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CBSE Class 10 Maths Chapter 13 Statistics Model Questions - 1 Marks - Part 1

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SECTION A — Multiple Choice Questions [1 Mark Each]

• 1 Mark Q1. Which of the following is a measure of central tendency?
  (a) Range

  (b) Standard deviation

  (c) Median

  (d) Variance
  Solution:
  Median is a measure of central tendency, whereas Range, Standard deviation, and Variance are measures of dispersion.
  Answer: (c) Median
• 1 Mark Q2. The formula for finding the mean of grouped data by the direct method is:
  (a) $\frac{\sum f_i x_i}{\sum f_i}$

  (b) $a + \frac{\sum f_i d_i}{\sum f_i}$

  (c) $a + \left(\frac{\sum f_i u_i}{\sum f_i}\right)h$

  (d) None of these
  Solution:
  The direct method formula for the mean of grouped data is $\frac{\sum f_i x_i}{\sum f_i}$.
  Answer: (a) $\frac{\sum f_i x_i}{\sum f_i}$
• 1 Mark Q3. If the mean of observations $x_1, x_2, \dots, x_n$ is $\bar{x}$, then the mean of $x_1+a, x_2+a, \dots, x_n+a$ is:
  (a) $\bar{x}$

  (b) $\bar{x} + a$

  (c) $a\bar{x}$

  (d) $\bar{x} - a$
  Solution:
  Adding a constant $a$ to each observation increases the mean by $a$, resulting in $\bar{x} + a$.
  Answer: (b) $\bar{x} + a$
• 1 Mark Q4. If each observation of the raw data is increased by 5, then the mean of the data:
  (a) remains same

  (b) increases by 5

  (c) increases by $5n$

  (d) is multiplied by 5
  Solution:
  When each observation is increased by a fixed value, the mean also increases by that same value.
  Answer: (b) increases by 5
• 1 Mark Q5. If each observation of a data is multiplied by 2, then their mean:
  (a) remains same

  (b) is doubled

  (c) is halved

  (d) increases by 2
  Solution:
  Multiplying each observation by a constant multiplies the mean by that same constant.
  Answer: (b) is doubled
• 1 Mark Q6. The mean of the first 10 natural numbers is:
  (a) 5.5

  (b) 5

  (c) 6

  (d) 6.5
  Solution:
  Sum of first 10 natural numbers = $\frac{10 \times 11}{2} = 55$. Mean = $\frac{55}{10} = 5.5$.
  Answer: (a) 5.5
• 1 Mark Q7. The mean of the first 5 prime numbers is:
  (a) 4.8

  (b) 5

  (c) 5.6

  (d) 5.2
  Solution:
  First 5 primes: $2, 3, 5, 7, 11$. Sum = $28$. Mean = $\frac{28}{5} = 5.6$.
  Answer: (c) 5.6
• 1 Mark Q8. If the mean of five observations $x, x+2, x+4, x+6,$ and $x+8$ is 11, then the mean of the first three observations is:
  (a) 9

  (b) 11

  (c) 13

  (d) 10
  Solution:
  Sum = $5x + 20 = 5 \times 11 = 55 \implies 5x = 35 \implies x = 7$.
  First three observations: $7, 9, 11$. Mean = $\frac{7 + 9 + 11}{3} = \frac{27}{3} = 9$.
  Answer: (a) 9
• 1 Mark Q9. In the formula $x_i = a + h u_i$ for the step-deviation method, $h$ represents:
  (a) frequency

  (b) class size

  (c) mid-point

  (d) total frequency
  Solution:
  In the step-deviation method, $h$ represents the class size (or class width).
  Answer: (b) class size
• 1 Mark Q10. The mid-point of a class interval is given by:
  (a) $\frac{\text{Upper limit} + \text{Lower limit}}{2}$

  (b) $\frac{\text{Upper limit} - \text{Lower limit}}{2}$

  (c) $\text{Upper limit} \times \text{Lower limit}$

  (d) $\text{Upper limit} + \text{Lower limit}$
  Solution:
  $\text{Class mark (mid-point)} = \frac{\text{Upper limit} + \text{Lower limit}}{2}$.
  Answer: (a) $\frac{\text{Upper limit} + \text{Lower limit}}{2}$
• 1 Mark Q11. If the mean of $2, 4, 6, 8, x$ is 6, then the value of $x$ is:
  (a) 10

