JEE Main & Advanced JEE Main Paper (Held on 08-4-2019 Afternoon)

  • question_answer
    A student scores the following marks in five tests : 45,54,41,57,43. His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six tests is                         [JEE Main 8-4-2019 Afternoon]

    A) \[\frac{10}{\sqrt{3}}\]           

    B) \[\frac{100}{\sqrt{3}}\]

    C) \[\frac{100}{3}\]                    

    D)   \[\frac{10}{3}\]

    Correct Answer: A

    Solution :

    Let x be the 6th observation \[\Rightarrow 45+54+41+57+43+x=48\times 6=288\] \[\Rightarrow x=48\] variance \[=\left( \frac{\Sigma X_{i}^{2}}{6}-{{(\bar{X})}^{2}} \right)\] \[\Rightarrow \] variance \[=\frac{14024}{6}-{{(48)}^{2}}\]\[=\frac{100}{3}\] \[\Rightarrow \] standard deviation\[=\frac{10}{\sqrt{3}}\]             


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