Class 10 Mathematics Practice Assignments: CBSE Class 10 Mathematics Statistics Assignment Set 18
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Board Level Exercise: Very Short Answer Questions (1 Mark Each)
Question 1. Construction of a cumulative frequency table is useful in determining which measure of central tendency?
Answer:
The cumulative frequency table gives the class in which the middle observation lies.
It is useful in determining the median.
Teacher's Note:
1. The median needs cumulative frequencies, the mean and mode do not.
| Monthly income range (in Rs.) | Number of families |
|---|---|
| Income more than Rs. 10000 | 100 |
| Income more than Rs. 13000 | 85 |
| Income more than Rs. 16000 | 69 |
| Income more than Rs. 19000 | 50 |
| Income more than Rs. 22000 | 33 |
| Income more than Rs. 25000 | 15 |
Answer:
Families with income more than 16000 = 69 and with income more than 19000 = 50.
Families in the range 16000 - 19000 = 69 − 50 = 19.
Teacher's Note:
1. For a more-than table, subtract the next value from the current one.
Question 3. If xᵢ’s are the mid points of the class intervals of grouped data, fᵢ’s are the corresponding frequencies and \( \bar{x} \) is the mean, then find the value of \( \sum f_i (x_i - \bar{x}) \).
Answer:
\( \sum f_i x_i - \bar{x}\sum f_i = \sum f_i x_i - \sum f_i x_i \) = 0 (the sum of deviations from the mean is zero).
Teacher's Note:
1. This is the basic property of the mean.
Question 4. In the formula \( \bar{x} = a + h\left( \dfrac{\sum f_i u_i}{\sum f_i} \right) \), for finding the mean of a grouped frequency distribution, what is the value of uᵢ?
Answer:
uᵢ = \( \dfrac{x_i - a}{h} \), where a is the assumed mean and h is the class size.
Teacher's Note:
1. uᵢ is the deviation divided by the class width.
| Class | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 |
| Frequency | 10 | 15 | 12 | 20 | 9 |
Answer:
N = 66, so N/2 = 33. Cumulative frequencies: 10, 25, 37, … so the median class is 10 - 15.
The highest frequency is 20, so the modal class is 15 - 20.
Sum of lower limits = 10 + 15 = 25.
Teacher's Note:
1. Median class from cumulative frequency, modal class from the highest frequency.
| Class | 0-5 | 6-11 | 12-17 | 18-23 | 24-29 |
| Frequency | 13 | 10 | 15 | 8 | 11 |
Answer:
N = 57, N/2 = 28.5. Cumulative frequencies: 13, 23, 38, … so the median class is 12 - 17.
Converting to continuous classes (subtract 0.5 from lower and add 0.5 to upper limits) it is 11.5 - 17.5.
Upper limit of the median class = 17.5.
Teacher's Note:
1. For inclusive classes, use the class boundaries.
| Class | 13.8-14 | 14-14.2 | 14.2-14.4 | 14.4-14.6 | 14.6-14.8 | 14.8-15 |
| Frequency | 2 | 4 | 5 | 71 | 48 | 20 |
Answer:
Athletes who took less than 14.6 seconds = 2 + 4 + 5 + 71 = 82.
Teacher's Note:
1. Add the frequencies of the classes below 14.6.
| Marks obtained | Number of students |
|---|---|
| More than or equal to 0 | 63 |
| More than or equal to 10 | 58 |
| More than or equal to 20 | 55 |
| More than or equal to 30 | 51 |
| More than or equal to 40 | 48 |
| More than or equal to 50 | 42 |
Answer:
Frequency of 30 - 40 = (number with marks ≥ 30) − (number with marks ≥ 40) = 51 − 48 = 3.
Teacher's Note:
1. Subtract consecutive more-than values.
Short Answer Questions (2 Marks Each)
| Height (in cm) | Number of students |
|---|---|
| 150-155 | 15 |
| 155-160 | 13 |
| 160-165 | 10 |
| 165-170 | 8 |
| 170-175 | 9 |
| 175-180 | 5 |
Answer:
The highest frequency is 15, so the modal class is 150 - 155 and its lower limit is 150.
N/2 = 30. Cumulative frequencies: 15, 28, 38, … so the median class is 160 - 165, with upper limit 165.
Sum = 150 + 165 = 315.
Teacher's Note:
1. Find the two classes separately before adding the limits.
| Marks | Number of students |
|---|---|
| Below 10 | 3 |
| Below 20 | 12 |
| Below 30 | 27 |
| Below 40 | 57 |
| Below 50 | 75 |
| Below 60 | 80 |
Answer:
Frequencies: 3, 9, 15, 30, 18, 5. The highest frequency is 30, for the class 30 - 40.
