Let the mean and the variance of 20 observati | vedclass

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

Question

(D) Given that the mean $\bar{x} = 15$ and variance $\sigma^{2} = 9$ for $n = 20$ observations.$\sum x_{i} = 15 	imes 20 = 300$$\sigma^{2} = \frac{\s

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) Given that the mean $\bar{x} = 15$ and variance $\sigma^{2} = 9$ for $n = 20$ observations.$\sum x_{i} = 15 	imes 20 = 300$$\sigma^{2} = \frac{\sum x_{i}^{2}}{n} - (\bar{x})^{2} \Rightarrow 9 = \frac{\sum x_{i}^{2}}{20} - 225$$\frac{\sum x_{i}^{2}}{20} = 234 \Rightarrow \sum x_{i}^{2} = 4680$Now,the mean of $(x_{i} + \alpha)^{2}$ is $178$:$\frac{1}{20} \sum (x_{i} + \alpha)^{2} = 178$$\sum (x_{i}^{2} + 2\alpha x_{i} + \alpha^{2}) = 178 	imes 20 = 3560$$\sum x_{i}^{2} + 2\alpha \sum x_{i} + 20\alpha^{2} = 3560$$4680 + 2\alpha(300) + 20\alpha^{2} = 3560$$20\alpha^{2} + 600\alpha + 1120 = 0$Dividing by $20$:$\alpha^{2} + 30\alpha + 56 = 0$$(\alpha + 28)(\alpha + 2) = 0$So,$\alpha = -28$ or $\alpha = -2$.The maximum value of $\alpha$ is $-2$.The square of the maximum value is $(-2)^{2} = 4$.

Let the mean and the variance of $20$ observations $x_{1}, x_{2}, \ldots, x_{20}$ be $15$ and $9$,respectively. For $\alpha \in R$,if the mean of $(x_{1}+\alpha)^{2}, (x_{2}+\alpha)^{2}, \ldots, (x_{20}+\alpha)^{2}$ is $178$,then the square of the maximum value of $\alpha$ is equal to $...........$

Similar Questions

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?

For a distribution with equal class widths,the median of $100$ observations is $25$. If the median class is $20-30$ and the number of observations less than $20$ is $45$,what is the frequency of the median class?

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

The mortality in a town during $4$ quarters of a year due to various causes is given in the bar graph below. Based on this data,calculate the percentage increase in mortality in the third quarter compared to the second quarter.

The mean of five observations is $4$ and their variance is also $4$. If three of the five observations are $1, 3, 4$,then the product of the other two is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module