If the variance of the first n natural number | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The variance of the first $n$ natural numbers is given by the formula $\sigma^2 = \frac{n^2 - 1}{12}$.Given $\frac{n^2 - 1}{12} = 10$,we have $n^2

Vedclass · Accepted Answer

$11:7$

Vedclass · Answer

(C) The variance of the first $n$ natural numbers is given by the formula $\sigma^2 = \frac{n^2 - 1}{12}$.Given $\frac{n^2 - 1}{12} = 10$,we have $n^2 - 1 = 120$,so $n^2 = 121$,which gives $n = 11$.The first $m$ even natural numbers are $2, 4, 6, \dots, 2m$. This is equivalent to $2 	imes (1, 2, 3, \dots, m)$.The variance of a set of numbers multiplied by a constant $k$ is $k^2$ times the original variance.Thus,the variance of the first $m$ even natural numbers is $2^2 	imes \frac{m^2 - 1}{12} = 4 	imes \frac{m^2 - 1}{12} = \frac{m^2 - 1}{3}$.Given $\frac{m^2 - 1}{3} = 16$,we have $m^2 - 1 = 48$,so $m^2 = 49$,which gives $m = 7$.Therefore,$n: m = 11: 7$.

If the variance of the first $n$ natural numbers is $10$ and the variance of the first $m$ even natural numbers is $16$,then $n: m=$

Similar Questions

The average marks of $10$ students in a class was $60$ with a standard deviation of $4$,while the average marks of another $10$ students was $40$ with a standard deviation of $6$. If all the $20$ students are taken together,their combined standard deviation will be:

For $10$ observations $x_1, x_2, ..., x_{10}$, if $\sum_{i=1}^{10} (x_i + 2)^2 = 180$ and $\sum_{i=1}^{10} (x_i - 1)^2 = 90$, then their standard deviation is:

If the variance of the terms in an increasing $A.P.$,$b_{1}, b_{2}, b_{3}, \ldots, b_{11}$ is $90$,then the common difference of this $A.P.$ is

The variance of the observations $112, 116, 120, 125, 132$ is $......$

The arithmetic mean of the observations $10, 8, 5, a, b$ is $6$ and their variance is $6.8$. Then $ab$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module