The Mean (Expected Value) is: μ = Σxp; The Variance is: Var(X) = Σx 2 p − μ 2… $\Rightarrow {{\left[ T-E\left[ T \right] \right]}^{2}}={{a}^{2}}{{\left[ X-E\left[ X \right] \right]}^{2}}$ For a discrete random variable the variance is calculated by summing the product of the square of the difference between the value of the random variable and the expected value, and the associated probability of the value of the random variable, taken over all of the values of the random variable. $i.e.,\sigma =\sqrt{V(X)}$ Solution: Mean, variance and standard deviation for discrete random variables in Excel. A Random Variable is a variable whose possible values are numerical outcomes of a random experiment. Compute V(X). $\Rightarrow T-E\left[ T \right]=a\left[ X-E\left[ X \right] \right]$ Variance is also denoted as σ2. $E\left( {{X}^{2}} \right)=\int\limits_{a}^{b}{{{x}^{2}}}\frac{1}{b-a}dx=\frac{1}{b-a}\left[ \frac{{{b}^{3}}-{{a}^{3}}}{3} \right]$ $\text{We have},\text{ E}\left( \text{X} \right)\text{ }=\frac{\left( a+b \right)}{2}$ Variance of a random variable is discussed in detail here on. 1. $f(x)=0,~~otherwise$ Statement. $P(X=1)=1-2p,for~0\le p\le \frac{1}{2}$     = 0.89 Properties 1. $\Rightarrow E\left[ {{X}^{2}} \right]=(4)(0.1)+(1)(0.1)+(1)(0.2)+(4)(0.3)+9(0.1)$ $\Rightarrow E\left[ X \right]=\frac{1}{m}\left\{ \frac{m\left( m+1 \right)}{2} \right\}=\frac{m+1}{2}$ We'll start with a few definitions. $E\left[ X \right]=\sum\nolimits_{x=1}^{m}{xf(x)=\frac{1}{m}\sum\nolimits_{x=1}^{m}{x}}$ $\Rightarrow T-E\left[ T \right]=aX+b-aE\left[ X \right]-b$ In probability and statistics, a random variable, random quantity, aleatory variable, or stochastic variable is described informally as a variable whose values depend on outcomes of a random phenomenon. $\Rightarrow T-E\left[ T \right]=aX+b-E\left[ aX+b \right]$ Then we have E(X) = (a+b)/2. $\Rightarrow V(X)=\frac{\left( m+1 \right)\left( 2m+1 \right)}{6}-\frac{{{\left( m+1 \right)}^{2}}}{4}$ So, for 0 ≤ p ≤ ½ Var(X) is maximum when p = ½ $\therefore E\left( x \right)=0.8$ The Variance of a random variable X is also denoted by σ;2 but when sometimes can be written as Var (X). Here, X = 0, 1, 2 Find the variance of X. $V(X)=E\left[ {{X}^{2}} \right]-E{{\left[ X \right]}^{2}}$ $i.e.,V(X)={{\sum\limits_{x}{\left( x-\mu \right)}}^{2}}f(x),\text{Where X is discrete}$ The Variance of a random variable is defined as.         = 4.5, Curriculum achievement objectives reference Formally, the expected value of a (discrete) random variable X is defined by: The variance of X is defined in terms of the expected value as: From this we can also obtain: 1. $E\left( X \right)=0(p)+1\left( 1-2p \right)+2(p)=1$ An alternative formula for the variance of a random variable (equation (3)): The binomial coefficient property (equation (4)): Using these identities, as well as a few simple mathematical tricks, we derived the binomial distribution mean and variance formulas. If X is a random variable, then V(aX+b) = a2V(X), where a and b are constants. The variance of that sum or the difference, the variability will increase. A measure of spread for a distribution of a random variable that determines the degree to which the values of a random variable differ from the expected value.         = 1.9, Var(X)    = (0 – 1.9)2x 0.1 + (1 – 1.9)2x 0.2 + (2 – 1.9)2x 0.4 + (3 – 1.9)2x 0.3 therefore, k = 0.1 The home of mathematics education in New Zealand. therefore has the nice interpretation of being the probabilty of X taking on the value 1. Therefore, variance of random variable is defined to measure the spread and scatter in data. Why not reach little more and connect with me directly on Facebook, Twitter or Google Plus. $E\left[ {{X}^{2}} \right]=\sum\limits_{x=1}^{m}{{{x}^{2}}f(x)=\frac{1}{m}\sum\limits_{x=1}^{m}{{{x}^{2}}}}$ Save my name, email, and website in this browser for the next time I comment. Be able to compute the variance and standard deviation of a random variable. 2 Spread E... 2. $\Rightarrow V(X)=\frac{1}{b-a}\left[ \frac{{{b}^{3}}-{{a}^{3}}}{3} \right]-\frac{{{\left( a+b \right)}^{2}}}{4}$ An introduction to the concept of the expected value of a discrete random variable. $\therefore V(X)=E\left[ {{X}^{2}} \right]-E{{\left[ X \right]}^{2}}$ Using Var(X) = E(X2) – [E(X)]2, E(X)    = 0 x 0.1 + 1 x 0.2 + 2 x 0.4 + 3 x 0.3 $\therefore E\left[ {{X}^{2}} \right]=\frac{\left( m+1 \right)\left( 2m+1 \right)}{6}$ $V(X)=E\left[ {{X}^{2}} \right]-E{{\left[ X \right]}^{2}}$   Cumulant-generating function [ edit ] For n ≥ 2 , the n th cumulant of the uniform distribution on the interval [−1/2, 1/2] is B n / n , where B n is the n th Bernoulli number . Solution: $\therefore V(X)=\frac{{{m}^{2}}-1}{12}$, A random variable X has the following probability function. $E\left[ T \right]=E\left[ aX+b \right]$ $V(X)=E\left[ {{\left( X-\mu \right)}^{2}} \right]=E\left[ {{X}^{2}}+{{\mu }^{2}}-2\mu X \right]$ $Let,T=aX+b,then$ For a given set of data the mean and variance random variable is calculated by the formula. $V(X)=\frac{{{\left( b-a \right)}^{2}}}{12}$, Your email address will not be published. In the last two sections below, I’m going to give a summary of these derivations. Random variable X has the following probability function: µ    = 0 x 0.1 + 1 x 0.2 + 2 x 0.4 + 3 x 0.3 One of the important measures of variability of a random variable is variance.