Q.In a probability distribution, the mean of the random variable $X$ is $\dfrac{6}{5}$ and mean of $X^2$ is 2, then the standard deviation of $X$ is

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 $E(X^2)=2$:

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

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) s …

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

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

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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.

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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. $10k^2+9k-1=0$

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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) …