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.
x
​P(x)
 
0
0.034
1
0.156
2
0.310
3
0.310
4
0.156
5
0.034
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Part 1
Does the table show a probability​ distribution? Select all that apply.
A.
​Yes, the table shows a probability distribution.
B.
​No, the sum of all the probabilities is not equal to 1.
C.
​No, not every probability is between 0 and 1 inclusive.
D.
​No, the random variable x is categorical instead of numerical.
E.
​No, the random variable​ x's number values are not associated with probabilities.

To determine whether the given table represents a probability distribution, we need to verify two key requirements:

1. **The sum of all probabilities must equal 1**: For a set of probabilities to define a distribution, the total probability of all possible outcomes must add up to 1.
2. **Each individual probability must be between 0 and 1 inclusive**: Probabilities represent chances of events, so each probability must fall within this range.

Let's evaluate the given table with these criteria in mind:

| \( x \) | \( P(x) \) |
|--------|-----------|
| 0      | 0.034     |
| 1      | 0.156     |
| 2      | 0.310     |
| 3      | 0.310     |
| 4      | 0.156     |
| 5      | 0.034     |

### Step 1: Check if the sum of probabilities equals 1
We sum the probabilities:
\[
0.034 + 0.156 + 0.310 + 0.310 + 0.156 + 0.034 = 1.000
\]
Since the sum is 1, the first condition is satisfied.

### Step 2: Check if all probabilities are between 0 and 1 inclusive
We observe that all probabilities (\( P(x) \)) are within the range \( 0 \leq P(x) \leq 1 \). This condition is also satisfied.

### Conclusion:
Since both conditions are satisfied, **the table does represent a valid probability distribution**.

**Answer:**
- The correct choice is **A**: "Yes, the table shows a probability distribution."

---

### Part 2: Finding the Mean and Standard Deviation

The formulas for the **mean** \( \mu \) and **standard deviation** \( \sigma \) of a probability distribution are as follows:

1. **Mean (Expected Value)** \( \mu \):
\[
\mu = \sum [x \cdot P(x)]
\]
2. **Variance** \( \sigma^2 \):
\[
\sigma^2 = \sum [(x - \mu)^2 \cdot P(x)]
\]
3. **Standard Deviation** \( \sigma \):
\[
\sigma = \sqrt{\sigma^2}
\]

Let's calculate the mean \( \mu \):

\[
\mu = (0 \cdot 0.034) + (1 \cdot 0.156) + (2 \cdot 0.310) + (3 \cdot 0.310) + (4 \cdot 0.156) + (5 \cdot 0.034)
\]
\[
\mu = 0 + 0.156 + 0.620 + 0.930 + 0.624 + 0.170 = 2.5
\]

So, the mean is \( \mu = 2.5 \).

Next, we'll calculate the variance \( \sigma^2 \):
\[
\sigma^2 = (0 - 2.5)^2 \cdot 0.034 + (1 - 2.5)^2 \cdot 0.156 + (2 - 2.5)^2 \cdot 0.310 + (3 - 2.5)^2 \cdot 0.310 + (4 - 2.5)^2 \cdot 0.156 + (5 - 2.5)^2 \cdot 0.034
\]
\[
\sigma^2 = (6.25 \cdot 0.034) + (2.25 \cdot 0.156) + (0.25 \cdot 0.310) + (0.25 \cdot 0.310) + (2.25 \cdot 0.156) + (6.25 \cdot 0.034)
\]
\[
\sigma^2 = 0.2125 + 0.351 + 0.0775 + 0.0775 + 0.351 + 0.2125 = 1.282
\]

Thus, the variance is \( \sigma^2 = 1.282 \), and the standard deviation is:
\[
\sigma = \sqrt{1.282} \approx 1.132
\]

### Final Answer:
- The **mean** is \( \mu = 2.5 \).
- The **standard deviation** is \( \sigma \approx 1.132 \).

Would you like further details or have any questions?

### Here are 5 related questions you might explore:
1. How does the mean change if the probability distribution shifts?
2. Can we calculate the mode for this distribution?
3. What is the interpretation of standard deviation in a probability distribution?
4. How would a change in one probability affect the overall distribution?
5. How does variance help in understanding the spread of a distribution?

**Tip**: The mean of a probability distribution is often called the "expected value," representing the average outcome over many trials.

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.

This problem examines a probability distribution for a genetic disorder inheritance model. Learn how to validate a probability distribution, calculate the mean (expected value), and find the variance and standard deviation using the provided probability data. Suitable for Grades 11-12 students.

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