The frequency distribution of the age of students in a class of 40 students is given below.
| Age | 15 | 16 | 17 | 18 | 19 | 20 |
|---|---|---|---|---|---|---|
| No of Students | 5 | 8 | 5 | 12 | $x$ | $y$ |
If the mean deviation about the median is 1.25, then $4x+5y$ is equal to :
Solution
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<colgroup>
<col style="width: 95px">
<col style="width: 200px">
<col style="width: 153px">
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<thead>
<tr>
<th class="tg-amwm">Age</th>
<th class="tg-amwm">No. of Students</th>
<th class="tg-amwm">CF</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">15</td>
<td class="tg-baqh">5</td>
<td class="tg-baqh">5</td>
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<tr>
<td class="tg-baqh">16</td>
<td class="tg-baqh">8</td>
<td class="tg-baqh">13</td>
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<td class="tg-baqh">17</td>
<td class="tg-baqh">5</td>
<td class="tg-baqh">18</td>
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<td class="tg-baqh">18</td>
<td class="tg-baqh">12</td>
<td class="tg-baqh">30</td>
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<td class="tg-baqh">19</td>
<td class="tg-baqh">$x$</td>
<td class="tg-baqh">$30+x$</td>
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<td class="tg-baqh">20</td>
<td class="tg-baqh">$y$</td>
<td class="tg-baqh">$30+x+y$</td>
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<p>$$\begin{aligned}
& 30+x+y=40 \\
& x+y=10
\end{aligned}$$</p>
<p>Median $=\left(\frac{n+1}{2}\right)^{\text {th }}$ observation</p>
<p>$=\frac{40+1}{2}=\frac{41}{2}$</p>
<p>Median $=18$</p>
<p>Mean deviation about median</p>
<p>$$\begin{aligned}
& 5.3+8.2+5.1+12.0+x \cdot 1+y \cdot 2=1.25 \times 40 \\
& 15+16+5+x+2 y=50 \\
& x+2 y=14 \\
& \quad x+y=10 \\
& \Rightarrow \quad x=4 \\
& \Rightarrow \quad y=6 \\
& \begin{aligned}
4 x+5 y & =24+20 \\
& =44
\end{aligned}
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Statistics · Topic: Measures of Central Tendency
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