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Vocational training complements… (Class 10 Maths 2024, Q37)

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MathsCBSE Class 10 · Standard · Board exam 2024, set 30/3/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Standard 2024: question 37

Set in the CBSE Class 10 board exam of 2024, paper 30/3/1, section E, question 37, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q37 · 4 marks · Case-basedCompetency-basedStatistics

Vocational training complements traditional education by providing practical skills and hands-on experience. While education equips individuals with a broad knowledge base, vocational training focuses on job-specific skills, enhancing employability thus making the student self-reliant. Keeping this in view, a teacher made the following table giving the frequency distribution of students/adults undergoing vocational training from the training institute.

Age (in years)15-1920-2425-2930-3435-3940-4445-4950-54
Number of participants6213296371311104

From the above answer the following questions:

  1. (i)

    What is the lower limit of the modal class of the above data?

    1 mark
  2. (ii)

    (a) Find the median class of the above data.

    2 marks

  3. OR

    (ii)

    (b) Find the number of participants of age less than 50 years who undergo vocational training.

    2 marks
  4. (iii)

    Give the empirical relationship between mean, median and mode.

    1 mark

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 30/3/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

The modal class is the class with the largest frequency. The classes have gaps, so make them continuous first.

Hint 2

For the median class, find n/2 and look for the first cumulative frequency that exceeds it.

Hint 3

Less than 50 years means every class except 50–54.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

  1. (i) The highest frequency is 132, which belongs to the class 20–24. Make the classes continuous by subtracting 0.5 from each lower limit and adding 0.5 to each upper limit. The modal class becomes 19.5–24.5, so its lower limit is 19.5.

    1 mark

  2. (ii)(a) Cumulative frequency table, using continuous classes:

    Age14.5–19.519.5–24.524.5–29.529.5–34.534.5–39.539.5–44.544.5–49.549.5–54.5
    f6213296371311104
    cf62194290327340351361365
    1 mark

  3. (ii)(a) Here , so .

    ½ mark

  4. (ii)(a) The first cumulative frequency greater than 182.5 is 194. So the median class is 19.5–24.5 (that is, 20–24).

    ½ mark

  5. (ii)(b) Participants younger than 50 years are those in all classes up to 45–49. Their number . (Equivalently, .)

    2 marks

  6. (iii) Empirical relationship:

    1 mark

Answer: (i) 19.5; (ii)(a) 19.5–24.5, (ii)(b) 361; (iii) 3 Median = Mode + 2 Mean

Where the marks are

  • (i) Identifies the modal class 19.5–24.5 with lower limit 19.5 (1 mark)
  • (ii)(a) Correct cumulative frequency table (1 mark), n/2 = 182.5 (½ mark), median class 19.5–24.5 (½ mark)
  • (ii)(b) Adds the frequencies of the classes below 50 to get 361 (2 marks)
  • (iii) States 3 Median = Mode + 2 Mean (1 mark)

Revise the chapter

Practise this chapterBoard-style questions from Statistics, one at a time, marked against the answer key and marking points.Back to the full paper2024 board paper, set 30/3/1 (Standard): every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.