Question Bank — Statistics

Q1 · MCQ

The class mark of the class interval 25–35 is:

Q2 · MCQ

For grouped data, the median is found using the position:

Q3 · MCQ

The modal class of a frequency distribution is the class with the:

Q4 · MCQ

The empirical relationship between the three averages is:

Q5 · MCQ

In the step-deviation method, uᵢ equals:

Q6 · MCQ

A "less than" ogive is drawn by plotting:

Q7 · MCQ

If Mean = 30 and Median = 28, then using the empirical relation the Mode is:

Q8 · MCQ

The cumulative frequency of the last class of a distribution is always equal to:

Q9 · MCQ

In the median formula, "cf" refers to the cumulative frequency of the:

Q10 · MCQ

Which measure of central tendency is most affected by extreme values?

Q11 · A-R

Assertion: For grouped data the median is found using n/2, not (n+1)/2.
Reason: Grouped data is treated as continuous, so the median corresponds to the n/2-th observation.

Both true, Reason explains Assertion. Because grouped data is continuous, the middle is at the n/2 position, so we locate the class whose cf first reaches n/2.

Q12 · A-R

Assertion: The three mean methods (direct, assumed, step-deviation) may give different answers.
Reason: They use different reference points a and scaling h.

Assertion false, Reason true. Although the methods use different a and h, all three give the same mean — only the arithmetic differs.

Q13 · A-R

Assertion: The median can be read off the point where the two ogives intersect.
Reason: At that point the "less than" and "more than" cumulative frequencies are equal (= n/2).

Both true, Reason explains Assertion. The intersection occurs at cf = n/2, and its x-coordinate is the median.

Q14 · Short

Find the class mark and class size of the interval 40–55.

Class mark = (40 + 55)/2 = 47.5; class size h = 55 − 40 = 15.

Q15 · Short

The mean of Σfᵢxᵢ = 1860 and Σfᵢ = 60. Find the mean.

x̄ = Σfx/Σf = 1860/60 = 31.

Q16 · Short

If the mode is 25 and the mean is 28, estimate the median.

3 Median = Mode + 2 Mean = 25 + 56 = 81 → Median = 27.

Q17 · Short

A modal class 30–40 has f₁ = 20, f₀ = 12, f₂ = 8, h = 10. Find the mode.

Mode = 30 + [(20−12)/(40−12−8)]×10 = 30 + (8/20)×10 = 30 + 4 = 34.

Q18 · Short

A median class 50–60 has l = 50, n/2 = 25, cf = 18, f = 14, h = 10. Find the median.

Median = 50 + [(25−18)/14]×10 = 50 + (7/14)×10 = 50 + 5 = 55.

Q19 · Short

Using assumed mean a = 25, Σfᵢdᵢ = 120 and Σfᵢ = 40. Find the mean.

x̄ = a + Σfd/Σf = 25 + 120/40 = 25 + 3 = 28.

Q20 · Short

Build the cumulative frequency for classes 0–10, 10–20, 20–30 with f = 5, 8, 7.

cf: 5, 5+8 = 13, 13+7 = 20. So cf column = 5, 13, 20 (and n = 20).

Q21 · Short

Using step-deviation with a = 45, h = 10, Σfᵢuᵢ = −6, Σfᵢ = 30. Find the mean.

x̄ = a + h·(Σfu/Σf) = 45 + 10×(−6/30) = 45 − 2 = 43.

Q22 · Short

Why must class intervals be continuous before finding the mode or median?

The mode and median formulas assume no gaps between classes. Discontinuous classes (e.g. 10–19, 20–29) are converted to continuous ones (9.5–19.5, 19.5–29.5) so the boundaries meet.

Q23 · Short

Explain how to read the median from a single "less than" ogive.

Locate y = n/2 on the cf-axis, draw a horizontal line to the curve, then drop a vertical line to the x-axis. That x-value is the median.

Q24 · Short

The mean and median of a distribution are 35.4 and 34.8. Estimate the mode.

Mode = 3 Median − 2 Mean = 3(34.8) − 2(35.4) = 104.4 − 70.8 = 33.6.

Q25 · Short

The total frequency n = 50. Which n/2 value locates the median class, and what does it mean?

n/2 = 25. The median class is the first class whose cumulative frequency ≥ 25.

Q26 · Long

Find the mean of: classes 0–10, 10–20, 20–30, 30–40 with frequencies 7, 10, 15, 8 (direct method).

Class marks: 5, 15, 25, 35. Σf = 40. Σfx = 35 + 150 + 375 + 280 = 840. Mean = 840/40 = 21.

Q27 · Long

Find the median of: classes 0–10, 10–20, 20–30, 30–40, 40–50 with f = 5, 8, 12, 10, 5.

n = 40, n/2 = 20. cf: 5, 13, 25, 35, 40 → median class 20–30 (cf first ≥ 20). l = 20, cf = 13, f = 12, h = 10. Median = 20 + [(20−13)/12]×10 = 20 + 5.83 = 25.83.

Q28 · Long

Find the mode of: classes 10–20, 20–30, 30–40, 40–50 with f = 6, 14, 9, 5.

Modal class 20–30 (highest f = 14). l = 20, f₁ = 14, f₀ = 6, f₂ = 9, h = 10. Mode = 20 + [(14−6)/(28−6−9)]×10 = 20 + (8/13)×10 = 20 + 6.15 = 26.15.

Q29 · Long

Find the mean of: classes 100–120, 120–140, 140–160, 160–180 with f = 12, 14, 8, 6 using the step-deviation method (a = 130, h = 20).

Class marks: 110, 130, 150, 170 → u = −1, 0, 1, 2. Σf = 40. Σfu = −12 + 0 + 8 + 12 = 8. Mean = 130 + 20×(8/40) = 130 + 4 = 134.

Q30 · Long

For a distribution, Mean = 45 and Mode = 42. Find the median and state which average resists outliers.

3 Median = Mode + 2 Mean = 42 + 90 = 132 → Median = 44. The median resists extreme values (outliers), whereas the mean is most affected by them.