For a group of 100 students,the mean bar{x}_1 | vedclass

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

Question

(C) Given $n=100$,$\bar{x}_1=40$,and $\sigma_1=15$. The sum of observations is $\sum x_i = n 	imes \bar{x}_1 = 100 	imes 40 = 4000$. The corrected s

Vedclass · Accepted Answer

$\bar{x}_1=\bar{x}_2 ; \sigma_1 > \sigma_2$

Vedclass · Answer

(C) Given $n=100$,$\bar{x}_1=40$,and $\sigma_1=15$. The sum of observations is $\sum x_i = n 	imes \bar{x}_1 = 100 	imes 40 = 4000$. The corrected sum is $\sum x_i^{\prime} = 4000 - 30 - 60 + 40 + 50 = 4000$. Thus,the corrected mean $\bar{x}_2 = \frac{4000}{100} = 40$. So,$\bar{x}_1 = \bar{x}_2$. Now,$\sigma_1^2 = \frac{\sum x_i^2}{n} - (\bar{x}_1)^2$ $\Rightarrow 225 = \frac{\sum x_i^2}{100} - 1600$ $\Rightarrow \sum x_i^2 = 182500$. The corrected sum of squares is $\sum x_i^{\prime 2} = 182500 - 30^2 - 60^2 + 40^2 + 50^2 = 182500 - 900 - 3600 + 1600 + 2500 = 182100$. The corrected variance is $\sigma_2^2 = \frac{182100}{100} - (40)^2 = 1821 - 1600 = 221$. Since $\sigma_1^2 = 225$ and $\sigma_2^2 = 221$,we have $\sigma_1^2 > \sigma_2^2$,which implies $\sigma_1 > \sigma_2$. Therefore,$\bar{x}_1 = \bar{x}_2$ and $\sigma_1 > \sigma_2$.

Daily Expenditure (	$0-50$	$50-100$	$100-150$	$150-200$	$200-250$
Number of Families $(f)$	$24$	$33$	$37$	$b$	$25$

Similar Questions

The mean and variance of $7$ observations are $8$ and $16$ respectively. If five of the observations are $2, 4, 10, 12, 14$,find the remaining two observations.

The mean and standard deviation of $20$ observations are found to be $10$ and $2$ respectively. On rechecking,it was found that an observation $8$ was incorrect. If the incorrect observation $8$ is replaced by $12$,calculate the correct mean and standard deviation.

The frequency distribution of the daily household expenditure of some families in a residential area is given below. If the mode of the distribution is $140$,what is the value of $b$?

Daily Expenditure ( $0-50$ $50-100$ $100-150$ $150-200$ $200-250$

Number of Families $(f)$ $24$ $33$ $37$ $b$ $25$

Let $a, b \in R$. Let the mean and the variance of $6$ observations $-3, 4, 7, -6, a, b$ be $2$ and $23$,respectively. The mean deviation about the mean of these $6$ observations is:

Let the mean and variance of $8$ numbers $x, y, 10, 12, 6, 12, 4, 8$ be $9$ and $9.25$ respectively. If $x > y$,then $3x - 2y$ is equal to $...........$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module