Modal class = 30 - 40.
Teacher's Note:
1. First convert the less-than table into class frequencies.
| Class | 65-85 | 85-105 | 105-125 | 125-145 | 145-165 | 165-185 | 185-205 |
| Frequency | 4 | 5 | 13 | 20 | 14 | 7 | 4 |
Answer:
N = 67, N/2 = 33.5. Cumulative frequencies: 4, 9, 22, 42, … so the median class is 125 - 145 (upper limit 145).
The modal class is also 125 - 145 (frequency 20), with lower limit 125.
Difference = 145 − 125 = 20.
Teacher's Note:
1. Here the median class and the modal class are the same.
| Number of pages written per day | 16-18 | 19-21 | 22-24 | 25-27 | 28-30 |
| Number of days | 1 | 3 | 4 | 9 | 13 |
Answer:
Class marks xᵢ: 17, 20, 23, 26, 29. Σfᵢ = 30.
Σfᵢxᵢ = 17 + 60 + 92 + 234 + 377 = 780.
Mean = \( \dfrac{780}{30} \) = 26 pages.
Teacher's Note:
1. Use the class mark of each class as xᵢ.
| Marks | Number of students |
|---|---|
| Below 10 | 10 |
| Below 20 | 50 |
| Below 30 | 130 |
| Below 40 | 270 |
| Below 50 | 440 |
| Below 60 | 570 |
| Below 70 | 670 |
| Below 80 | 740 |
| Below 90 | 780 |
| Below 100 | 800 |
Answer:
Subtract each cumulative frequency from the next one.
| Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 | 90-100 |
| Number of students | 10 | 40 | 80 | 140 | 170 | 130 | 100 | 70 | 40 | 20 |
Teacher's Note:
1. The first class frequency is the first cumulative value itself.
| Marks (out of 90) | Number of candidates |
|---|---|
| More than or equal to 80 | 4 |
| More than or equal to 70 | 6 |
| More than or equal to 60 | 11 |
| More than or equal to 50 | 17 |
| More than or equal to 40 | 23 |
| More than or equal to 30 | 27 |
| More than or equal to 20 | 30 |
| More than or equal to 10 | 32 |
| More than or equal to 0 | 34 |
Answer:
Subtract each more-than value from the one above it; the top class is 80 - 90 with frequency 4.
| Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 |
| Number of candidates | 2 | 2 | 3 | 4 | 6 | 6 | 5 | 2 | 4 |
Teacher's Note:
1. The class 80 - 90 keeps the value 4.
| Height (in cm) | Frequency | Cumulative frequency |
|---|---|---|
| 150-155 | 12 | a |
| 155-160 | b | 25 |
| 160-165 | 10 | c |
| 165-170 | d | 43 |
| 170-175 | e | 48 |
| 175-180 | 2 | f |
| Total | 50 |
Answer:
a = 12 (first cumulative frequency). b = 25 − 12 = 13. c = 25 + 10 = 35. d = 43 − 35 = 8. e = 48 − 43 = 5. f = 48 + 2 = 50 (equal to the total).
a = 12, b = 13, c = 35, d = 8, e = 5, f = 50.
Teacher's Note:
1. The last cumulative frequency must equal the total frequency.
| Age (in years) | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
| Number of patients | 60 | 42 | 55 | 70 | 53 | 20 |
Answer:
(i) Less than type:
| Age | Cumulative frequency |
|---|---|
| Less than 20 | 60 |
| Less than 30 | 102 |
| Less than 40 | 157 |
| Less than 50 | 227 |
| Less than 60 | 280 |
| Less than 70 | 300 |
(ii) More than type:
| Age | Cumulative frequency |
|---|---|
| More than or equal to 10 | 300 |
| More than or equal to 20 | 240 |
| More than or equal to 30 | 198 |
| More than or equal to 40 | 143 |
| More than or equal to 50 | 73 |
| More than or equal to 60 | 20 |
Teacher's Note:
1. Less than adds from the top, more than adds from the bottom.
| Marks | Below 20 | Below 40 | Below 60 | Below 80 | Below 100 |
| Number of students | 17 | 22 | 29 | 37 | 50 |
Answer:
Differences of consecutive values: 17, 5, 7, 8, 13.
| Marks | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 |
| Number of students | 17 | 5 | 7 | 8 | 13 |
Teacher's Note:
1. Subtract consecutive cumulative frequencies.
Long Answer Questions (3 Marks Each)
| Number of seats | 100-104 | 104-108 | 108-112 | 112-116 | 116-120 |
| Frequency | 15 | 20 | 32 | 18 | 15 |
Answer:
Class marks: 102, 106, 110, 114, 118. Σfᵢ = 100.