  (b) 12

  (c) 8

  (d) 14
  Solution:
  $\frac{2 + 4 + 6 + 8 + x}{5} = 6 \implies 20 + x = 30 \implies x = 10$.
  Answer: (a) 10
• 1 Mark Q12. The mean of 10 numbers is 20. If 5 is subtracted from every number, the new mean is:
  (a) 20

  (b) 15

  (c) 10

  (d) 25
  Solution:
  Subtracting 5 from each observation decreases the mean by 5 ($20 - 5 = 15$).
  Answer: (b) 15
• 1 Mark Q13. If the arithmetic mean of $x, x+3, x+6, x+9,$ and $x+12$ is 15, then $x$ equals:
  (a) 6

  (b) 9

  (c) 12

  (d) 15
  Solution:
  $\frac{5x + 30}{5} = 15 \implies x + 6 = 15 \implies x = 9$.
  Answer: (b) 9
• 1 Mark Q14. For a given data with 50 observations, $\sum f_i x_i = 1520$. The mean of the data is:
  (a) 30.4

  (b) 31.4

  (c) 29.4

  (d) 30.2
  Solution:
  $\text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{1520}{50} = 30.4$.
  Answer: (a) 30.4
• 1 Mark Q15. In the step-deviation method, $u_i$ is equal to:
  (a) $\frac{x_i - a}{h}$

  (b) $\frac{a - x_i}{h}$

  (c) $x_i - a$

  (d) $\frac{x_i}{h}$
  Solution:
  $u_i = \frac{x_i - a}{h}$, where $a$ is the assumed mean and $h$ is the class size.
  Answer: (a) $\frac{x_i - a}{h}$
• 1 Mark Q16. The sum of deviations of all observations of a data from their mean is always:
  (a) 1

  (b) -1

  (c) 0

  (d) Depends on data
  Solution:
  $\sum (x_i - \bar{x}) = 0$ is a fundamental property of the arithmetic mean.
  Answer: (c) 0
• 1 Mark Q17. If the mean of a frequency distribution is 7.5 and $\sum f_i x_i = (120 + 3k), \sum f_i = 20$, then $k$ is:
  (a) 10

  (b) 20

  (c) 30

  (d) 40
  Solution:
  $\frac{120 + 3k}{20} = 7.5 \implies 120 + 3k = 150 \implies 3k = 30 \implies k = 10$.
  Answer: (a) 10
• 1 Mark Q18. Which method is most suitable when class sizes are unequal and large values of data are involved?
  (a) Direct method

  (b) Assumed mean method

  (c) Step-deviation method

  (d) Any of these
  Solution:
  Since step-deviation requires equal class widths ($h$), the assumed mean method is preferred when class sizes are unequal and values are large.
  Answer: (b) Assumed mean method
• 1 Mark Q19. If the mean of $x_1, x_2, x_3$ is 20 and the mean of $x_4, x_5$ is 30, then the mean of all five numbers is:
  (a) 24

  (b) 25

  (c) 26

  (d) 22
  Solution:
  Sum of first 3 numbers = $3 \times 20 = 60$. Sum of last 2 numbers = $2 \times 30 = 60$. Total sum = $120$. Mean = $\frac{120}{5} = 24$.
  Answer: (a) 24
• 1 Mark Q20. The mean of 5 observations is 7. If 2 of the observations are 4 and 8, the mean of the remaining 3 observations is:
  (a) 7.67

  (b) 8

  (c) 7

  (d) 6.5
  Solution:
  Total sum = $5 \times 7 = 35$. Sum of 2 observations = $4 + 8 = 12$. Remaining sum = $35 - 12 = 23$. Mean = $\frac{23}{3} \approx 7.67$.
  Answer: (a) 7.67
• 1 Mark Q21. If the weights of 4 students are $45\text{ kg}, 48\text{ kg}, 52\text{ kg},$ and $55\text{ kg}$, their mean weight is:
  (a) 50 kg

