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JEE Main 20198 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
8 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Statistics

JEE Main 2019 · 8 April · Evening Shift · Mathematics · Statistics

A student scores the following marks in five tests: 45,54,41,57,43...

A student scores the following marks in five tests: 45,54,41,57,4345,54,41,57,43. His score is not known for the sixth test. If the mean score is 4848 in the six tests, then the standard deviation of the marks in six tests is:

Options

  1. A103\dfrac{10}{3}
  2. B1003\dfrac{100}{3}
  3. C103\dfrac{10}{\sqrt{3}}
  4. D1003\dfrac{100}{\sqrt{3}}

Correct answer

C. 103\dfrac{10}{\sqrt{3}}

Step-by-step solution

Given mean of 66 observations is 4848

⇒41+45+54+57+43+x6=48⇒ \dfrac{41 + 45 + 54 + 57 + 43 + x}{6} = 48

⇒x=48⇒ x = 48
Now variance σ2=∑(x−x)2n{\sigma}^{2} = \dfrac{∑ {\left(x - x\right)}^{2}}{n}

=72+32+62+92+52+026= \dfrac{{7}^{2} + {3}^{2} + {6}^{2} + {9}^{2} + {5}^{2} + {0}^{2}}{6}
=1003= \dfrac{100}{3}
Hence standard deviation =103= \dfrac{10}{\sqrt{3}}.

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