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Chapter 13: Statistics

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This page provides comprehensive Chapter 13: Statistics - MCQ Worksheet - SJMaths. Multiple Choice Questions (MCQ) worksheet for Class 10 Statistics. Practice Mean, Median, and Mode questions for CBSE Board Exams.

MCQ Worksheet

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  1. Question 1: The relationship between mean, median and mode for a moderately skewed distribution is:
    (A) Mode = 3 Median - 2 Mean
    (B) Mode = 2 Median - 3 Mean
    (C) Mode = 2 Mean - 3 Median
    (D) Mode = 3 Mean - 2 Median
    Solution: (A) Mode = 3 Median - 2 Mean
    Reason: This is the empirical relationship between the three measures of central tendency.
  2. Question 2: Construction of a cumulative frequency table is useful in determining the:
    (A) Mean
    (B) Median
    (C) Mode
    (D) All of the above
    Solution: (B) Median
    Reason: Cumulative frequencies are specifically used to find the median class and calculate the median.
  3. Question 3: In the formula $\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$ for finding the mean of grouped data, $d_i$'s are deviations from $a$ of:
    (A) lower limits of the classes
    (B) upper limits of the classes
    (C) mid-points of the classes
    (D) frequencies of the class marks
    Solution: (C) mid-points of the classes
    Reason: $d_i = x_i - a$, where $x_i$ are the class marks (mid-points).
  4. Question 4: If the arithmetic mean of $x, x+3, x+6, x+9$ and $x+12$ is 10, then $x$ is equal to:
    (A) 1
    (B) 2
    (C) 6
    (D) 4
    Solution: (D) 4
    Step 1: Sum $= 5x + 30$. Number of terms $= 5$.
    Step 2: Mean $= \frac{5x+30}{5} = x+6$.
    Step 3: $x+6 = 10 \Rightarrow x = 4$.
  5. Question 5: While computing mean of grouped data, we assume that the frequencies are:
    (A) evenly distributed over all the classes
    (B) centred at the classmarks of the classes
    (C) centred at the upper limits of the classes
    (D) centred at the lower limits of the classes
    Solution: (B) centred at the classmarks of the classes
    Reason: We use the class mark $x_i$ to represent the entire class interval.
  6. Question 6: If $x_i$'s are the mid-points of the class intervals of grouped data, $f_i$'s are the corresponding frequencies and $\bar{x}$ is the mean, then $\sum (f_i x_i - \bar{x})$ is equal to:
    (A) 0
    (B) -1
    (C) 1
    (D) 2
    Solution: (A) 0
    Step 1: $\sum (f_i x_i - \bar{x})$ is not standard notation, usually it implies $\sum f_i(x_i - \bar{x})$.
    Step 2: $\sum (f_i x_i - f_i \bar{x}) = \sum f_i x_i - \bar{x} \sum f_i = N\bar{x} - \bar{x}(N) = 0$.
  7. Question 7: For the following distribution:
    Class0-55-1010-1515-2020-25
    Frequency101512209
    The sum of lower limits of the median class and modal class is:
    (A) 15
    (B) 25
    (C) 30
    (D) 35
    Solution: (B) 25
    Step 1: Modal class is 15-20 (max freq 20). Lower limit = 15.
    Step 2: Total freq $N = 10+15+12+20+9 = 66$. $N/2 = 33$.
    Step 3: CF: 10, 25, 37... 33 lies in 10-15. Median class is 10-15. Lower limit = 10.
    Step 4: Sum $= 15 + 10 = 25$.
  8. Question 8: Consider the following frequency distribution:
    Class0-56-1112-1718-2324-29
    Frequency131015811
    The upper limit of the median class is:
    (A) 17
    (B) 17.5
    (C) 18
    (D) 18.5
    Solution: (B) 17.5
    Step 1: Classes are discontinuous. Convert to continuous: 0.5-5.5, 5.5-11.5, 11.5-17.5...
    Step 2: $N = 13+10+15+8+11 = 57$. $N/2 = 28.5$.
    Step 3: CF: 13, 23, 38. 28.5 lies in 12-17 (continuous 11.5-17.5).
    Step 4: Upper limit is 17.5.
  9. Question 9: The mean of first $n$ odd natural numbers is:
    (A) $\frac{n+1}{2}$
    (B) $\frac{n}{2}$
    (C) $n$
    (D) $n^2$
    Solution: (C) $n$
    Step 1: Sum of first $n$ odd numbers is $n^2$.
    Step 2: Mean $= \frac{\text{Sum}}{n} = \frac{n^2}{n} = n$.
  10. Question 10: If the mode of a distribution is 8 and its mean is 8, then its median is:
    (A) 6
    (B) 7
    (C) 8
    (D) 24
    Solution: (C) 8
    Step 1: 3 Median = Mode + 2 Mean.
    Step 2: 3 Median $= 8 + 2(8) = 24$.
    Step 3: Median $= 24/3 = 8$.
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