  (b) 50.25 kg

  (c) 49.5 kg

  (d) 51 kg
  Solution:
  $\text{Mean} = \frac{45 + 48 + 52 + 55}{4} = \frac{200}{4} = 50\text{ kg}$.
  Answer: (a) 50 kg
• 1 Mark Q22. If the mean of $a, b, c$ is $\frac{a+b+c}{3}$, what is the mean of $a+b, b+c, c+a$?
  (a) $\frac{2(a+b+c)}{3}$

  (b) $a+b+c$

  (c) $\frac{a+b+c}{3}$

  (d) $2(a+b+c)$
  Solution:
  $\text{Mean} = \frac{(a+b) + (b+c) + (c+a)}{3} = \frac{2a + 2b + 2c}{3} = \frac{2(a+b+c)}{3}$.
  Answer: (a) $\frac{2(a+b+c)}{3}$
• 1 Mark Q23. In a grouped frequency distribution, to find the mean, the classes must be:
  (a) Continuous

  (b) Discontinuous

  (c) Equal in width always

  (d) Ascending only
  Solution:
  Class intervals must be continuous (inclusive/exclusive adjusted properly) to correctly find mid-points and calculate the mean.
  Answer: (a) Continuous
• 1 Mark Q24. If the mean of observations $1, 2, 3, \dots, n$ is $\frac{n+1}{2}$, what is the mean of squares of these numbers?
  (a) $\frac{(n+1)(2n+1)}{6}$

  (b) $\frac{n(n+1)}{2}$

  (c) $\frac{n+1}{3}$

  (d) $\frac{2n+1}{3}$
  Solution:
  Sum of squares of first $n$ natural numbers is $\frac{n(n+1)(2n+1)}{6}$. Dividing by $n$ gives $\frac{(n+1)(2n+1)}{6}$.
  Answer: (a) $\frac{(n+1)(2n+1)}{6}$
• 1 Mark Q25. The arithmetic mean of a set of 10 numbers is 20. If each number is multiplied by $\lambda$ and then decreased by 2, the new mean becomes 58. The value of $\lambda$ is:
  (a) 2

  (b) 3

  (c) 4

  (d) 5
  Solution:
  New mean = $20\lambda - 2 = 58 \implies 20\lambda = 60 \implies \lambda = 3$.
  Answer: (b) 3
• 1 Mark Q26. The formula for finding the median of grouped data is:
  (a) $L + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h$

  (b) $L + \left(\frac{n - cf}{f}\right) \times h$

  (c) $L + \left(\frac{\frac{n}{2} + cf}{f}\right) \times h$

  (d) $L - \left(\frac{\frac{n}{2} - cf}{f}\right) \times h$
  Solution:
  The standard formula for the median of a grouped frequency distribution is $L + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h$.
  Answer: (a) $L + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h$
• 1 Mark Q27. In the median formula, $cf$ stands for:
  (a) Cumulative frequency of the median class

  (b) Cumulative frequency of preceding median class

  (c) Class frequency

  (d) Total frequency
  Solution:
  In the median formula, $cf$ denotes the cumulative frequency of the class preceding the median class.
  Answer: (b) Cumulative frequency of preceding median class
• 1 Mark Q28. In the median formula, $f$ stands for:
  (a) Frequency of the median class

  (b) Total frequency

  (c) Frequency of preceding class

  (d) Cumulative frequency
  Solution:
  Here, $f$ represents the frequency of the median class itself.
  Answer: (a) Frequency of the median class
• 1 Mark Q29. In the median formula, $L$ stands for:
  (a) Upper limit of median class

  (b) Lower limit of median class

  (c) Length of median class

  (d) Last value
  Solution:
  $L$ stands for the lower limit of the median class.
  Answer: (b) Lower limit of median class
• 1 Mark Q30. The empirical relationship between three measures of central tendency is:
  (a) $3 \text{ Median} = \text{Mode} + 2 \text{ Mean}$

