CA Foundation P3 · Chapter 14

Measures of Central Tendency and Dispersion MCQs with Answers

10 multiple-choice questions on Measures of Central Tendency and Dispersion for CA Foundation P3 Quantitative Aptitude. Try each one before revealing the answer and explanation.

1. Question 1

   A student scores 60, 75 and 80 in three papers carrying weights 2, 3 and 5 respectively. The weighted mean score is:

   • A) 71.67
   • B) 76.0
   • C) 74.5
   • D) 72.5
   Show answer & explanation

   Answer: C) 74.5

   Weighted mean = Σwx/Σw = (2 x 60 + 3 x 75 + 5 x 80)/(2 + 3 + 5) = (120 + 225 + 400)/10 = 745/10 = 74.5. The simple mean (60 + 75 + 80)/3 = 71.67 ignores the weights.

2. Question 2

   The median of 23, 11, 37, 19, 42, 28 is:

   • A) 25.5
   • B) 23
   • C) 28
   • D) 26.67
   Show answer & explanation

   Answer: A) 25.5

   Arranged in ascending order: 11, 19, 23, 28, 37, 42. With n = 6 (even), the median is the average of the 3rd and 4th values: (23 + 28)/2 = 25.5. 26.67 is the mean (160/6).

3. Question 3

   In a moderately skewed distribution, the mean is 42 and the median is 40. Using the empirical relationship, the mode is approximately:

   • A) 38
   • B) 44
   • C) 36
   • D) 46
   Show answer & explanation

   Answer: C) 36

   Empirical relation: Mode = 3 Median − 2 Mean = 3(40) − 2(42) = 120 − 84 = 36. Reversing the coefficients (3 Mean − 2 Median = 46) is a common error.

4. Question 4

   For the distribution below, the median is (to two decimals): Class: 0–10, 10–20, 20–30, 30–40, 40–50 Frequency: 5, 8, 12, 10, 5

   • A) 25.00
   • B) 25.83
   • C) 24.17
   • D) 26.67
   Show answer & explanation

   Answer: B) 25.83

   N = 40, so N/2 = 20. Cumulative frequencies are 5, 13, 25, 35, 40, so the median class is 20–30 with L = 20, cf = 13, f = 12, h = 10. Median = L + [(N/2 − cf)/f] x h = 20 + (7/12) x 10 = 20 + 5.833 = 25.83. 26.67 is the mode of this distribution.

5. Question 5

   For the distribution below, the mode is (to two decimals): Class: 0–10, 10–20, 20–30, 30–40, 40–50 Frequency: 5, 8, 12, 10, 5

   • A) 26.67
   • B) 25.00
   • C) 23.33
   • D) 28.00
   Show answer & explanation

   Answer: A) 26.67

   The modal class is 20–30 with f₁ = 12, f₀ = 8, f₂ = 10, L = 20, h = 10. Mode = L + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] x h = 20 + [4/(24 − 8 − 10)] x 10 = 20 + (4/6) x 10 = 26.67. Using (f₁ − f₂) in the numerator gives 23.33.

6. Question 6

   The largest and smallest values of a data set are 80 and 20. The coefficient of range is:

   • A) 60
   • B) 0.75
   • C) 4
   • D) 0.6
   Show answer & explanation

   Answer: D) 0.6

   Coefficient of range = (L − S)/(L + S) = (80 − 20)/(80 + 20) = 60/100 = 0.6. 60 is the range itself, and 0.75 is (L − S)/L.

7. Question 7

   The standard deviation of the observations 2, 4, 6, 8, 10, taking them as the whole population (divisor n), is (to two decimals):

   • A) 8.00
   • B) 3.16
   • C) 2.83
   • D) 2.40
   Show answer & explanation

   Answer: C) 2.83

   Mean = 30/5 = 6. Squared deviations: 16, 4, 0, 4, 16, total 40. Variance = 40/5 = 8 and SD = √8 = 2.83. 8 is the variance, 3.16 uses divisor n − 1 (√10), and 2.40 is the mean deviation about the mean.

8. Question 8

   A distribution has mean 50 and standard deviation 8. The coefficient of variation is:

   • A) 6.25%
   • B) 0.16%
   • C) 62.5%
   • D) 16%
   Show answer & explanation

   Answer: D) 16%

   CV = (SD/Mean) x 100 = (8/50) x 100 = 16%. Inverting the ratio (50/8) gives 6.25, and treating the unscaled ratio 0.16 as a percentage gives 0.16%.

9. Question 9

   The standard deviation of x is 5. If y = 3 − 2x, the standard deviation of y is:

   • A) 10
   • B) −10
   • C) 7
   • D) 13
   Show answer & explanation

   Answer: A) 10

   Standard deviation is unaffected by change of origin but affected by change of scale in absolute value: SD(y) = |−2| x SD(x) = 2 x 5 = 10. SD can never be negative, so −10 is impossible.

10. Question 10

    Group I has 40 items with mean 50 and SD 6. Group II has 60 items with mean 60 and SD 8. The combined standard deviation is (to two decimals):

    • A) 7.20
    • B) 8.76
    • C) 7.27
    • D) 76.80
    Show answer & explanation

    Answer: B) 8.76

    Combined mean = (40 x 50 + 60 x 60)/100 = 5,600/100 = 56. d₁ = 50 − 56 = −6, d₂ = 60 − 56 = 4. Combined variance = [40(36 + 36) + 60(64 + 16)]/100 = (2,880 + 4,800)/100 = 76.8. Combined SD = √76.8 = 8.76. Ignoring the d² terms gives √52.8 = 7.27, and the weighted average of SDs is 7.20.