Easy MCQ +4 / -1 PYQ · JEE Mains 2024

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

  1. A 5
  2. B 8
  3. C 6
  4. D 7 Correct answer

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} </style> <table class="tg" style="undefined;table-layout: fixed; width: 339px"> <colgroup> <col style="width: 82px"> <col style="width: 83px"> <col style="width: 86px"> <col style="width: 88px"> </colgroup> <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> </tr> </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> <tr> <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> <tr> <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> </tr> </tbody> </table></p> <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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