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Statistics question

2019 · 8 Apr · Shift 2 · Q39
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Statistics question

2019 · 8 Apr · Shift 2 · Q39

JEE MainMathematicsStatisticsMCQ+4 / −1
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
  1. A
    1003100 \over {\sqrt 3}3​100​
  2. B
    10310 \over {\sqrt 3}3​10​
  3. C
    10310 \over3310​
  4. D
    1003100 \over33100​
View written solutionFree

Correct answer: B

  1. Find the sixth score using the mean

Given five scores: 45, 54, 41, 57, 4345,\ 54,\ 41,\ 57,\ 4345, 54, 41, 57, 43

Their sum is: 45+54+41+57+43=24045+54+41+57+43=24045+54+41+57+43=240

Let the sixth score be xxx.

The mean of six tests is 484848, so: 240+x6=48\frac{240+x}{6}=486240+x​=48

Thus, 240+x=288240+x=288240+x=288 x=48x=48x=48

So the six scores are: 45, 54, 41, 57, 43, 4845,\ 54,\ 41,\ 57,\ 43,\ 4845, 54, 41, 57, 43, 48

  1. Compute deviations from the mean

Mean xˉ=48\bar{x}=48xˉ=48.

Now calculate each deviation and its square:

45−48=−3⇒(−3)2=945-48=-3 \quad \Rightarrow \quad (-3)^2=945−48=−3⇒(−3)2=9 54−48=6⇒62=3654-48=6 \quad \Rightarrow \quad 6^2=3654−48=6⇒62=36 41−48=−7⇒(−7)2=4941-48=-7 \quad \Rightarrow \quad (-7)^2=4941−48=−7⇒(−7)2=49 57−48=9⇒92=8157-48=9 \quad \Rightarrow \quad 9^2=8157−48=9⇒92=81 43−48=−5⇒(−5)2=2543-48=-5 \quad \Rightarrow \quad (-5)^2=2543−48=−5⇒(−5)2=25 48−48=0⇒02=048-48=0 \quad \Rightarrow \quad 0^2=048−48=0⇒02=0

Sum of squared deviations: 9+36+49+81+25+0=2009+36+49+81+25+0=2009+36+49+81+25+0=200

  1. Find the standard deviation

For the six observations, standard deviation is: σ=∑(xi−xˉ)2n\sigma=\sqrt{\frac{\sum (x_i-\bar{x})^2}{n}}σ=n∑(xi​−xˉ)2​​

Here n=6n=6n=6, so: σ=2006\sigma=\sqrt{\frac{200}{6}}σ=6200​​ σ=1003\sigma=\sqrt{\frac{100}{3}}σ=3100​​ σ=103\sigma=\frac{10}{\sqrt{3}}σ=3​10​

  1. Match with the options

The correct option is: B 103\boxed{\text{B }\frac{10}{\sqrt{3}}}B 3​10​​

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