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Statistics

CBSE Class 10 · Maths · Chapter 13

Past board exam questions

CBSE 2027 sample paper · 1 mark

For the following distribution | Class | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 | |---|---|---|---|---|---| | Frequency | 10 | 15 | 12 | 20 | 9 | The lower limit of the median class is

  1. 5
  2. 10
  3. 15
  4. 20
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10

CBSE 2027 sample paper · 1 mark

Mean and median of certain data are 32 and 30 respectively. Using empirical formula, the value of mode is

  1. 36
  2. 26
  3. 30
  4. 20
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26

CBSE 2027 sample paper · 1 mark

The mean age of a combined group of men and women is 35 years. If the mean ages of the men and women are 38 years and 30 years respectively, then the percentage of women in the group is:

  1. 15
  2. 25.5
  3. 35
  4. 37.5
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37.5

CBSE 2027 sample paper · 1 mark

The mean and median of the data are 35.5 and 32 respectively. The value of mode for this data is:

  1. 23
  2. 26
  3. 25
  4. 30
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25

CBSE 2027 sample paper · 3 marks

The mean of the following distribution is 48 and sum of all the frequencies is 50. Find the missing frequencies x and y. | Class | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 | |---|---|---|---|---|---| | Frequency | 8 | 6 | x | 11 | y |

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Assumed mean A = 45, h = 10, ui = (xi - 45)/(10). | Class | fi | xi | ui | fiui | |---|---|---|---|---| | 20-30 | 8 | 25 | −2 | −16 | | 30-40 | 6 | 35 | −1 | −6 | | 40-50 | x | 45 | 0 | 0 | | 50-60 | 11 | 55 | 1 | 11 | | 60-70 | y | 65 | 2 | 2y | | Total | 25 + x + y | | | 2y − 11 | 25 + x + y = 50 ⇒ x + y = 25 ...(1) Mean = 45 + (2y - 11)/(50) × 10 ⇒ 48 = 45 + (2y - 11)/(5) ⇒ 2y - 11 = 15 ⇒ y = 13. From (1), x = 12.

CBSE 2027 sample paper · 4 marks

A school conducted a weekly test for Class X students, before the commencement of the pre-board examination and recorded their scores (out of 50). To analyse performance patterns, the academic coordinator grouped the marks into intervals. The grouped frequency distribution is as below: | Marks Obtained | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | |---|---|---|---|---|---| | Number of students | 3 | 6 | 12 | 15 | 14 | The academic coordinator computed the central tendencies to judge overall learning level, the most common performance range and to understand consistency across the batch for planning the remedial sessions. Based on the above information, answer the following questions: (i) Identify the class with the most common performance range. (1) (ii) Find the class interval containing the median. (1) (iii) (A) What is the average performance of the students? (2) OR (iii) (B) Find the mode of the data. (2)

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(i) The highest frequency is 15, so the most common performance range (modal class) is 30 – 40. (ii) Cumulative frequencies: 3, 9, 21, 36, 50. N = 50, (N)/(2) = 25, which lies in the class with cf 36, so the median class is 30 – 40. (iii)(A) Class marks xi: 5, 15, 25, 35, 45; fixi: 15, 90, 300, 525, 630. Σ fi = 50, Σ fixi = 1560. Mean = (1560)/(50) = 31.2. The average performance is 31.2 marks. OR (iii)(B) Modal class 30 – 40: l = 30, f0 = 12, f1 = 15, f2 = 14, h = 10. Mode = l + (f1 - f0)/(2f1 - f0 - f2) × h = 30 + (15 - 12)/(30 - 12 - 14) × 10 = 30 + (3)/(4) × 10 = 37.5.

CBSE 2027 sample paper · 3 marks

The distribution below gives the weight of 50 students of class X. Find the Modal weight of the students. | Weight ( in kg ) | 35 – 45 | 45 – 55 | 55 – 65 | 65 – 75 | 75 – 85 | |---|---|---|---|---|---| | Number of students | 5 | 10 | 20 | 12 | 3 |

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The maximum frequency is 20, so the modal class is 55 – 65. l = 55, h = 10, f1 = 20, f0 = 10, f2 = 12. Mode = l + ((f1 - f0)/(2f1 - f0 - f2))h = 55 + (20 - 10)/(40 - 10 - 12) × 10 = 55 + (100)/(18) = 60.555... Hence the modal weight of the students is 60.56 kg (approx.).

CBSE 2026 board paper · 1 mark

If mean of a frequency distribution is 7.5, Sigma fi xi = 120 + 3k and Sigma fi = 30, then value of k is

  1. 45
  2. 50
  3. 35
  4. 40
Show answer
35