The problem requires determining if the table provided is a probability distribution, and if it is, calculating the mean and standard deviation of the distribution.

Step 1: Checking for a Probability Distribution

For a set of values to represent a valid probability distribution, two conditions must be satisfied:

1. The probabilities must sum to 1.
2. Each probability $P(x)$ must be between 0 and 1 (inclusive).

Let's check these conditions:

• The probabilities given are: $P(0) = 0.033, \quad P(1) = 0.162, \quad P(2) = 0.305, \quad P(3) = 0.305, \quad P(4) = 0.162, \quad P(5) = 0.033$

• Summing the probabilities: $0.033 + 0.162 + 0.305 + 0.305 + 0.162 + 0.033 = 1.000$ So, the sum of probabilities is 1, which satisfies the first condition.

• Since all probabilities are between 0 and 1, the second condition is also satisfied.

Therefore, this is a valid probability distribution.

Step 2: Finding the Mean (Expected Value)

The formula for the mean $\mu$ of a probability distribution is: $\mu = \sum [x \cdot P(x)]$ Substitute the values from the table: $\mu = (0 \cdot 0.033) + (1 \cdot 0.162) + (2 \cdot 0.305) + (3 \cdot 0.305) + (4 \cdot 0.162) + (5 \cdot 0.033)$ $\mu = 0 + 0.162 + 0.61 + 0.915 + 0.648 + 0.165 = 2.5$

Thus, the mean $\mu = 2.5$.

Step 3: Finding the Standard Deviation

The standard deviation $\sigma$ is given by the square root of the variance. The variance $\sigma^2$ is calculated as: $\sigma^2 = \sum [(x - \mu)^2 \cdot P(x)]$ First, calculate $(x - \mu)^2$ for each $x$: $(x - 2.5)^2 = (0 - 2.5)^2 = 6.25, \quad (1 - 2.5)^2 = 2.25, \quad (2 - 2.5)^2 = 0.25$ $(3 - 2.5)^2 = 0.25, \quad (4 - 2.5)^2 = 2.25, \quad (5 - 2.5)^2 = 6.25$ Now, calculate the variance: $\sigma^2 = (6.25 \cdot 0.033) + (2.25 \cdot 0.162) + (0.25 \cdot 0.305) + (0.25 \cdot 0.305) + (2.25 \cdot 0.162) + (6.25 \cdot 0.033)$ $\sigma^2 = 0.20625 + 0.3645 + 0.07625 + 0.07625 + 0.3645 + 0.20625 = 1.294$ Finally, the standard deviation $\sigma$ is: $\sigma = \sqrt{1.294} \approx 1.137$

Summary:

• Mean (Expected Value): $\mu = 2.5$
• Standard Deviation: $\sigma \approx 1.137$

Would you like more details or have any questions?

Here are 5 additional questions to consider:

1. How does the expected value help in understanding the distribution?
2. What is the significance of the standard deviation in this context?
3. Can this distribution be modeled using a binomial distribution? Why or why not?
4. How would the probabilities change if there were 6 males instead of 5?
5. What are some real-world applications of probability distributions in genetic studies?

Tip: Always check both conditions (sum of probabilities and individual probability values) when verifying if a set of values forms a valid probability distribution.