IMO Maths Olympiad Class 10 Chapter 14 Statistics MCQs with Answers

Answer: (a) 20.22

Explanation:
1. The highest frequency is 35, so the modal class is 18−24, with l = 18, h = 6, f₁ = 35, f₀ = 25 and f₂ = 18.
2. Mode = \( l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 18 + \dfrac{35-25}{70-25-18} \times 6 \).
3. = 18 + \( \dfrac{10}{27} \times 6 \) = 18 + 2.22 = 20.22.

Teacher's Note:
1. First find the modal class (the class with the greatest frequency).
2. f₀ and f₂ are the frequencies of the classes just before and just after it.

Answer: (c) 6.27

Explanation:
1. Σf = 2 + 3 + 6 + 4 = 15.
2. Σfx = 7 × 2 + 2 × 3 + 9 × 6 + 5 × 4 = 14 + 6 + 54 + 20 = 94.
3. Mean = \( \dfrac{94}{15} \) = 6.27 (approximately).

Teacher's Note:
1. Mean = Σfx ÷ Σf.
2. Do not divide by 4, the number of distinct values; divide by the total frequency.

Answer: (d) Standard deviation

Explanation:
1. Mean, median and mode are the three measures of central tendency, since each gives a central value of the data.
2. Standard deviation measures how spread out the data is (a measure of dispersion), so the answer is standard deviation.

Teacher's Note:
1. Central tendency tells where the data is centred; dispersion tells how widely it is spread.
2. Range and standard deviation are measures of dispersion.

Answer: (d) Ogive

Explanation:
1. The median is found from the cumulative frequency curve, called an ogive.
2. The x-coordinate of the point where the less than and more than ogives intersect (or the point of the less than ogive at n/2) is the median.
3. So the answer is Ogive.

Teacher's Note:
1. A histogram is used to locate the mode graphically.
2. The ogive is the graph of cumulative frequencies.

Answer: (b) 1

Explanation:
1. Σf = 2 + 3 + 4 + 5p + 6 = 15 + 5p.
2. Σfx = 30 + 51 + 76 + 5p(20 + p) + 138 = 295 + 100p + 5p².
3. Mean = 20 gives 20(15 + 5p) = 295 + 100p + 5p², so 300 + 100p = 295 + 100p + 5p².
4. Then 5p² = 5, p = ±1. Since p ≠ −1 (a frequency 5p cannot be negative), p = 1.

Teacher's Note:
1. Form Σfx and Σf first, then use the mean formula.
2. A frequency cannot be negative, so reject p = −1.

Answer: (b) 52

Explanation:
1. Let the number of candidates from school III be x. Then Σf = 60 + 48 + x + 40 = 148 + x.
2. Σfx = 60 × 75 + 48 × 80 + 55x + 40 × 50 = 4500 + 3840 + 55x + 2000 = 10340 + 55x.
3. \( \dfrac{10340 + 55x}{148 + x} = 66 \) gives 10340 + 55x = 9768 + 66x, so 11x = 572 and x = 52.

Teacher's Note:
1. This is a weighted mean, so multiply each average by its number of candidates.
2. Collect the x terms on one side and solve.

Answer: (c) f₁ = 9, f₂ = 15

Explanation:
1. The known frequencies add up to 2 + 5 + 12 + 17 + 20 + 9 + 7 + 4 = 76, so 76 + f₁ + f₂ = 100, i.e. f₁ + f₂ = 24.
2. The median 525 lies in the class 500−600, so l = 500, f = 20, h = 100 and the cumulative frequency before it is cf = 2 + 5 + f₁ + 12 + 17 = 36 + f₁.
3. 525 = \( 500 + \dfrac{50 - (36 + f_1)}{20} \times 100 \), so 25 = 5(14 − f₁), giving 14 − f₁ = 5 and f₁ = 9.
4. Then f₂ = 24 − 9 = 15. So f₁ = 9, f₂ = 15.

Teacher's Note:
1. Use the total frequency to get one equation and the median formula to get the other.
2. The median class is the class that contains the given median.

Answer: (c) Mean 62.4, Median 61.67, Mode 65

Explanation:
1. Σf = 50. Using class marks 10, 30, 50, 70, 90, 110, 130: Σfx = 60 + 240 + 500 + 840 + 540 + 550 + 390 = 3120, so mean = 3120 ÷ 50 = 62.4.
2. n/2 = 25. The cumulative frequencies are 6, 14, 24, 36, …, so the median class is 60−80 with l = 60, cf = 24, f = 12 and h = 20: median = \( 60 + \dfrac{25-24}{12} \times 20 \) = 61.67.
3. The modal class is 60−80 (frequency 12): mode = \( 60 + \dfrac{12-10}{24-10-6} \times 20 \) = 60 + 5 = 65.

Teacher's Note:
1. Use class marks (mid-points) for the mean.
2. Mean, median and mode of the same data are usually close but not equal.

Answer: (d) 47.58, 48.22

Explanation:
1. Frequencies of the classes 0−10, 10−20, …: 7, 32, 56, 106, 180, 164, 86, 44 (differences of the cumulative frequencies).
2. n/2 = 337.5, which lies in the class 40−50 (cf before it = 201, f = 180, h = 10): median = \( 40 + \dfrac{337.5-201}{180} \times 10 \) = 47.58.
3. The modal class is also 40−50 (greatest frequency 180): mode = \( 40 + \dfrac{180-106}{2 \times 180 - 106 - 164} \times 10 = 40 + \dfrac{74}{90} \times 10 \) = 48.22.

Teacher's Note:
1. A ‘less than’ table gives cumulative frequencies, so subtract to get class frequencies.
2. Median uses the cumulative frequency before the median class; mode uses the neighbouring frequencies.

Answer: (b) 6.84

Explanation:
1. Cumulative frequencies: 6, 59, 144, 230, 251, 267, 271, 275, so n = 275 and n/2 = 137.5.
2. The median class is 5−7, with l = 5, cf = 59, f = 85 and h = 2.
3. Median = \( 5 + \dfrac{137.5-59}{85} \times 2 = 5 + \dfrac{78.5}{85} \times 2 \) ≈ 5 + 1.85 = 6.85, which is closest to option (b), 6.84.

Teacher's Note:
1. The median class is the first class whose cumulative frequency is at least n/2.
2. The book truncates 1.847 to 1.84, so the closest option is chosen.