Question

A random variable X, accept any one integer out of 1 to 100 and its probabilities are equal. Then find E(X), E $ \left( {{\text{X}}^{2}} \right) $ and Var (X) of X.

Answer 1

Hint: It is given to us that X can accept any one integer out of 1 to 100 and its probabilities are equal. Thus, we will find the probability of each substitution of the value of X. Then we will find the Mean E(X) as the sum of the products of the value of X and its probabilities. We can find E $ \left( {{\text{X}}^{2}} \right) $ as the sum of the product of the squares of the value of X and its probability. To find the Var(X), we need to find the difference of E $ \left( {{\text{X}}^{2}} \right) $ and square of E(X).

Complete step-by-step answer:
The random variable X can take any value out of 1 to 100. That is, it can take 1 or 2 or 3 and so on up to 100 and the probability of accepting any value is equal.
Thus, there are 100 values that X can take.
This means, the probability of taking the value 1 will be P(X = 1) = $ \dfrac{1}{100} $ . Similarly, probability of taking the value 2 will be P(X = 2) = $ \dfrac{1}{100} $ .
Now, we can find the mean E(X) as the sum of the products of the value of X and its probabilities.
 $ \Rightarrow $ E(X) = $ \sum{\text{XP(X)}} $
 $ \Rightarrow $ E(X) = $ 1\left( \dfrac{1}{100} \right)+2\left( \dfrac{1}{100} \right)+3\left( \dfrac{1}{100} \right)+...+100\left( \dfrac{1}{100} \right) $
 $ \Rightarrow $ E(X) = \[\dfrac{1}{100}\left( 1+2+3+....+100 \right)\]
As we know, the sum of the first n natural number is given $ \dfrac{n\left( n+1 \right)}{2} $ .
So, 1 + 2 + 3 + …. + 100 = $ \dfrac{100\left( 100+1 \right)}{2} $
 $ \Rightarrow $ 1 + 2 + 3 +…. + 100 = 5050
 $ \Rightarrow $ E(X) = \[\dfrac{1}{100}\left( 5050 \right)\]
 $ \Rightarrow $ E(X) = 50.5
Now, to find E $ \left( {{\text{X}}^{2}} \right) $ , we will find the sum of the product of the squares of the value of X and its probability.
\[\begin{align}
  & \Rightarrow \text{E}\left( {{\text{X}}^{\text{2}}} \right)=\sum{{{\text{X}}^{2}}\text{P}\left( \text{X} \right)} \\
 & \Rightarrow \text{E}\left( {{\text{X}}^{\text{2}}} \right)={{1}^{2}}\left( \dfrac{1}{100} \right)+{{2}^{2}}\left( \dfrac{1}{100} \right)+{{3}^{2}}\left( \dfrac{1}{100} \right)+....+{{\left( 100 \right)}^{2}}\left( \dfrac{1}{100} \right) \\
 & \Rightarrow \text{E}\left( {{\text{X}}^{\text{2}}} \right)=\left( \dfrac{1}{100} \right)\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+....+{{100}^{2}} \right) \\
\end{align}\]
Now, sum of squares of n natural numbers is given by $ \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} $ . $ \begin{align}
  & \Rightarrow {{1}^{2}}+{{2}^{2}}+{{3}^{3}}+....+{{100}^{2}}=\dfrac{100\left( 101 \right)\left( 201 \right)}{6} \\
 & \Rightarrow {{1}^{2}}+{{2}^{2}}+{{3}^{3}}+....+{{100}^{2}}=338350 \\
\end{align} $
So, \[\text{E}\left( {{\text{X}}^{\text{2}}} \right)=\left( \dfrac{1}{100} \right)\left( 338350 \right)\]
Thus, E $ \left( {{\text{X}}^{2}} \right) $ = 3,383.5
Thus, Variance Var(X) is given as the difference of E $ \left( {{\text{X}}^{2}} \right) $ and E(X).
Thus, Var(X) = E $ \left( {{\text{X}}^{2}} \right) $ ─ $ {{\left( \text{E}\left( \text{X} \right) \right)}^{2}} $
 $ \Rightarrow $ Var(X) = 3,383.5 ─ $ {{\left( 50.5 \right)}^{2}} $
 $ \Rightarrow $ Var(X) = 3,383.5 ─ 2550.25
 $ \Rightarrow $ Var(X) = 833.25

Note: Another method to find the variance is to find the sum of the product of the square of the difference of the value of X and the mean and the probabilities. Thus Var(X) = $ \sum{{{\left[ \text{X}-\text{E}\left( \text{X} \right) \right]}^{2}}\text{P}\left( \text{X} \right)} $