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 :
Solution
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<colgroup>
<col style="width: 141px">
<col style="width: 153px">
<col style="width: 162px">
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<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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