Let the Mean and Variance of five observation | vedclass

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

Question

(D) Given the mean $\overline{x} = \frac{\sum x_i}{5} = 5$,we have $1+3+a+7+b = 25$,which implies $a+b = 14$.Given the variance $\sigma^2 = \frac{\sum

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(D) Given the mean $\overline{x} = \frac{\sum x_i}{5} = 5$,we have $1+3+a+7+b = 25$,which implies $a+b = 14$.Given the variance $\sigma^2 = \frac{\sum x_i^2}{5} - (\overline{x})^2 = 10$,we have $\frac{1^2+3^2+a^2+7^2+b^2}{5} - 25 = 10$,so $1+9+a^2+49+b^2 = 175$,which gives $a^2+b^2 = 116$.Since $a+b=14$ and $a^2+b^2=116$,we solve $(a+b)^2 - 2ab = 116$ $\Rightarrow 196 - 2ab = 116$ $\Rightarrow ab = 40$.The values $a$ and $b$ are roots of $t^2 - 14t + 40 = 0$,so $(t-10)(t-4) = 0$. Given $a > b$,we have $a=10$ and $b=4$.The new observations $y_n = n+x_n$ are $1+1=2, 2+3=5, 3+10=13, 4+7=11, 5+4=9$.The new mean $\overline{y} = \frac{2+5+13+11+9}{5} = \frac{40}{5} = 8$.The new variance is $\frac{2^2+5^2+13^2+11^2+9^2}{5} - 8^2 = \frac{4+25+169+121+81}{5} - 64 = \frac{400}{5} - 64 = 80 - 64 = 16$.

Let the Mean and Variance of five observations $x_1=1, x_2=3, x_3=a, x_4=7$ and $x_5=b$,where $a > b$,be $5$ and $10$ respectively. Then the Variance of the observations $n+x_n$ for $n=1, 2, 3, 4, 5$ is:

Similar Questions

Variance of first $2n$ natural numbers is

The coefficient of variation of $9, 3, 11, 5, 7$ is

The variance is independent of the change of which of the following?

Given that the total of $16$ values is $528$ and the sum of the squares of deviations from $33$ is $9158$. The variance 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?

Vedclass Test Series

Exam Paper Generator

Online Exam Module