CBSE Class 10 Mathematics Statistics MCQs Set 16

Answer: (a) an ogive

Explanation:
1. Plotting cumulative frequencies against the class boundaries and joining the points gives an ogive (cumulative frequency curve).

Teacher's Note:
1. Recall the graphs used for grouped data.
2. Less-than and more-than ogives are the two types.

Answer: (b) geometric mean

Explanation:
1. The geometric mean is the nth root of the product of the observations, so a zero observation makes it zero (and its logarithm undefined).
2. The arithmetic mean, median and mode are unaffected by a zero value.

Teacher's Note:
1. Think about the product of the observations.
2. Zero makes the product vanish.

Answer: (d) The mean, median and mode of A are all the same

Explanation:
1. A sorted: 1, 2, 2, 3, 3, 3, 7. Mean = 21/7 = 3, median = 3, mode = 3.
2. B sorted: 3, 5, 5, 7, 8, 9, 12: median 7. C sorted: 2, 3, 4, 4, 4, 7, 11: mean 5, mode 4.
3. Only the statement about A holds (mean A = 3 ≠ mode C = 4, mean C = 5 ≠ median B = 7, median B = 7 ≠ mode A = 3).

Teacher's Note:
1. Compute the three measures for each set.
2. Compare them one pair at a time.

Answer: (d) \( \frac{2n + 1}{3} \)

Explanation:
1. Mean = Σ k(kx)/Σ kx = Σk²/Σk.
2. Σk² = n(n + 1)(2n + 1)/6 and Σk = n(n + 1)/2.
3. Mean = (2n + 1)/3.

Teacher's Note:
1. The common factor x cancels.
2. Use the sum formulas for k and k².

Answer: (b) a bar diagram

Explanation:
1. Discrete data are shown with separate bars (a bar diagram); a histogram is for continuous class intervals.

Teacher's Note:
1. Histograms need continuous classes.
2. Discrete values have gaps between them.

Answer: (c) \( \frac{n + 1}{2}, \frac{2n + 1}{3} \)

Explanation:
1. Simple mean = (1 + 2 + ... + n)/n = (n + 1)/2.
2. Weighted mean = Σ k·k/Σ k = [n(n + 1)(2n + 1)/6]/[n(n + 1)/2] = (2n + 1)/3.

Teacher's Note:
1. Weighted mean = Σwx/Σw with w = x.
2. Use the sum formulas.

Answer: (b) \( \frac{15}{4} \)

Explanation:
1. Total = 10 × 4.5 + 30 × 3.5 = 45 + 105 = 150.
2. Average = 150/40 = 15/4.

Teacher's Note:
1. Add the two group totals.
2. Divide by 40.

Answer: (d) remains the same as that of the original set

Explanation:
1. The median of 9 observations is the 5th value in order.
2. The largest four are the 6th to 9th values, so the 5th value does not change.

Teacher's Note:
1. Identify which position the median occupies.
2. Only the largest 4 values are changed.

Answer: (c) \( \frac{1}{2a}\left\{(a^2 - 1)M_1 + 2M_2\right\} \)

Explanation:
1. x₁ + x₂ = 2M₁ and x₃ + x₄ = 4M₂ - 2M₁.
2. Mean = [a(2M₁) + (4M₂ - 2M₁)/a]/4 = aM₁/2 + M₂/a - M₁/(2a).
3. = (1/2a)[(a² - 1)M₁ + 2M₂].

Teacher's Note:
1. Express x₁ + x₂ and x₃ + x₄ using M₁ and M₂.
2. Combine over the denominator 2a.

Answer: (b) proportional to the frequency densities if the classes are not of equal size

Explanation:
1. For unequal class widths the area of a rectangle must represent the frequency, so the height must be the frequency density (frequency ÷ class width).
2. For equal widths the frequency density is proportional to the frequency.

Teacher's Note:
1. Area represents frequency in a histogram.
2. Width differs, so height must adjust.

Answer: (d) \( \frac{(n + 1)m}{n + 2} \)

Explanation:
1. Original total = nm.
2. New total = nm + 0 + m = (n + 1)m with n + 2 observations.
3. New mean = (n + 1)m/(n + 2).

Teacher's Note:
1. Find the new total and the new number of observations.
2. The number of observations increases by 2.

