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JEE Main 202313 AprilMorning Shift
Exam
JEE Main
Year
2023
Date
13 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Statistics

JEE Main 2023 · 13 April · Morning Shift · Mathematics · Statistics

Let the mean of the data x 1 3 5 7 9 Frequency (f) 4 24 28 α 8 be 5...

Let the mean of the data

xx 11 33 55 77 99
Frequency \left(f\right)</td><tdstyle="text−align:center;"></td> <td style="text-align: center;">4</td><tdstyle="text−align:center;"></td> <td style="text-align: center;">24</td><tdstyle="text−align:center;"></td> <td style="text-align: center;">28</td><tdstyle="text−align:center;"></td> <td style="text-align: center;">\alpha</td><tdstyle="text−align:center;"></td> <td style="text-align: center;">8</td></tr></tbody></table><p>be</td> </tr> </tbody></table><p>be5 .IfmandIf m and{\sigma}^{2} are respectively the mean deviation about the mean and the variance of the data, then 3αm+σ2\dfrac{3\alpha}{m + {\sigma}^{2}} is equal to _______ .</p>

Correct answer

0

Step-by-step solution

Given,

The mean of the below data is 55

xx 11 33 55 77 99
Frequency \left(f\right)</td><tdstyle="text−align:center;"></td> <td style="text-align: center;">4</td><tdstyle="text−align:center;"></td> <td style="text-align: center;">24</td><tdstyle="text−align:center;"></td> <td style="text-align: center;">28</td><tdstyle="text−align:center;"></td> <td style="text-align: center;">\alpha</td><tdstyle="text−align:center;"></td> <td style="text-align: center;">8</td></tr></tbody></table><p>Nowfindingmeanweget,</p><p></td> </tr> </tbody></table><p>Now finding mean we get,</p><p>\dfrac{4 + 72 + 28 \times 5 + 7\alpha + 72}{64 + \alpha} = 5</p><p></p><p>⇒ \alpha = 16</p><p>So,totalnumberofobservationwillbe,</p><p>So, total number of observation will be,∑ {f}_{i} = 80</p><p>Nowfinding,meandeviationweget,</p><p></p><p>Now finding, mean deviation we get,</p><p>M . D = \dfrac{∑ {f}_{i}\left|{x}_{i} - 5\right|}{∑ {f}_{i}}</p><p></p><p>= \dfrac{4 \times 4 + 24 \times 2 + 0 + 16 \times 2 + 8 \times 4}{80}= \dfrac{8}{5}</p><p>So,</p><p>So,m = \dfrac{8}{5}</p><p>Nowfinding,varianceweget,</p><p></p><p>Now finding, variance we get, </p><p>{\sigma}^{2} = \dfrac{∑ {f}_{i}{\left|{x}_{i} - 5\right|}^{2}}{∑ {f}_{i}}</p><p></p><p>⇒ {\sigma}^{2} = \dfrac{4 \times 16 + 24 \times 4 + 0 + 16 \times 4 + 8 \times 16}{80}</p><p></p><p>⇒ {\sigma}^{2} = \dfrac{22}{5}</p><p>Hence,thevalueof</p><p>Hence, the value of\dfrac{3\alpha}{m + {\sigma}^{2}} = \dfrac{3 \times 16}{\dfrac{8}{5} + \dfrac{22}{5}} = 8

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