Question

How do you find the mean of the random variable \[X\]?
\[X=0,1,2,3\]
\[P(x)=0.15,\text{ }0.35,\text{ 0}\text{.45, 0}\text{.05}\]
What are the variance and standard deviation of the random variable\[X\]?

Answer 1

Hint:First calculate the mean that is the summation of all, a product of variable and its probability then calculate the product of the square of variable and its probability and then add all of them and finally subtract the square of the mean from other calculated stuff that is our variance and the square root of the variance is the standard deviation.

Complete step by step solution:
Given that, the distribution of
Random variable \[(X)=0,1,2,3\]
Probability of random variable \[P(X)=\text{0}\text{.15, 0}\text{.35, 0}\text{.45, 0}\text{.05}\]
Variance \[=?\]
Standard Deviation \[(\sigma )\]\[=?\]
Now,
Mean \[(\mu )\]: When we know the random variable \[X\] and their respective probability then the mean is calculated as
\[\Rightarrow (\mu )=\sum{XP(X)}\] where Σ is Sigma Notation, and means to sum up.
Then the variance \[Var\left( X \right)\] is given by
\[\Rightarrow Var\left( X \right)\text{ }=\text{ }\sum{{{X}^{2}}\text{P(X)}}\text{ }-\text{ }{{\mu }^{2}}\]

Random Variable \[\left( X \right)\]	Probability \[P\left( X \right)\]	\[\mu =XP\left( X \right)\]	\[{{X}^{2}}P(X)\]
\[0\]	\[0.15\]	\[0\]	\[0\]
\[1\]	\[0.35\]	\[0.35\]	\[0.35\]
\[2\]	\[0.45\]	\[0.90\]	\[1.80\]
\[3\]	\[0.05\]	\[0.15\]	\[0.45\]
\[\sum{X}=6\]		\[\mu =1.40\]	\[\sum{{{X}^{2}}P(X)=2.60}\]

From the above calculated data,
\[Mean(\mu )=1.40\] , \[\sum{{{X}^{2}}P(X)=2.60}\]
Now calculating the variance
\[Var\left( X \right)\text{ }=\text{ }\sum{{{X}^{2}}\text{P(X)}}\text{ }-\text{ }{{\mu }^{2}}\]
\[\Rightarrow 2.60-{{(1.40)}^{2}}\]
\[\Rightarrow 2.60-1.96\]
\[\Rightarrow 0.64\]
Thus, the variance of random variable \[X\] has been calculated and that is \[Var\left( X \right)\text{ }=\text{ 0}\text{.64}\]
Now, we know that the standard deviation \[(\sigma )\] is the square root of variance \[Var\left( X \right)\]
\[\Rightarrow \sigma =\sqrt{Var\left( X \right)}\]
\[\Rightarrow \sigma =\sqrt{0.64}\]
Now the square root of \[0.64\] is \[0.8\]\[\Rightarrow \sqrt{0.64}=0.8\]
\[\Rightarrow \sigma =0.8\]
Hence, the variance and standard deviation \[(\sigma )\] of the random variable \[X\] \[Var\left( X \right)\] is
\[0.64\] and \[0.8\] respectively.
Note: During solving the problem, properly do the calculations with the corresponding data. The mean or expected value that we have calculated is a weighted mean which means values with higher probability have a higher contribution to the mean.