RowQ
The Vault
RowQ
The Vault
CBSE Class 10 Maths · 11 questions · 26 marks
Statistics in Class 10 is about squeezing a whole table of grouped data into one representative number — a mean, a median or a mode. The arithmetic is not hard, but the marks live in the table you build: class marks for the mean, cumulative frequencies for the median, and the neighbouring frequencies for the mode. Set out the table neatly first, and the formula almost fills itself in.
For a grouped frequency distribution the mean is 32 and the median is 30. Using the empirical relationship, the mode is:
Answer
The mode is 26. The empirical relationship is 3 × Median = Mode + 2 × Mean. So Mode = 3 × Median - 2 × Mean. = 3(30) - 2(32) = 90 - 64 = 26.
The class mark of the class interval 25-40 is:
Answer
The class mark is 32.5. Class mark = (lower limit + upper limit)/2. = (25 + 40)/2 = 65/2 = 32.5. The value 15 is the class size (40 - 25), not the class mark.
The mean of the first eight positive multiples of 5 is:
Answer
The mean is 22.5. The first eight multiples of 5 are 5, 10, 15, 20, 25, 30, 35 and 40. Sum = 5(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) = 5 × 36 = 180. Mean = sum/number of observations = 180/8 = 22.5.
In the formula for the mode of a grouped frequency distribution, mode = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] × h, the symbol f₁ stands for the:
Answer
The correct option is: frequency of the modal class. In this formula l is the lower limit of the modal class and h is the class size. f₁ is the frequency of the modal class itself, which is the largest frequency in the table. f₀ is the frequency of the class just before it and f₂ is the frequency of the class just after it. Cumulative frequency does not appear in the mode formula at all — it belongs to the median formula.
Assertion (A): For a grouped frequency distribution, the mode is calculated from the modal class. Reason (R): The modal class of a grouped frequency distribution is the class interval that has the highest frequency.
Answer
Both A and R are true and R is the correct explanation of A. Reason R gives the correct definition: the modal class is the class with the greatest frequency. The mode is the value that occurs most often, so it must lie inside the class where observations are most crowded — that is, inside the modal class identified by R. That is why the formula mode = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] × h takes l and h from the modal class and then adjusts the position using its neighbouring frequencies. So A is true and R is exactly the reason for it.
Find the mean of the following distribution by the direct method: Class 0-10, frequency 4; class 10-20, frequency 6; class 20-30, frequency 8; class 30-40, frequency 2.
Answer
Class marks xᵢ are 5, 15, 25 and 35. Products fᵢxᵢ: 4 × 5 = 20 6 × 15 = 90 8 × 25 = 200 2 × 35 = 70 Σfᵢ = 4 + 6 + 8 + 2 = 20 and Σfᵢxᵢ = 20 + 90 + 200 + 70 = 380. Mean = Σfᵢxᵢ/Σfᵢ = 380/20 = 19. The mean of the distribution is 19.
Find the mode of the following distribution: Class 0-20, frequency 5; class 20-40, frequency 12; class 40-60, frequency 20; class 60-80, frequency 9; class 80-100, frequency 4.
Answer
The highest frequency is 20, so the modal class is 40-60. From this class: l = 40, f₁ = 20, f₀ = 12 (class 20-40), f₂ = 9 (class 60-80) and h = 20. Mode = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] × h = 40 + [(20 - 12)/(40 - 12 - 9)] × 20 = 40 + (8/19) × 20 = 40 + 160/19 = 40 + 8.42 = 48.42 (correct to two decimal places).
The mean of a distribution is 25 and its mode is 22. Find the median using the empirical relationship.
Answer
The empirical relationship is 3 × Median = Mode + 2 × Mean. 3 × Median = 22 + 2(25) 3 × Median = 22 + 50 = 72 Median = 72/3 = 24. The median of the distribution is 24.
The daily earnings (in ₹) of 50 shopkeepers in a market are given below. Find the mean by the step deviation method and also find the median. Class 100-150, frequency 7; class 150-200, frequency 10; class 200-250, frequency 14; class 250-300, frequency 11; class 300-350, frequency 8.
Answer
Step 1 — Class marks: 125, 175, 225, 275 and 325. The class size is h = 50. Step 2 — Take the assumed mean a = 225 and form uᵢ = (xᵢ - 225)/50. u values are -2, -1, 0, 1 and 2 respectively. Step 3 — Compute fᵢuᵢ: 7 × (-2) = -14 10 × (-1) = -10 14 × 0 = 0 11 × 1 = 11 8 × 2 = 16 Σfᵢ = 7 + 10 + 14 + 11 + 8 = 50 and Σfᵢuᵢ = -14 - 10 + 0 + 11 + 16 = 3. Step 4 — Mean = a + h × (Σfᵢuᵢ/Σfᵢ) = 225 + 50 × (3/50) = 225 + 3 = ₹228. Step 5 — Build the cumulative frequencies for the median: 100-150 → 7 150-200 → 17 200-250 → 31 250-300 → 42 300-350 → 50 Step 6 — Here n = 50, so n/2 = 25. The cumulative frequency first crosses 25 in the class 200-250, so that is the median class. l = 200, cf = 17, f = 14, h = 50. Step 7 — Median = l + [(n/2 - cf)/f] × h = 200 + [(25 - 17)/14] × 50 = 200 + (8/14) × 50 = 200 + 400/14 = 200 + 28.57 = ₹228.57 (correct to two decimal places). So the mean daily earning is ₹228 and the median daily earning is about ₹228.57.
The mean of the following frequency distribution is 50. Find the missing frequency x, and then find the mode of the completed distribution. Class 0-20, frequency 8; class 20-40, frequency x; class 40-60, frequency 12; class 60-80, frequency 10; class 80-100, frequency 6.
Answer
Step 1 — Class marks are 10, 30, 50, 70 and 90. Step 2 — Write the products fᵢxᵢ: 8 × 10 = 80 x × 30 = 30x 12 × 50 = 600 10 × 70 = 700 6 × 90 = 540 Σfᵢ = 8 + x + 12 + 10 + 6 = 36 + x. Σfᵢxᵢ = 80 + 30x + 600 + 700 + 540 = 1920 + 30x. Step 3 — Use the given mean: (1920 + 30x)/(36 + x) = 50 1920 + 30x = 50(36 + x) 1920 + 30x = 1800 + 50x 1920 - 1800 = 50x - 30x 120 = 20x x = 6. Step 4 — The completed frequencies are 8, 6, 12, 10 and 6, with total 42. Step 5 — The highest frequency is 12, so the modal class is 40-60. Here l = 40, f₁ = 12, f₀ = 6, f₂ = 10 and h = 20. Step 6 — Mode = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] × h = 40 + [(12 - 6)/(24 - 6 - 10)] × 20 = 40 + (6/8) × 20 = 40 + 15 = 55. So the missing frequency is x = 6 and the mode is 55.
Read the following and answer the questions that follow: A mathematics teacher gave a test of maximum 50 marks to the 40 students of her class and grouped the scores as shown. Marks 0-10: 3 students; 10-20: 7 students; 20-30: 12 students; 30-40: 13 students; 40-50: 5 students. (a) Write the cumulative frequencies and identify the median class. (b) Find the median score. (c) Identify the modal class and find the mode. (d) Find the mean score of the class.
Answer
(a) Cumulative frequencies: 0-10 → 3 10-20 → 3 + 7 = 10 20-30 → 10 + 12 = 22 30-40 → 22 + 13 = 35 40-50 → 35 + 5 = 40 Here n = 40, so n/2 = 20. The cumulative frequency first reaches or crosses 20 in the class 20-30, so the median class is 20-30. (b) For the median class: l = 20, cf = 10, f = 12, h = 10. Median = l + [(n/2 - cf)/f] × h = 20 + [(20 - 10)/12] × 10 = 20 + (10/12) × 10 = 20 + 100/12 = 20 + 8.33 = 28.33 marks (correct to two decimal places). (c) The greatest frequency is 13, so the modal class is 30-40. Here l = 30, f₁ = 13, f₀ = 12, f₂ = 5 and h = 10. Mode = 30 + [(13 - 12)/(26 - 12 - 5)] × 10 = 30 + (1/9) × 10 = 30 + 1.11 = 31.11 marks (correct to two decimal places). (d) Class marks are 5, 15, 25, 35 and 45. Products fᵢxᵢ: 3 × 5 = 15, 7 × 15 = 105, 12 × 25 = 300, 13 × 35 = 455, 5 × 45 = 225. Σfᵢxᵢ = 15 + 105 + 300 + 455 + 225 = 1100. Mean = 1100/40 = 27.5 marks.
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