Let's solve each problem step by step.

Problem 1: Mean, Variance, and Standard Deviation of a Probability Distribution

Given the probability distribution:

Step 1: Mean (Expected Value)

The mean $E(X)$ is given by:

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

$E(X) = (0 \times 0.75) + (1 \times 0.17) + (2 \times 0.04) + (3 \times 0.025) + (4 \times 0.005)$

$E(X) = 0 + 0.17 + 0.08 + 0.075 + 0.02 = 0.345$

Step 2: Variance $\sigma^2$

Variance is given by:

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

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

$E(X^2) = \sum X^2 P(X)$

$E(X^2) = (0^2 \times 0.75) + (1^2 \times 0.17) + (2^2 \times 0.04) + (3^2 \times 0.025) + (4^2 \times 0.005)$

$E(X^2) = (0) + (1 \times 0.17) + (4 \times 0.04) + (9 \times 0.025) + (16 \times 0.005)$

$E(X^2) = 0 + 0.17 + 0.16 + 0.225 + 0.08 = 0.635$

Now, compute variance:

$Var(X) = 0.635 - (0.345)^2$

$Var(X) = 0.635 - 0.119 = 0.516$

Step 3: Standard Deviation $\sigma$

$\sigma = \sqrt{Var(X)} = \sqrt{0.516} \approx 0.718$

Problem 2: Mean of the Probability Distribution

Given values:

$X = \{3, 5, 7\}$

$P(3) = \frac{7}{30}, \quad P(5) = \frac{1}{3}, \quad P(7) = \frac{13}{30}$

Mean $E(X)$ is given by:

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

$E(X) = (3 \times \frac{7}{30}) + (5 \times \frac{1}{3}) + (7 \times \frac{13}{30})$

$E(X) = \frac{21}{30} + \frac{5}{3} + \frac{91}{30}$

Convert to common denominator (30):

$E(X) = \frac{21}{30} + \frac{50}{30} + \frac{91}{30}$

$E(X) = \frac{162}{30} = 5.4$

Final Answers

1. Problem 1:

   • Mean (E(X)) = 0.345
   • Variance (Var(X)) = 0.516
   • Standard Deviation (σ) = 0.718

2. Problem 2:

   • Mean (E(X)) = 5.4

Would you like more details or explanations?

Related Questions

1. How do you interpret the standard deviation in a probability distribution?
2. What is the significance of the expected value in probability theory?
3. How can variance help in understanding data spread?
4. What are real-world applications of probability distributions?
5. How do you calculate skewness for a given probability distribution?

Tip: The mean of a probability distribution represents the "long-term" expected outcome if the experiment is repeated many times.