Q.A random variable $X$ has the probability distribution $P(X=0)=0.1$, $P(X=1)=0.2$, $P(X=2)=0.4$, $P(X=3)=0.3$. Find the mean $E(X)$.

Variance $=E(X^2)-[E(X)]^2$; take the square root for the standard deviation.

Mean, variance and standard deviation of a random variable are CBSE/NCERT Class 12 probability topics.

Given $E(X)=\dfrac65$ and $E(X^2)=2$:

$\text{Var}(X) = E(X^2) - [E(X)]^2 = 2 - \left(\frac65\right)^2 = 2 - \frac{36}{25} = \frac{50-36}{25} = \frac{14}{25}.$

Standard deviation:

$\sigma = \sqrt{\frac{14}{25}} = \frac{\sqrt{14}}{5}.$

Adding a constant doesn't change variance; scaling by $a$ multiplies variance by $a^2$.

For a random variable $X$ and constants $a,b$: $\text{Var}(aX+b) = a^2\text{Var}(X)$. The additive constant $b$ shifts every outcome equally so it does not affect spread, while the multiplicative factor $a$ scales the spread by $a$, hence variance (a squared-deviation measure) scales by $a^2$.

Here $a=5, b=3$: $\text{Var}(5X+3) = 5^2\,\text{Var}(X) = 25\,\text{Var}(X)$.

Variance is the mean of the square minus the square of the mean.

For a random variable $X$ with probability distribution $p(x_i)$:

Mean: $E(X)=\sum x_ip(x_i)$

$E(X^2)=\sum x_i^2p(x_i)$

$\text{Var}(X)=E(X^2)-[E(X)]^2=\sum x_i^2p(x_i)-\left(\sum x_ip(x_i)\right)^2$

A constant has no spread at all, so its variance is always zero.

Step 1. Variance measures $E[(X-E[X])^2]$, the average squared deviation from the mean.

Step 2. If $X=k$ always (a constant), then $E[X]=k$ and every observation equals the mean, so $(X-E[X])^2=0$ always.

Step 3. Hence $\mathrm{var}(k)=0$ for any constant $k$.

Sum all probabilities to $1$ and solve the resulting quadratic in $k$.

Sum of all probabilities must equal 1:

$0+k+2k+2k+3k+k^2+2k^2+(7k^2+k)=1$

Collecting terms: $10k^2+9k=1$, i.e. $10k^2+9k-1=0$

By the quadratic formula: $k=\dfrac{-9\pm\sqrt{81+40}}{20}=\dfrac{-9\pm11}{20}$

So $k=\dfrac{1}{10}$ or $k=-1$ (rejected, probability can't be negative).

definition recall

If $X$ is a discrete random variable taking values $x_1,x_2,\dots,x_n$ with respective probabilities $p_1,p_2,\dots,p_n$ (where $\sum p_i=1$), then its mean (expected value) is

$E(X)=\mu=\sum_{i=1}^n x_ip_i$