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Question

In a class of 50 students, the marks obtained are

Marks

15

30

37

40

45

48

No. of students

2

8

14

12

10

4

Its median is

The correct answer is

40

To find the median marks for the 50 students, we first need to arrange the data and calculate the cumulative frequency.

Data and Cumulative Frequency Calculation

The given data is:

Marks No. of students (Frequency) Cumulative Frequency (CF)
15 2 2
30 8 10 (2 + 8)
37 14 24 (10 + 14)
40 12 36 (24 + 12)
45 10 46 (36 + 10)
48 4 50 (46 + 4)

The total number of students, N, is 50.

Median Position Determination

The median is the middle value in an ordered dataset. Since the total number of students (N) is 50, which is an even number, the median is the average of the two middle values.

The positions of these middle values are calculated as:

  • N/2 = 50 / 2 = 25th position
  • (N/2) + 1 = (50 / 2) + 1 = 25 + 1 = 26th position

So, the median is the average of the marks obtained by the 25th and 26th students.

Identifying Median Marks

We use the cumulative frequency (CF) column to find the marks corresponding to the 25th and 26th students:

  • The cumulative frequency up to marks 37 is 24. This means students ranked 1st to 24th scored 37 or less.
  • The cumulative frequency up to marks 40 is 36. This means students ranked 25th to 36th scored 40.

Therefore:

  • The 25th student falls within the group that scored 40 marks.
  • The 26th student also falls within the group that scored 40 marks.

Median Calculation

To find the median, we calculate the average of the marks for the 25th and 26th students:

$$ \text{Median} = \frac{\text{Marks of 25th student} + \text{Marks of 26th student}}{2} $$

$$ \text{Median} = \frac{40 + 40}{2} $$

$$ \text{Median} = \frac{80}{2} $$

$$ \text{Median} = 40 $$

Thus, the median marks obtained by the students is 40.

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Important Questions from Statistical Variables

  1. Let X be a normal random variable with mean zero and variance 9. If a = P(X ≥ 3) then P(|X| ≤ 3) equals:

  2. Find the median if the given data set is:

    3, 3, 7, 8, 12, 13, 16, 19

  3. A machine produces 0, 1 or 2 defective pieces in a day with an associated probability of  \(\frac{{1}}{{6}}\)\(\frac{{2}}{{3}}\) and \(\frac{{1}}{{6}}\) respectively. The mean value and the variance of the number of defective pieces produced by the machine in a day, respectively, are

  4. The median of 7, 5, 8, x, 12, 17 is 10, then what is the value of x?

  5. Which one is parameter from population?

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