Math Problem Statement
Solution
The problem describes a discrete random variable with the probability distribution:
\frac{m}{36}, & m = 1, 2, 3, \dots, 8 \\ 0, & \text{elsewhere} \end{cases}$$ ### Step 1: Finding the Mean (Expected Value) $$\mu$$ The expected value $$\mu$$ of a discrete random variable is given by: $$\mu = E[M] = \sum_{m} m \cdot f(m)$$ Using the given distribution, $$f(m) = \frac{m}{36}$$ for $$m = 1, 2, \dots, 8$$, the mean is: $$\mu = \sum_{m=1}^{8} m \cdot \frac{m}{36} = \frac{1}{36} \sum_{m=1}^{8} m^2$$ First, compute the sum of squares: $$\sum_{m=1}^{8} m^2 = 1^2 + 2^2 + 3^2 + \dots + 8^2 = 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 = 204$$ So the expected value is: $$\mu = \frac{1}{36} \times 204 = \frac{204}{36} = 5.67$$ ### Step 2: Finding the Variance $$\sigma^2$$ The variance is given by: \[ \sigma^2 = E[M^2] - (E[M])^2 \] First, we compute $$E[M^2]$$: $$E[M^2] = \sum_{m=1}^{8} m^2 \cdot f(m) = \sum_{m=1}^{8} m^2 \cdot \frac{m}{36} = \frac{1}{36} \sum_{m=1}^{8} m^3$$ Now, compute the sum of cubes: $$\sum_{m=1}^{8} m^3 = 1^3 + 2^3 + 3^3 + \dots + 8^3 = 1 + 8 + 27 + 64 + 125 + 216 + 343 + 512 = 1296$$ Thus, $$E[M^2]$$ is: $$E[M^2] = \frac{1}{36} \times 1296 = 36$$ Finally, the variance is: $$\sigma^2 = 36 - (5.67)^2 = 36 - 32.1489 = 3.8511$$ ### Summary: - **Mean $$\mu$$ = 5.67** - **Variance $$\sigma^2$$ = 3.8511** Would you like further clarification or details on any step? Here are 5 related questions to expand on this: 1. How would the mean and variance change if the probability distribution were uniform? 2. What is the standard deviation of $$M$$ based on the variance we found? 3. How would the calculations differ if the range of $$m$$ values increased to 10? 4. Can you verify the calculation of $$E[M^2]$$ using a different method? 5. How would you interpret these results in terms of real-life applications? **Tip:** When calculating variance, always check both $$E[M^2]$$ and $$E[M]^2$$ separately to ensure accuracy.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Probability
Discrete Random Variables
Expected Value
Variance
Formulas
Mean (Expected Value): E[M] = Σ(m * f(m))
Variance: Var(M) = E[M^2] - (E[M])^2
Sum of squares: Σm^2 = 1^2 + 2^2 + ...
Sum of cubes: Σm^3 = 1^3 + 2^3 + ...
Theorems
The expected value theorem
Variance formula for discrete random variables
Suitable Grade Level
Undergraduate
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