For two data sets,each of size 5,the variance | vedclass

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

Question

(A) Given: $\sigma_{x}^{2} = 4$ and $\sigma_{y}^{2} = 5$ for two sets of size $n_1 = 5$ and $n_2 = 5$.Means are $\bar{x} = 2$ and $\bar{y} = 4$.Using

Vedclass · Accepted Answer

$\frac{11}{2}$

Vedclass · Answer

(A) Given: $\sigma_{x}^{2} = 4$ and $\sigma_{y}^{2} = 5$ for two sets of size $n_1 = 5$ and $n_2 = 5$.Means are $\bar{x} = 2$ and $\bar{y} = 4$.Using the formula $\sigma^2 = \frac{\Sigma x_i^2}{n} - (\bar{x})^2$,we have:$\Sigma x_i^2 = n(\sigma_x^2 + \bar{x}^2) = 5(4 + 2^2) = 5(8) = 40$.$\Sigma y_i^2 = n(\sigma_y^2 + \bar{y}^2) = 5(5 + 4^2) = 5(21) = 105$.Combined mean $\bar{z} = \frac{n_1\bar{x} + n_2\bar{y}}{n_1 + n_2} = \frac{5(2) + 5(4)}{10} = \frac{30}{10} = 3$.Combined variance $\sigma_z^2 = \frac{\Sigma x_i^2 + \Sigma y_i^2}{n_1 + n_2} - (\bar{z})^2$.$\sigma_z^2 = \frac{40 + 105}{10} - (3)^2$.$\sigma_z^2 = \frac{145}{10} - 9 = 14.5 - 9 = 5.5 = \frac{11}{2}$.

Income group	No. of firms
$150-300$	$300$
$300-500$	$500$
$500-800$	$900$
$800-1200$	$1000$
$1200-1800$	$1200$

For two data sets,each of size $5$,the variances are given to be $4$ and $5$ and the corresponding means are given to be $2$ and $4$,respectively. The variance of the combined data set is

Similar Questions

If $x_1, x_2, \ldots, x_n$ are $n$ observations such that $\sum_{i=1}^n x_i^2 = 400$ and $\sum_{i=1}^n x_i = 80$,then the least value of $n$ is

The mean of $9$ observations is $15$. If a new observation is added,the new mean becomes $16$. What is the value of the new observation?

If the median and the range of four numbers $\{x, y, 2x + y, x - y\}$,where $0 < y < x < 2y$,are $10$ and $28$ respectively,then the mean of the numbers is

If the coefficient of variation is $ 60 $ and the standard deviation is $ 24 $,then the arithmetic mean is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module