Question

For the following probability distribution determine standard deviation of the random variable X.

$X$	$2$	$3$	$4$
$P(X)$	$0.2$	$0.5$	$0.3$

Answer 1

Hint: The question is about finding the standard deviation in the above probability.
We are able to find it by the use of the variance formula, which is in the form of generalized exponential functions.
We need to find the multiplicative terms for the $XP(X)$ and also for the term ${X^2}P(X)$, asking the summation of the terms will give the values for the variance.
Formula used: \[VarX = E({X^2}) - {[E(X)]^2}\] which is the formula of the variance; where the exponential square minus the exponential whole square.

Complete step-by-step solution:
First, we need to find the values of $XP(X)$ and ${X^2}P(X)$, simply we are able to derive these values with the multiplication of X and P(X).
Thus, we get while first multiplying the terms $X,P(X)$we get $XP(X)$ the values are $0.4,1.5,1.2$
And also, we find the ${X^2}P(X)$ thus we get ${X^2}P(X) = 0.8,4.5,4.8$
The generalized form is in the table below;

$X$	$2$	$3$	$4$
$P(X)$	$0.2$	$0.5$	$0.3$
$XP(X)$	$0.4$	$1.5$	$1.2$
${X^2}P(X)$	$0.8$	$4.5$	$4.8$

Since the variance is given as \[VarX = E({X^2}) - {[E(X)]^2}\] of now we will apply the know values into the formula; \[VarX = E({X^2}) - {[E(X)]^2} \Rightarrow (0.8 + 4.5 + 4.8) - {(0.4 + 1.5 + 1.2)^2}\] (values of the summation from the given table with $XP(X)$ and ${X^2}P(X)$).
Further solving we get, \[VarX = 10.1 - 9.61 \Rightarrow 0.49\] which is the value of the variance
But we need to find the standard deviation thus we get; standard deviation $X = \sqrt {VarX} $
Hence, we get a standard deviation $X = \sqrt {VarX} \Rightarrow \sqrt {0.49} = 0.7$which is the required answer.

Note: : Since the given terms are in the form of probability, the probability is the number of terms of favorable events that divides the number of possible events.
Also, if we divide the probability terms, we get the value of the percentage.
The only difference between the variance and standard deviation is the square root of the variance is the standard deviation.