Find the expected values and the variance of (i) X (ii) 2X+1 (iii) 2X, given the following probability distribution.

Steps to Solve

1. Define the expected value for X

The expected value ( E(X) ) is calculated using the formula:

$$ E(X) = \sum (x \cdot P(X=x)) $$

Calculating ( E(X) ):

[ E(X) = 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{3} + 3 \cdot 0 + 4 \cdot \frac{1}{3} + 5 \cdot \frac{1}{6} ]

2. Calculate each term for E(X)

Breaking down each term:

[ 1 \cdot \frac{1}{6} = \frac{1}{6} ]

[ 2 \cdot \frac{1}{3} = \frac{2}{3} = \frac{4}{6} ]

[ 3 \cdot 0 = 0 ]

[ 4 \cdot \frac{1}{3} = \frac{4}{3} = \frac{8}{6} ]

[ 5 \cdot \frac{1}{6} = \frac{5}{6} ]

3. Sum the expected value terms

Now add them together to find ( E(X) ):

$$ E(X) = \frac{1}{6} + \frac{4}{6} + 0 + \frac{8}{6} + \frac{5}{6} $$

Summing gives:

$$ E(X) = \frac{1 + 4 + 0 + 8 + 5}{6} = \frac{18}{6} = 3 $$

4. Find variance of X

Variance ( Var(X) ) is given by:

$$ Var(X) = E(X^2) - (E(X))^2 $$

First, calculate ( E(X^2) ):

[ E(X^2) = \sum (x^2 \cdot P(X=x)) ]

Calculating ( E(X^2) ):

$$ E(X^2) = 1^2 \cdot \frac{1}{6} + 2^2 \cdot \frac{1}{3} + 3^2 \cdot 0 + 4^2 \cdot \frac{1}{3} + 5^2 \cdot \frac{1}{6} $$

5. Calculate E(X^2)

Calculating each term:

[ 1^2 \cdot \frac{1}{6} = \frac{1}{6} ]

[ 2^2 \cdot \frac{1}{3} = 4 \cdot \frac{1}{3} = \frac{4}{3} = \frac{8}{6} ]

[ 3^2 \cdot 0 = 0 ]

[ 4^2 \cdot \frac{1}{3} = 16 \cdot \frac{1}{3} = \frac{16}{3} = \frac{32}{6} ]

[ 5^2 \cdot \frac{1}{6} = 25 \cdot \frac{1}{6} = \frac{25}{6} ]

6. Sum E(X^2) terms

Now add them to find ( E(X^2) ):

$$ E(X^2) = \frac{1}{6} + \frac{8}{6} + 0 + \frac{32}{6} + \frac{25}{6} $$

Summing gives:

$$ E(X^2) = \frac{1 + 8 + 0 + 32 + 25}{6} = \frac{66}{6} = 11 $$

7. Calculate variance

Now substitute back to find ( Var(X) ):

$$ Var(X) = E(X^2) - (E(X))^2 = 11 - 3^2 = 11 - 9 = 2 $$

8. Calculate expected values for 2X + 1 and 2X

For ( E(2X + 1) ):

$$ E(2X + 1) = 2E(X) + 1 = 2 \cdot 3 + 1 = 6 + 1 = 7 $$

For ( E(2X) ):

$$ E(2X) = 2E(X) = 2 \cdot 3 = 6 $$