The random variable X takes the values 1, 2, | vedclass

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

Question

(B) The random variable $X$ follows a discrete uniform distribution on the set $\{1, 2, \ldots, m\}$.The mean $\bar{X} = E[X] = \sum_{n=1}^{m} n \cdot

Vedclass · Accepted Answer

$\frac{m^2-1}{12}$

Vedclass · Answer

(B) The random variable $X$ follows a discrete uniform distribution on the set $\{1, 2, \ldots, m\}$.The mean $\bar{X} = E[X] = \sum_{n=1}^{m} n \cdot P(X=n) = \frac{1}{m} \sum_{n=1}^{m} n = \frac{1}{m} \cdot \frac{m(m+1)}{2} = \frac{m+1}{2}$.The variance is given by $\operatorname{Var}(X) = E[X^2] - (E[X])^2$.$E[X^2] = \sum_{n=1}^{m} n^2 \cdot P(X=n) = \frac{1}{m} \sum_{n=1}^{m} n^2 = \frac{1}{m} \cdot \frac{m(m+1)(2m+1)}{6} = \frac{(m+1)(2m+1)}{6}$.Therefore,$\operatorname{Var}(X) = \frac{(m+1)(2m+1)}{6} - \left(\frac{m+1}{2}\right)^2$.$\operatorname{Var}(X) = \frac{m+1}{2} \left[ \frac{2(2m+1)}{3} - \frac{m+1}{2} \right] = \frac{m+1}{2} \left[ \frac{8m+4-3m-3}{6} \right] = \frac{m+1}{2} \cdot \frac{5m+1}{6}$ (Wait,let's re-calculate).$\operatorname{Var}(X) = \frac{(m+1)(2m+1)}{6} - \frac{(m+1)^2}{4} = \frac{m+1}{2} \left[ \frac{2(2m+1) - 3(m+1)}{6} \right] = \frac{m+1}{2} \left[ \frac{4m+2-3m-3}{6} \right] = \frac{m+1}{2} \cdot \frac{m-1}{6} = \frac{m^2-1}{12}$.

The random variable $X$ takes the values $1, 2, 3, \ldots, m$. If $P(X=n) = \frac{1}{m}$ for each $n$,then the variance of $X$ is

Similar Questions

Which of the following pairs of species is used in the preparation of antacid medicines?

$A$ slit of width $a$ is illuminated by red light of wavelength $6500 \,\, \mathring{A}$. The first minimum will fall at $\theta = 30^{\circ}$ if $a$ is equal to:

$A$ block of $200 \ g$ mass moves with a uniform speed in a horizontal circular groove,with vertical side walls of radius $20 \ cm$. If the block takes $40 \ s$ to complete one round,the normal force by the side walls of the groove is:

The value of $x$ for which $\sin (\cot^{-1} (1 + x)) = \cos(\tan^{-1} x)$ is

Which of the following gates will have an output of $1$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module