Three rotten apples are mixed accidently with seven good apples and four apples are drawn one by one without replacement. Let the random variable X denote the number of rotten apples. If $\mu$ and $\sigma^2$ represent mean and variance of X, respectively, then $10(\mu^2+\sigma^2)$ is equal to :
Solution
<p>3 rotten apples are mixed with 7 good apples.</p>
<p>$\therefore$ Total apples = 10</p>
<p>Among those 10 apples 4 are chosen randomly.</p>
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<th class="tg-7btt">${x_i}$</th>
<th class="tg-7btt">${p_i}$<br></th>
<th class="tg-7btt">${p_i}{x_i}$</th>
<th class="tg-7btt">${p_i}{({x_i})^2}$<br></th>
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<td class="tg-c3ow">0</td>
<td class="tg-c3ow">${{{}^7{C_4}} \over {{}^{10}{C_4}}} = {{35} \over {210}}$</td>
<td class="tg-c3ow">0</td>
<td class="tg-c3ow">0</td>
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<td class="tg-c3ow">1</td>
<td class="tg-c3ow">${{{}^3{C_1} \times {}^7{C_3}} \over {{}^{10}{C_4}}} = {{105} \over {210}}$</td>
<td class="tg-c3ow">${{105} \over {210}}$</td>
<td class="tg-c3ow">${{105} \over {210}}$</td>
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<td class="tg-c3ow">2</td>
<td class="tg-c3ow">${{{}^3{C_2} \times {}^7{C_2}} \over {{}^{10}{C_4}}} = {{63} \over {210}}$</td>
<td class="tg-c3ow">${{126} \over {210}}$</td>
<td class="tg-c3ow">${{252} \over {210}}$</td>
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<td class="tg-c3ow">3</td>
<td class="tg-c3ow">${{{}^3{C_3} \times {}^7{C_1}} \over {{}^{10}{C_4}}} = {7 \over {210}}$</td>
<td class="tg-c3ow">${{21} \over {210}}$</td>
<td class="tg-c3ow">${{63} \over {210}}$</td>
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<p>${x_i}$ = Number of rotten apples drawn.</p>
<p>${p_i}$ = Probability of rotten apple.</p>
<p>We know,</p>
<p>Mean $(\mu ) = \sum {{p_i}{x_i}}$</p>
<p>$= 0 + {{105} \over {210}} + {{126} \over {210}} + {{21} \over {210}}$</p>
<p>$= {{252} \over {210}} = {6 \over 5}$</p>
<p>Also,</p>
<p>Variance $({\sigma ^2}) = \left( {\sum {{p_i}{{({x_i})}^2}} } \right) - {\mu ^2}$</p>
<p>$$ = {{105} \over {210}} + {{252} \over {210}} + {{63} \over {210}} - {{36} \over {25}}$$</p>
<p>$$ = {1 \over 2} + {{12} \over {10}} + {3 \over {10}} - {{36} \over {25}} = {{14} \over {25}}$$</p>
<p>$\therefore$ $10(\mu^2 + {\sigma ^2})$</p>
<p>$= 10\left( {({6 \over 5})^2 + {{14} \over {25}}} \right)$</p>
<p>$= 10\left( {{{36 + 14} \over {25}}} \right)$</p>
<p>$= 10 \times {{50} \over {25}} = 20$</p>
About this question
Subject: Mathematics · Chapter: Statistics · Topic: Measures of Central Tendency
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