Question

A pair of dice is rolled. The random variable x is defined as the sum of the scores obtained. Find the probability function, mathematical expectation and variance

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Clarabelle

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1. The possible sums \(x\) that can be obtained when a pair of dice is rolled range from 2 to 12.

2. Calculate the probability function \( P(X = x) \):

P(X = x) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}

P(X = x) = \begin{cases} \frac{x-1}{36} & x = 2,3,4,5,6,7 \\\frac{13-x}{36} & x = 8,9,10,11,12 \\0 & \text{otherwise}\end{cases}

3. Calculate the mathematical expectation \( E(X) \):

E(X) = \sum_{x=2}^{12} x \cdot P(X = x)

E(X) = \sum_{x=2}^{7} x \cdot \frac{x-1}{36} + \sum_{x=8}^{12} x \cdot \frac{13-x}{36}

4. Simplify the summation to find \( E(X) \):

E(X) = \sum_{x=2}^{7} \frac{x(x-1)}{36} + \sum_{x=8}^{12} \frac{x(13-x)}{36}

E(X) = 7

5. Calculate the variance \( \text{Var}(X) \):

\text{Var}(X) = E(X^2) - [E(X)]^2

6. Calculate \( E(X^2) \):

E(X^2) = \sum_{x=2}^{12} x^2 \cdot P(X = x)

7. Simplify the summation to find \( E(X^2) \) and then \( \text{Var}(X) \):

E(X^2) = \sum_{x=2}^{7} \frac{x^2(x-1)}{36} + \sum_{x=8}^{12} \frac{x^2(13-x)}{36}

\text{Var}(X) = E(X^2) - [E(X)]^2

\text{Var}(X) = \frac{35}{6}

Therefore, the probability function, mathematical expectation, and variance are:

P(X = x) = \begin{cases} \frac{x-1}{36} & x = 2,3,4,5,6,7 \\\frac{13-x}{36} & x = 8,9,10,11,12 \\0 & \text{otherwise}\end{cases}

E(X) = 7

\text{Var}(X) = \frac{35}{6}

2. Calculate the probability function \( P(X = x) \):

3. Calculate the mathematical expectation \( E(X) \):

4. Simplify the summation to find \( E(X) \):

5. Calculate the variance \( \text{Var}(X) \):

6. Calculate \( E(X^2) \):

7. Simplify the summation to find \( E(X^2) \) and then \( \text{Var}(X) \):

Therefore, the probability function, mathematical expectation, and variance are:

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