If the variance of the frequency distribution
| $x$ | $c$ | $2c$ | $3c$ | $4c$ | $5c$ | $6c$ |
|---|---|---|---|---|---|---|
| $f$ | 2 | 1 | 1 | 1 | 1 | 1 |
is 160, then the value of $c\in N$ is
Solution
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<col style="width: 83px">
<col style="width: 86px">
<col style="width: 88px">
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<thead>
<tr>
<th class="tg-baqh">$x_i$</th>
<th class="tg-baqh">$f(x_i)$</th>
<th class="tg-baqh">$x(f(x)$</th>
<th class="tg-baqh">$x^2f(x)$</th>
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</thead>
<tbody>
<tr>
<td class="tg-baqh">C</td>
<td class="tg-baqh">2</td>
<td class="tg-baqh">2C</td>
<td class="tg-baqh">2C$^2$</td>
</tr>
<tr>
<td class="tg-baqh">2C</td>
<td class="tg-baqh">1</td>
<td class="tg-baqh">2C</td>
<td class="tg-baqh">4C$^2$</td>
</tr>
<tr>
<td class="tg-baqh">3C</td>
<td class="tg-baqh">1</td>
<td class="tg-baqh">3C</td>
<td class="tg-baqh">9C$^2$</td>
</tr>
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<td class="tg-baqh">4C</td>
<td class="tg-baqh">1</td>
<td class="tg-baqh">4C</td>
<td class="tg-baqh">16C$^2$</td>
</tr>
<tr>
<td class="tg-baqh">5C</td>
<td class="tg-baqh">1</td>
<td class="tg-baqh">5C</td>
<td class="tg-baqh">25C$^2$</td>
</tr>
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<td class="tg-baqh">6C</td>
<td class="tg-baqh">1</td>
<td class="tg-baqh">6C</td>
<td class="tg-baqh">36C$^2$</td>
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<p>$$\begin{aligned}
& \sigma^2=E\left(x^2\right)-[E(x)], \sum f\left(x_i\right)=7 \\
& E(x)=\sum x f(x)=22 C \\
& E\left(x^2\right)=\sum x^2 f(x)=92 C^2
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \sigma^2=160=\frac{92 C^2}{7}-\left(\frac{22 C}{7}\right)^2 \\
& \Rightarrow C= \pm 7 \text { but } C \in N \\
& \Rightarrow C=7
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Statistics · Topic: Measures of Central Tendency
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