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Find the variance of 2, 4, 6, 8, 10.

  • Option 1)

    2

  • Option 2)

    4

  • Option 3)

    8

  • Option 4)

    6

 

Answers (1)

best_answer

As we learned

 

Short cut Method for Variance -

In case of grouped frequency distribution 

\sigma ^{2}= h^{2}\left [ \frac{1}{N}\sum_{i=1}^{n}f_{i}d_{i}^{2}-\left ( \frac{1}{N} \sum_{i=1}^{n}f_{i}d_{i}\right )^{2} \right ]

 

 

- wherein

where 

d_{i}=\frac{x_{i}-A}{h},

and

\bar{x}=A+h\bar{d}=A+\frac{h}{N}\sum_{i=1}^{n}f_{i}d_{i}

 

 

 Variance of 1, 2, 3, 4, 5, is \frac{n^{2}-1}{12}=\frac{5^{2}-1}{12}=2

( \Theta variance of first n natural number is  (n^{2}-1)/12, when each item is doubled (i.e. 2, 4, 5, 8, 10) variance is multiplied by 2^{2}=4 Required variance  =4\times 2=8)


Option 1)

2

Option 2)

4

Option 3)

8

Option 4)

6

Posted by

Himanshu

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