  (b) $2 \text{ Median} = \text{Mode} + 3 \text{ Mean}$

  (c) $\text{Mode} = 3 \text{ Mean} - 2 \text{ Median}$

  (d) $\text{Median} = 3 \text{ Mode} - 2 \text{ Mean}$
  Solution:
  The empirical formula is $\text{Mode} = 3\text{Median} - 2\text{Mean}$, which can be rearranged as $3\text{Median} = \text{Mode} + 2\text{Mean}$.
  Answer: (a) $3 \text{ Median} = \text{Mode} + 2 \text{ Mean}$
• 1 Mark Q31. If the mode of a distribution is 80 and the mean is 110, then the median is:
  (a) 100

  (b) 90

  (c) 105

  (d) 95
  Solution:
  $3\text{Median} = \text{Mode} + 2\text{Mean} \implies 3\text{Median} = 80 + 2(110) = 80 + 220 = 300 \implies \text{Median} = 100$.
  Answer: (a) 100
• 1 Mark Q32. If the median of a distribution is 50 and the mean is 45, then its mode is:
  (a) 60

  (b) 50

  (c) 65

  (d) 55
  Solution:
  $\text{Mode} = 3(50) - 2(45) = 150 - 90 = 60$.
  Answer: (a) 60
• 1 Mark Q33. If the mode and median of a data are 12 and 15 respectively, then its mean is:
  (a) 16.5

  (b) 15.5

  (c) 14.5

  (d) 13.5
  Solution:
  $12 = 3(15) - 2(\text{Mean}) \implies 12 = 45 - 2\text{Mean} \implies 2\text{Mean} = 33 \implies \text{Mean} = 16.5$.
  Answer: (a) 16.5
• 1 Mark Q34. The median of the first 10 prime numbers is:
  (a) 11

  (b) 12

  (c) 13

  (d) 12.5
  Solution:
  First 10 primes: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29$. The 5th and 6th numbers are $11$ and $13$. Median = $\frac{11 + 13}{2} = 12$.
  Answer: (b) 12
• 1 Mark Q35. The median of the data $15, 14, 19, 21, 16, 13, 22, 18, 20$ is:
  (a) 17

  (b) 19

  (c) 16

  (d) 18
  Solution:
  Ascending order: $13, 14, 15, 16, 18, 19, 20, 21, 22$. Total 9 observations, so the 5th value is $18$.
  Answer: (d) 18
• 1 Mark Q36. If the median of observations $11, 12, 14, 18, x+2, x+4, 30, 32, 35, 41$ arranged in ascending order is 24, then $x$ is:
  (a) 21

  (b) 20

  (c) 23

  (d) 22
  Solution:
  $n = 10$ (even). Median is the average of 5th and 6th terms: $\frac{(x+2) + (x+4)}{2} = 24 \implies x + 3 = 24 \implies x = 21$.
  Answer: (a) 21
• 1 Mark Q37. For finding the median of grouped data, the first step is to find:
  (a) $\frac{n}{2}$

  (b) Cumulative frequencies

  (c) Mid-points

  (d) Class width
  Solution:
  The first step in calculating the median for grouped data is to compute the cumulative frequencies.
  Answer: (b) Cumulative frequencies
• 1 Mark Q38. The median class of a distribution is the class whose cumulative frequency is:
  (a) Less than $\frac{n}{2}$

  (b) Greater than $\frac{n}{2}$

  (c) Nearest to or greater than $\frac{n}{2}$

  (d) Equal to $n$
  Solution:
  The median class is the class whose cumulative frequency is nearest to or greater than $\frac{n}{2}$.
  Answer: (c) Nearest to or greater than $\frac{n}{2}$
• 1 Mark Q39. If $n = 60$ in a grouped frequency distribution, then the value used to locate the median class is:
  (a) 15

  (b) 60

  (c) 30

  (d) 45
  Solution:
  $\frac{n}{2} = \frac{60}{2} = 30$.
  Answer: (c) 30
• 1 Mark Q40. If the lower limit of the median class is 20, class size is 10, frequency is 15, cumulative frequency of preceding class is 12, and $\frac{n}{2} = 25$, the median is:
  (a) $20 + \frac{13}{15} \times 10$

