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  • If the variance of the first n natural numbers is 10 and the variance of the first m even natural numbers is 16, then m+n is equal to _____. Option: 1 12 Option: 2 18 Option: 3 16<

If the variance of the first n natural numbers is 10 and the variance of the first m even natural numbers is 16, then m+n is equal to _____.
Option: 1 12
Option: 2 18
Option: 3 16
Option: 4 20

Answers (1)

best_answer

 

 

Dispersion (Variance and Standard Deviation) -

Variance and Standard Deviation

The mean of the squares of the deviations from the mean is called the variance and is denoted by σ2 (read as sigma square).

Variance is a quantity which leads to a proper measure of dispersion. 

The variance of n observations x1 , x2 ,..., xn is given by

\sigma^{2}=\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}

Standard Deviation

The standard deviation is a number that measures how far data values are from their mean.

The positive square-root of the variance is called standard deviation. The standard deviation, usually denoted by σ  and it is given by

\sigma=\sqrt{\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}}

-

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

variance of the first n natural numbers is 10

n = 1,2,3,4........n

\sum x_i={1+2+3.....n}=\frac{n(n+1)}{2}

\sum x_i^2={1^2+2^2+3^2.....n^2}=\frac{n(n+1)(2n+1)}{6}

Using variance formula

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

\sigma ^{2}= \left ( \frac{n(n+1)(2n+1)}{6n} \right )-\left ( \frac{n(n+1)}{2n} \right )^{2}

\sigma ^{2}= \left ( \frac{(n+1)(2n+1)}{6} \right )-\left ( \frac{(n+1)}{2} \right )^{2}

\sigma ^{2}=\frac{n+1}{2}\left [ \left ( \frac{2n+1}{3} \right )-\left ( \frac{n+1}{2} \right ) \right ]

\sigma ^{2}=\frac{n+1}{2} \left ( \frac{4n+2-3n-3}{6} \right )

\\\sigma _n=\frac{n^2-1}{12}=10 \\

n^2=121\;\;\;\;\Rightarrow\;\;\;\;n=11

 

variance of the first m even natural numbers is 16

N = 2,4,6,8........2m

N = 2(1,2,3,4........m)

\sum x_j=2{(1+2+3.....m)}=2\times\frac{m(m+1)}{2}=m(m+1)

\sum x_j^2=2^2{(1^2+2^2+3^2.....m^2)}=4\times\frac{m(m+1)(2m+1)}{6}

Using variance formula

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

\sigma ^{2}= \left ( \frac{2m(m+1)(2m+1)}{3m} \right )-\left ( \frac{m(m+1)}{m} \right )^{2}

\sigma ^{2}= \left ( \frac{2(m+1)(2m+1)}{3} \right )-\left (m+1 \right )^{2}

\sigma ^{2}=\frac{m+1}{3} \left ( 4m+2-3m-3 \right )

\\ \sigma _{m(even)} = \frac{m^2-1}{3}=16

m^2=49\;\;\;\Rightarrow\;\;\;m=7

Hence

                n+m = 18

Posted by

Kuldeep Maurya

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