Answer: (c) Mode - Mean = 3(Median - Mean)

Explanation:
1. The empirical relation is Mode = 3 Median - 2 Mean.
2. Rewrite: Mode - Mean = 3(Median - Mean).

Teacher's Note:
1. Remember Mode = 3 Median - 2 Mean.
2. Subtract Mean from both sides.

Answer: (b) the frequencies of the respective classes

Explanation:
1. With equal class widths, area is proportional to height, so heights are proportional to frequencies.

Teacher's Note:
1. Widths are all equal.
2. Height then represents frequency directly.

Answer: (a) 1

Explanation:
1. Total = 18 + f₈ and the cumulative frequencies are 2, 5, 9, 10, 12, 16, 18.
2. The median is 4 only if the middle position(s) fall on the value 4, i.e. N/2 > 9 and the (N + 1)/2 th and N/2 th values are both 4, which requires N = 19.
3. So f₈ = 1.

Teacher's Note:
1. Work out the cumulative frequencies.
2. For N = 20 the median would be (4 + 5)/2 = 4.5.

Answer: (a) 47 and 37

Explanation:
1. Half the class size = 5. Upper limit = 42 + 5 = 47 and lower limit = 42 - 5 = 37.

Teacher's Note:
1. Class mark is the midpoint of the class.
2. Add and subtract half the class size.

Answer: (b) 46

Explanation:
1. The mode is the value with the highest frequency.
2. The highest frequency 16 occurs at 46 kg.

Teacher's Note:
1. Look for the largest frequency.
2. Ungrouped data need no formula.

Answer: (c) \( l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right)h \)

Explanation:
1. Mode = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] × h, where l is the lower limit of the modal class, f₁ its frequency, f₀ the frequency of the preceding class and f₂ of the next class.
2. The first option is the direct method for the mean, the second the step-deviation method and the last the median.

Teacher's Note:
1. The modal class has the largest frequency.
2. The denominator contains 2f₁ - f₀ - f₂.

Answer: (c) \( l + \left(\frac{\frac{N}{2} - cf}{f}\right)h \)

Explanation:
1. Median = l + [(N/2 - cf)/f] × h, where l is the lower limit of the median class, cf the cumulative frequency of the preceding class, f the frequency of the median class and h the class size.

Teacher's Note:
1. The median class contains the N/2 th observation.
2. cf is the cumulative frequency before the median class.

Answer: (a) 44.5 - 49.5

Explanation:
1. The fourth class is 45-49. The gap between classes is 1, so subtract 0.5 from the lower limit and add 0.5 to the upper limit.
2. Boundaries: 44.5 - 49.5.

Teacher's Note:
1. Find the gap between consecutive classes.
2. Adjust each limit by half the gap.

Answer: (c) 42

Explanation:
1. The third class is 40-44.
2. Class mark = (40 + 44)/2 = 42.

Teacher's Note:
1. Class mark is the midpoint.
2. Use the stated limits.

Answer: (d) 5

Explanation:
1. The boundaries of the third class are 39.5 and 44.5, so the size is 44.5 - 39.5 = 5.

Teacher's Note:
1. Size = upper boundary - lower boundary.
2. Use the actual class boundaries.

Answer: (b) 54.5 - 59.5

Explanation:
1. Subtract 0.5 from the lower limit and add 0.5 to the upper limit: 54.5 - 59.5.

Teacher's Note:
1. The gap between classes is 1.
2. Half of the gap is 0.5.

Answer: (c) 7

Explanation:
1. Σfx = 6 + 8 + 18 + 10 + 2(p + 5) = 2p + 52 and Σf = 11.
2. Mean = 6 gives 2p + 52 = 66, so p = 7.

Teacher's Note:
1. Use mean = Σfx/Σf.
2. Multiply out and solve.

Answer: (a) 76 and 38

Explanation:
1. f₁ + f₂ = 200 - 46 - 25 - 10 - 5 = 114.
2. Σfx = f₁ + 2f₂ + 75 + 40 + 25 = 292 (mean 1.46 × 200), so f₁ + 2f₂ = 152.
3. Subtracting: f₂ = 38 and f₁ = 76.

Teacher's Note:
1. Form two equations: one for the total and one for the mean.
2. Check: 76 + 76 + 75 + 40 + 25 = 292.