The variance of the first 10 natural numbers | vedclass

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

Question

(D) The first $10$ natural numbers that are multiples of $3$ are: $3, 6, 9, 12, 15, 18, 21, 24, 27, 30$.The mean $\mu$ is calculated as: $\mu = \frac{

Vedclass · Accepted Answer

$74.25$

Vedclass · Answer

(D) The first $10$ natural numbers that are multiples of $3$ are: $3, 6, 9, 12, 15, 18, 21, 24, 27, 30$.The mean $\mu$ is calculated as: $\mu = \frac{3+6+9+12+15+18+21+24+27+30}{10} = \frac{165}{10} = 16.5$.The variance $\sigma^2$ is given by the formula $\sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \mu)^2$.Calculating the sum of squares of deviations: $\sum (x_i - 16.5)^2 = (3-16.5)^2 + (6-16.5)^2 + \dots + (30-16.5)^2 = 182.25 + 110.25 + 56.25 + 20.25 + 2.25 + 2.25 + 20.25 + 56.25 + 110.25 + 182.25 = 742.5$.Therefore,$\sigma^2 = \frac{742.5}{10} = 74.25$.

The variance of the first $10$ natural numbers which are multiples of $3$ is

Similar Questions

The variance of the data $1001, 1003, 1006, 1007, 1009, 1010$ is:

Consider three observations $a, b,$ and $c$ such that $b = a + c$. If the standard deviation of $a + 2, b + 2, c + 2$ is $d$,then which of the following holds true?

The variance of the data $1, 2, 3, 5, 8, 13, 17$ is approximately

$A$ student scored 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 the six tests is:

In a series of $2n$ observations,half of them are equal to $a$ and the remaining half are equal to $-a$. If the standard deviation of these observations is $2$,then $|a|$ equals:

Vedclass Test Series

Exam Paper Generator

Online Exam Module