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  • Tell me? 5 students of a class have an average height of 150 cm and variance of 18cm2 .A new student whose height is 156 cm, joined them. The variance (in cm2) of the height of these six students is:

5 students of a class have an average height of 150 cm and variance of 18cm2 . A new student whose height is 156 cm, joined them. The variance (in cm2) of the height of these six students is:

  • Option 1)

    16

  • Option 2)

    22

  • Option 3)

    20

  • Option 4)

    18

Answers (1)

best_answer

 

Variance -

In case of discrete data 

\dpi{100} \sigma ^{2}= \left ( \frac{\sum x_{i}^{2}}{n} \right )-\left ( \frac{\sum x_{i}}{n} \right )^{2}

-

 

 

 

ARITHMETIC Mean -

For the values x1, x2, ....xn of the variant x the arithmetic mean is given by 

\bar{x}= \frac{x_{1}+x_{2}+x_{3}+\cdots +x_{n}}{n}

in case of discrete data.

-

from the concept we have learned .

mean=\bar{x}=\frac{\sum xi}{5}=150

\Rightarrow \sum_{i=1}^{S}xi=750

and , 

\frac{\sum xi^{2}}{5}-\left ( \bar{x} \right )^{2}=18

or, 

\frac{\sum xi^{2}}{5}-\left ( 150 \right )^{2}=18\Rightarrow \sum xi^{2}=112590

New student join 

let x_{6}=156

Now, \bar{X_{n}}=\frac{\sum_{i=1}^{6}xi}{6}=\frac{750+156}{6}=151

New, variance

=\frac{\sum_{i=1}^{6}xi^{2}}{6}-\left ( \bar{x_{n}} \right )^{2}=\frac{112590+\left ( 156 \right )^{2}}{6}-(156)^{2}

=20

 

 


Option 1)

16

Option 2)

22

Option 3)

20

Option 4)

18

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