Σfᵢxᵢ = 1530 + 2120 + 3520 + 2052 + 1770 = 10992.
Mean = \( \dfrac{10992}{100} \) = 109.92 seats.
Teacher's Note:
1. Use the mid-point of each class.
| Weekly income (in Rs.) | Number of families |
|---|---|
| 0-1000 | 250 |
| 1000-2000 | 190 |
| 2000-3000 | 100 |
| 3000-4000 | 40 |
| 4000-5000 | 15 |
| 5000-6000 | 5 |
| Total | 600 |
Answer:
Cumulative frequencies: 250, 440, 540, 580, 595, 600. N/2 = 300, so the median class is 1000 - 2000.
ℓ = 1000, C = 250, f = 190, h = 1000.
Median = \( 1000 + \dfrac{300 - 250}{190} \times 1000 \) = Rs. 1263.16.
Teacher's Note:
1. The median class is the first class with cumulative frequency above N/2.
| Weight (in g) | Number of packets |
|---|---|
| 200-201 | 12 |
| 201-202 | 26 |
| 202-203 | 20 |
| 203-204 | 9 |
| 204-205 | 2 |
| 205-206 | 1 |
Answer:
The modal class is 201 - 202: ℓ = 201, f₁ = 26, f₀ = 12, f₂ = 20, h = 1.
Mode = \( 201 + \dfrac{26 - 12}{52 - 12 - 20} \times 1 = 201 + 0.7 = \) 201.7 g.
Teacher's Note:
1. Mode formula: ℓ + (f₁ − f₀)/(2f₁ − f₀ − f₂) × h.
| Class interval | 11-13 | 13-15 | 15-17 | 17-19 | 19-21 | 21-23 | 23-25 |
| Frequency | 3 | 6 | 9 | 13 | f | 5 | 4 |
Answer:
Class marks: 12, 14, 16, 18, 20, 22, 24. Σfᵢ = 40 + f and Σfᵢxᵢ = 36 + 84 + 144 + 234 + 20f + 110 + 96 = 704 + 20f.
\( \dfrac{704 + 20f}{40 + f} = 18 \Rightarrow 704 + 20f = 720 + 18f \Rightarrow \) f = 8.
Teacher's Note:
1. Form the equation from mean = Σfx ÷ Σf.
Very Long Answer Questions (5 Marks Each)
| Class | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 |
| Frequency | 17 | f₁ | 32 | f₂ | 19 |
Answer:
Σfᵢ = 17 + f₁ + 32 + f₂ + 19 = 120, so f₁ + f₂ = 52 …(i)
Class marks 10, 30, 50, 70, 90: Σfᵢxᵢ = 170 + 30f₁ + 1600 + 70f₂ + 1710 = 3480 + 30f₁ + 70f₂.
Mean = 50 gives 3480 + 30f₁ + 70f₂ = 6000, so 3f₁ + 7f₂ = 252 …(ii)
From (i) and (ii): 3(52 − f₂) + 7f₂ = 252, so 4f₂ = 96. f₂ = 24 and f₁ = 28.
Teacher's Note:
1. Two unknowns need two equations: one from the total, one from the mean.
| Marks | Frequency |
|---|---|
| 20-30 | p |
| 30-40 | 15 |
| 40-50 | 25 |
| 50-60 | 20 |
| 60-70 | q |
| 70-80 | 8 |
| 80-90 | 10 |
Answer:
p + 15 + 25 + 20 + q + 8 + 10 = 90, so p + q = 12 …(i)
N/2 = 45 and the median 50 lies in the class 50 - 60, with ℓ = 50, f = 20, h = 10 and C = p + 15 + 25 = p + 40.
\( 50 = 50 + \dfrac{45 - (p + 40)}{20} \times 10 \Rightarrow 45 - p - 40 = 0 \Rightarrow p = 5 \).
From (i), q = 7. p = 5, q = 7.
Teacher's Note:
1. A median of exactly 50, the lower limit of the class, means C = N/2.
Free study material for Mathematics
Chapter 13 Statistics Printable Assignments & Solutions for Class 10 Mathematics
Chapter Practice Questions for Class 10 Mathematics
Access structured practice assignments for Chapter 13 Statistics designed in alignment with the latest CBSE curriculum for Class 10 Mathematics. These printable sets cover objective and descriptive problem types to support thorough revision.
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Effective Strategy for Class 10 Mathematics Assignments
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