  (b) $20 + \frac{25}{15} \times 10$

  (c) $20 + \frac{12}{15} \times 10$

  (d) $20 + \frac{25-12}{12} \times 10$
  Solution:
  $\text{Median} = L + \left(\frac{\frac{n}{2} - cf}{f}\right)h = 20 + \left(\frac{25 - 12}{15}\right) \times 10 = 20 + \frac{13}{15} \times 10$.
  Answer: (a) $20 + \frac{13}{15} \times 10$
• 1 Mark Q41. The median of a set of 9 distinct observations is 20. If each of the 4 largest observations is increased by 2, the median of the new set is:
  (a) 18

  (b) 22

  (c) 20

  (d) 24
  Solution:
  For 9 observations, the median is the 5th observation. Since only the 4 largest observations are changed, the first 5 observations remain unaffected. Thus, the median remains 20.
  Answer: (c) 20
• 1 Mark Q42. The median of $10, 12, 14, 16, 18, 20$ is:
  (a) 15

  (b) 15.5

  (c) 16.5

  (d) 16
  Solution:
  $n = 6$. The 3rd and 4th terms are 14 and 16. Median = $\frac{14 + 16}{2} = 15$.
  Answer: (a) 15
• 1 Mark Q43. If the median of $\frac{x}{5}, \frac{x}{4}, \frac{x}{3}, \frac{x}{2}, x$ is 8 (where $x > 0$), then $x$ is:
  (a) 12

  (b) 20

  (c) 24

  (d) 16
  Solution:
  The numbers in ascending order are $\frac{x}{5}, \frac{x}{4}, \frac{x}{3}, \frac{x}{2}, x$. The 3rd term (median) is $\frac{x}{3}$. Given $\frac{x}{3} = 8 \implies x = 24$.
  Answer: (c) 24
• 1 Mark Q44. Which of the following can be determined graphically from ogives?
  (a) Mean

  (b) Median

  (c) Mode

  (d) Variance
  Solution:
  The median can be determined graphically using ogives (cumulative frequency curves).
  Answer: (b) Median
• 1 Mark Q45. The abscissa of the point of intersection of the 'less than type' and 'more than type' cumulative frequency curves gives:
  (a) Mean

  (b) Median

  (c) Mode

  (d) Sum
  Solution:
  The x-coordinate (abscissa) of the intersection point of both ogives gives the median.
  Answer: (b) Median
• 1 Mark Q46. If total frequency $n = 100$, the point on the y-axis corresponding to which the median is read from the less than ogive is:
  (a) 100

  (b) 50

  (c) 25

  (d) 75
  Solution:
  $\frac{n}{2} = \frac{100}{2} = 50$.
  Answer: (b) 50
• 1 Mark Q47. In a continuous frequency distribution, median is computed using:
  (a) Class mark

  (b) Lower limit of median class

  (c) Upper limit of median class

  (d) Class width directly without limits
  Solution:
  The formula uses the lower limit ($L$) of the median class.
  Answer: (b) Lower limit of median class
• 1 Mark Q48. If the difference between mode and median is 2, then the difference between median and mean is:
  (a) 4

  (b) 2

  (c) 1

  (d) 3
  Solution:
  $\text{Mode} - \text{Median} = 2(\text{Median} - \text{Mean}) \implies 2 = 2(\text{Median} - \text{Mean}) \implies \text{Median} - \text{Mean} = 1$.
  Answer: (c) 1
• 1 Mark Q49. If mode = 24 and mean = 27, median is:
  (a) 25.5

  (b) 26

  (c) 25

  (d) 28
  Solution:
  $3\text{Median} = \text{Mode} + 2\text{Mean} \implies 3\text{Median} = 24 + 2(27) = 24 + 54 = 78 \implies \text{Median} = 26$.
  Answer: (b) 26
• 1 Mark Q50. If median = 15 and mean = 12, mode is:
  (a) 18

  (b) 20

  (c) 21

  (d) 24
  Solution:
  $\text{Mode} = 3(15) - 2(12) = 45 - 24 = 21$.
  Answer: (c) 21

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