If the mean and variance of five observations | vedclass

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

Question

(C) Let the five observations be $x_1, x_2, x_3, x_4, x_5$. Given mean $\bar{X} = \frac{24}{5}$ and variance $\sigma^2 = \frac{194}{25}$. Sum of five

Vedclass · Accepted Answer

$\frac{5}{4}$

Vedclass · Answer

(C) Let the five observations be $x_1, x_2, x_3, x_4, x_5$. Given mean $\bar{X} = \frac{24}{5}$ and variance $\sigma^2 = \frac{194}{25}$. Sum of five observations: $\sum_{i=1}^5 x_i = 5 	imes \frac{24}{5} = 24$. Mean of first four observations is $\frac{7}{2}$,so $\sum_{i=1}^4 x_i = 4 	imes \frac{7}{2} = 14$. Thus,$x_5 = 24 - 14 = 10$. Using the variance formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{X})^2$: $\frac{194}{25} = \frac{\sum_{i=1}^5 x_i^2}{5} - (\frac{24}{5})^2$. $\frac{194}{25} = \frac{\sum_{i=1}^5 x_i^2}{5} - \frac{576}{25}$ $\Rightarrow \frac{\sum_{i=1}^5 x_i^2}{5} = \frac{770}{25} = \frac{154}{5}$. $\sum_{i=1}^5 x_i^2 = 154$. Since $x_5 = 10$,$x_5^2 = 100$. $\sum_{i=1}^4 x_i^2 = 154 - 100 = 54$. Variance of first four observations: $	ext{Var} = \frac{\sum_{i=1}^4 x_i^2}{4} - (	ext{mean of first four})^2$. $	ext{Var} = \frac{54}{4} - (\frac{7}{2})^2 = \frac{27}{2} - \frac{49}{4} = \frac{54 - 49}{4} = \frac{5}{4}$.

If the mean and variance of five observations are $\frac{24}{5}$ and $\frac{194}{25}$ respectively,and the mean of the first four observations is $\frac{7}{2}$,then the variance of the first four observations is equal to:

Similar Questions

The coefficient of variation of the first $5$ prime numbers is

The outcome of each of $30$ items was observed; $10$ items gave an outcome $\frac{1}{2} - d$ each,$10$ items gave an outcome $\frac{1}{2}$ each,and the remaining $10$ items gave an outcome $\frac{1}{2} + d$ each. If the variance of this outcome data is $\frac{4}{3}$,then $|d|$ equals:

For a set of observations,if the coefficient of variation is $25$ and the mean is $44$,then the variance is:

Statement $1$: The variance of the first $n$ odd natural numbers is $\frac{n^2 - 1}{3}$.
Statement $2$: The sum of the first $n$ odd natural numbers is $n^2$ and the sum of the squares of the first $n$ odd natural numbers is $\frac{n(4n^2 - 1)}{3}$.

Find the standard deviation of the first $n$ natural numbers.

Vedclass Test Series

Exam Paper Generator

Online Exam Module