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Statistics question

2024 · 30 Jan · Shift 1 · Q42
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Statistics question

2024 · 30 Jan · Shift 1 · Q42

JEE MainMathematicsStatisticsMCQ+4 / −1

Let M denote the median of the following frequency distribution

Class 0 - 4 4 - 8 8 - 12 12 - 16 16 - 20
Frequency 3 9 10 8 6

Then 20M is equal to :

  1. A
    104
  2. B
    52
  3. C
    208
  4. D
    416
View written solutionFree

Correct answer: C

  1. Write the class intervals and frequencies

    ClassFrequency0−434−898−121012−16816−206\begin{array}{c|c} \text{Class} & \text{Frequency} \\ \hline 0-4 & 3 \\ 4-8 & 9 \\ 8-12 & 10 \\ 12-16 & 8 \\ 16-20 & 6 \end{array}Class0−44−88−1212−1616−20​Frequency391086​​

  2. Find the total frequency

    N=3+9+10+8+6=36N = 3+9+10+8+6 = 36N=3+9+10+8+6=36

  3. Find N2\dfrac{N}{2}2N​

    N2=362=18\frac{N}{2} = \frac{36}{2} = 182N​=236​=18

  4. Compute cumulative frequencies

    ClassfCumulative Frequency0−4334−89128−12102212−1683016−20636\begin{array}{c|c|c} \text{Class} & f & \text{Cumulative Frequency} \\ \hline 0-4 & 3 & 3 \\ 4-8 & 9 & 12 \\ 8-12 & 10 & 22 \\ 12-16 & 8 & 30 \\ 16-20 & 6 & 36 \end{array}Class0−44−88−1212−1616−20​f391086​Cumulative Frequency312223036​​

    Since 181818 lies in the cumulative frequency interval from 121212 to 222222, the median class is:

    8−128-128−12

  5. Use the median formula for grouped data

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

    where:

    • l=8l = 8l=8 (lower limit of median class)
    • cf=12c_f = 12cf​=12 (cumulative frequency before median class)
    • f=10f = 10f=10 (frequency of median class)
    • h=4h = 4h=4 (class width)

    Substitute:

    M=8+(18−1210)4M = 8 + \left(\frac{18-12}{10}\right)4M=8+(1018−12​)4 M=8+(610)4=8+2.4=10.4M = 8 + \left(\frac{6}{10}\right)4 = 8 + 2.4 = 10.4M=8+(106​)4=8+2.4=10.4

  6. Find 20M20M20M

    20M=20×10.4=20820M = 20 \times 10.4 = 20820M=20×10.4=208

  7. Match with the options

    208208208

    So the correct option is C.

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