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Statistics question

2025 · 23 Jan · Shift 1 · Q37
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Statistics question

2025 · 23 Jan · Shift 1 · Q37

JEE MainMathematicsStatisticsMCQ+4 / −1
Marks obtains by all the students of class 12 are presented in a freqency distribution with classes of equal width. Let the median of this grouped data be 14 with median class interval 12-18 and median class frequency 12. If the number of students whose marks are less than 12 is 18 , then the total number of students is :
  1. A
    52
  2. B
    44
  3. C
    40
  4. D
    48
View written solutionFree

Correct answer: B

  1. Use the median formula for grouped data

For grouped frequency distribution,

Median=l+(N2−cff)h\text{Median} = l + \left(\frac{\frac{N}{2}-c_f}{f}\right)hMedian=l+(f2N​−cf​​)h

where:

  • lll = lower limit of median class
  • NNN = total frequency
  • cfc_fcf​ = cumulative frequency before median class
  • fff = frequency of median class
  • hhh = class width
  1. Identify the given values

Median =14=14=14

Median class interval is 12−1812-1812−18, so

l=12,h=18−12=6l=12, \quad h=18-12=6l=12,h=18−12=6

Median class frequency is 121212, so

f=12f=12f=12

Number of students with marks less than 121212 is 181818, hence cumulative frequency before median class:

cf=18c_f=18cf​=18
  1. Substitute into the formula
14=12+(N2−1812)614=12+\left(\frac{\frac{N}{2}-18}{12}\right)614=12+(122N​−18​)6
  1. Solve step-by-step

Subtract 121212 from both sides:

2=(N2−1812)62=\left(\frac{\frac{N}{2}-18}{12}\right)62=(122N​−18​)6

Simplify 612=12\frac{6}{12}=\frac{1}{2}126​=21​:

2=12(N2−18)2=\frac{1}{2}\left(\frac{N}{2}-18\right)2=21​(2N​−18)

Multiply by 222:

4=N2−184=\frac{N}{2}-184=2N​−18

So,

N2=22\frac{N}{2}=222N​=22

Hence,

N=44N=44N=44
  1. Check the options

The total number of students is

44\boxed{44}44​

which corresponds to Option B.

  1. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.

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