A die is tossed once. If the random variable X is defined as X = …, then the variance of X is

When a die is rolled, the outcomes are: 1, 2, 3, 4, 5, or 6

Odd numbers (1, 3, 5) give $X = 1$X=1

Even numbers (2, 4, 6) give $X = -1$X=−1

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Out of 6 possible outcomes:

3 are odd, so $P(X = 1) = \dfrac{3}{6} = \dfrac{1}{2}$P(X=1)=63​=21​

3 are even, so $P(X = -1) = \dfrac{3}{6} = \dfrac{1}{2}$P(X=−1)=63​=21​

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The expected value is:

$$E(X) = (1) \times \dfrac{1}{2} + (-1) \times \dfrac{1}{2}$$E(X)=(1)×21​+(−1)×21​

$$E(X) = \dfrac{1}{2} - \dfrac{1}{2}$$E(X)=21​−21​

$$E(X) = 0$$E(X)=0

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The expected value of $X^2$X2 is:

$$E(X^2) = (1)^2 \times \dfrac{1}{2} + (-1)^2 \times \dfrac{1}{2}$$E(X2)=(1)2×21​+(−1)2×21​

$$E(X^2) = 1 \times \dfrac{1}{2} + 1 \times \dfrac{1}{2}$$E(X2)=1×21​+1×21​

$$E(X^2) = \dfrac{1}{2} + \dfrac{1}{2}$$E(X2)=21​+21​

$$E(X^2) = 1$$E(X2)=1

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Using the variance formula:

$$Var(X) = E(X^2) - [E(X)]^2$$Var(X)=E(X2)−[E(X)]2

$$Var(X) = 1 - (0)^2$$Var(X)=1−(0)2

$$Var(X) = 1 - 0$$Var(X)=1−0

$$Var(X) = 1$$Var(X)=1

Therefore, the variance of $X$X is $1$1.