Easy MCQ +4 / -1 PYQ · JEE Mains 2024

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 Correct answer
  4. D 416

Solution

<p><style type="text/css"> .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-baqh{text-align:center;vertical-align:top} .tg .tg-amwm{font-weight:bold;text-align:center;vertical-align:top} </style> <table class="tg" style="undefined;table-layout: fixed; width: 456px"> <colgroup> <col style="width: 141px"> <col style="width: 153px"> <col style="width: 162px"> </colgroup> <thead> <tr> <th class="tg-amwm">Class</th> <th class="tg-amwm">Frequency</th> <th class="tg-amwm">Cumulative<br>frequency</th> </tr> </thead> <tbody> <tr> <td class="tg-baqh">0-4</td> <td class="tg-baqh">3</td> <td class="tg-baqh">3</td> </tr> <tr> <td class="tg-baqh">4-8</td> <td class="tg-baqh">9</td> <td class="tg-baqh">12</td> </tr> <tr> <td class="tg-baqh">8-12</td> <td class="tg-baqh">10</td> <td class="tg-baqh">22</td> </tr> <tr> <td class="tg-baqh">12-16</td> <td class="tg-baqh">8</td> <td class="tg-baqh">30</td> </tr> <tr> <td class="tg-baqh">16-20</td> <td class="tg-baqh">6</td> <td class="tg-baqh">36</td> </tr> </tbody> </table></p> <p>$$\begin{aligned} & \mathrm{M}=1+\left(\frac{\frac{\mathrm{N}}{2}-\mathrm{C}}{\mathrm{f}}\right) \mathrm{h} \\ & \mathrm{M}=8+\frac{18-12}{10} \times 4 \\ & \mathrm{M}=10.4 \\ & 20 \mathrm{M}=208 \end{aligned}$$</p>

About this question

Subject: Mathematics · Chapter: Statistics · Topic: Measures of Central Tendency

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