https://file.aiquickdraw.com/mathbot/file/17286750257929cxtgs1c.jpg

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.

Five males with an X-linked genetic disorder have one child each. The random variable x is the number of children among the five who inherit the X-linked genetic disorder. Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviation. If a probability distribution is not given, identify the requirements that are not satisfied.

Determine if the given table forms a probability distribution for an X-linked genetic disorder. This problem covers key probability concepts, including checking for valid distributions, calculating the mean, and finding the standard deviation. Suitable for Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation