 |
| 6000 | 8 | 48000 |
| 7000 | 6 | 42000 |
| 8000 | 4 | 32000 |
| 9000 | 3 | 27000 |
| 10000 | 1 | 10000 |
| Total | 60 | 305000 |
Mean = ∑fx / ∑f = 305000 / 60 = Rs 5083.33.
Marks: 0, 1, 2, 3, 4, 5, 6, 7, 8; Frequency: 3, 8, 12, 16, 13, 10, 6, 4, 3.
∑f = 3+8+12+16+13+10+6+4+3 = 75.
∑fx =
(0×3)+(1×8)+(2×12)+(3×16)+(4×13)+(5×10)+(6×6)+(7×4)+(8×3)
= 0+8+24+48+52+50+36+28+24 = 270.
Mean = 270/75 = 3.6.
Chapter 14: Statistics (Practice Questions)
RD Sharma / Extra Practice
First 10 natural numbers: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
Sum = n(n+1)/2 = 10 × 11 / 2 = 55.
Mean = 55 / 10 = 5.5.
Sum of 6 numbers = 6 × 30 = 180.
Sum of remaining 5 numbers = 5 × 28 = 140.
Excluded number = 180 - 140 = 40.
Arrange in order: 20, 22, 23, 25, 26, 28, 31, 32, 34, 35.
n = 10 (even). Median = average of 5th and 6th term = (26 + 28)/2 = 27.
Sum of 5 numbers = 5 × 18 = 90.
Sum of 6 numbers = 6 × 20 = 120.
Included number = 120 - 90 = 30.
Count of each value: 12 appears 5 times. 14 appears 5 times. Others appear fewer times.
Since both 12 and 14 appear 5 times, the data is bimodal. Mode = 12 and 14.
Mean = (x + x+2 + x+4 + x+6 + x+8) / 5 = (5x + 20) / 5 = x + 4.
x + 4 = 11 ⇒ x = 7.
Arrange: 45, 56, 72, 87, 88, 97, 101.
n = 7 (odd). Median = [(7+1)/2]th = 4th term = 87.
Total marks of 100 students = 100 × 40 = 4000.
Total marks of top 60 = 60 × 45 = 2700.
Total marks of bottom 40 = 4000 - 2700 = 1300.
Mean of bottom 40 = 1300 / 40 = 32.5.
x: 5, 10, 15, 20, 25; f: 6, 8, 15, 9, 7.
∑f = 6+8+15+9+7 = 45.
∑fx = (5×6)+(10×8)+(15×15)+(20×9)+(25×7) = 30+80+225+180+175 =
690.
Mean = 690/45 = 15.33.
Sum = 25+27+32+38+29+33+45+37 = 266.
Mean = 266 / 8 = 33.25 years.
Arrange: 15, 20, 25, 30, 35, 40, 45, 66, 70.
n = 9 (odd). Median = 5th term = 35.
New Mean = (3 × Original Mean) + 5 = (3 × 12) + 5 = 36 + 5 = 41.
Sum = 55+60+62+58+53+65+57+68+52+70 = 600.
Mean = 600 / 10 = 60 kg.
Arrange: 52, 53, 55, 57, 58, 60, 62, 65, 68, 70.
Median = (58 + 60)/2 = 59 kg.
∑f = 3+4+f₃+4+3 = 14 + f₃.
∑fx = (1×3)+(2×4)+(3×f₃)+(4×4)+(5×3) = 3+8+3f₃+16+15 = 42 +
3f₃.
Mean = (42 + 3f₃)/(14 + f₃) = 3.
42 + 3f₃ = 42 + 3f₃. This is always true, confirming that any value satisfies this if data is
symmetric. Re-checking: 42 + 3f₃ = 3(14 + f₃) = 42 + 3f₃. Hence f₃ cannot be uniquely determined
from mean alone; additional data needed. However, for the distribution to make sense with mean =
3 (middle), by symmetry f₃ = 8 (any central value that preserves symmetry).
Possible values: 1, 2, 3, 4, 5, 6, 7, 8, 9.
Mean = (1+2+3+4+5+6+7+8+9)/9 = 45/9 = 5.
Chapter 14: Statistics (Concepts & Formulas)
1. Key Definitions
- Data: A collection of facts, numbers, measurements, or observations.
- Primary Data: Data collected directly by the investigator.
- Secondary Data: Data already collected and available from other sources.
- Class Interval: A range of values in a grouped frequency distribution (e.g., 10-20).
- Class Size (Width): Difference between upper and lower class limits.
- Class Mark: Midpoint of a class interval = (Upper limit + Lower limit) / 2.
2. Measures of Central Tendency
Mean (¯x) — Arithmetic Average:
- Ungrouped: Mean = (Sum of all observations) / (Number of observations)
- Grouped (Direct Method): Mean = ∑fx / ∑f
Median — Middle Value:
- Arrange data in ascending order first.
- If n is odd: Median = [(n+1)/2]th observation.
- If n is even: Median = average of (n/2)th and (n/2 + 1)th observations.
Mode — Most Frequent Value:
- The value that appears the most number of times.
- A data set can have one mode (unimodal), two modes (bimodal), or more.
3. Graphical Representations
- Bar Graph: Used for discrete or categorical data. Bars are of equal width with gaps between them.
- Histogram: Similar to bar graph but for continuous grouped data. No gaps between bars.
- Frequency Polygon: Line graph connecting the midpoints of the tops of histogram bars. Also extends to the x-axis on both sides.
4. Important Properties
- If each observation is multiplied by k: New Mean = k × Old Mean.
- If k is added to each observation: New Mean = Old Mean + k.
- The sum of deviations from the mean is always zero: ∑(x - ¯x) = 0.