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Let's go through the problem step by step:

1. **Does the table show a probability distribution?**
   - A valid probability distribution must satisfy two conditions:
     - All probabilities \( P(x) \) must be between 0 and 1.
     - The sum of the probabilities must equal 1.
   
   The given table has the following values:
   - \( x = 0 \), \( P(x) = 0.033 \)
   - \( x = 1 \), \( P(x) = 0.151 \)
   - \( x = 2 \), \( P(x) = 0.316 \)
   - \( x = 3 \), \( P(x) = 0.316 \)
   - \( x = 4 \), \( P(x) = 0.151 \)
   - \( x = 5 \), \( P(x) = 0.033 \)

   Let's check the sum of these probabilities:

\[
0.033 + 0.151 + 0.316 + 0.316 + 0.151 + 0.033 = 1.0
\]

   - All probabilities are between 0 and 1.
   - The sum of probabilities is exactly 1.

Therefore, the table **does show a valid probability distribution**.

### Selection for first part:  
- **A. Yes, the table shows a probability distribution.**

The other options are incorrect because:
- **B.** All probabilities are between 0 and 1, so this is false.
- **C.** The sum of probabilities is 1, so this is false.
- **D.** The variable \( x \) is numerical (number of children), so this is false.
- **E.** The values of \( x \) are properly associated with probabilities, so this is false.

---

2. **Finding the Mean (Expected Value) \( \mu \):**

The mean of a probability distribution is calculated using the formula:

\[
\mu = \sum (x \cdot P(x))
\]

Let's calculate this:

\[
\mu = (0 \cdot 0.033) + (1 \cdot 0.151) + (2 \cdot 0.316) + (3 \cdot 0.316) + (4 \cdot 0.151) + (5 \cdot 0.033)
\]

\[
\mu = 0 + 0.151 + 0.632 + 0.948 + 0.604 + 0.165 = 2.5
\]

### Mean \( \mu = 2.5 \) children.

---

3. **Finding the Standard Deviation \( \sigma \):**

The standard deviation is calculated using the formula:

\[
\sigma = \sqrt{\sum (x^2 \cdot P(x)) - \mu^2}
\]

We already have \( \mu = 2.5 \). Now, let's calculate \( \sum (x^2 \cdot P(x)) \):

\[
\sum (x^2 \cdot P(x)) = (0^2 \cdot 0.033) + (1^2 \cdot 0.151) + (2^2 \cdot 0.316) + (3^2 \cdot 0.316) + (4^2 \cdot 0.151) + (5^2 \cdot 0.033)
\]

\[
= 0 + 0.151 + 1.264 + 2.844 + 2.416 + 0.825 = 7.5
\]

Now, using the formula for standard deviation:

\[
\sigma = \sqrt{7.5 - (2.5)^2} = \sqrt{7.5 - 6.25} = \sqrt{1.25} \approx 1.1
\]

### Standard Deviation \( \sigma \approx 1.1 \).

---

### Final answers:
1. **Yes, the table shows a probability distribution.**
2. **Mean \( \mu = 2.5 \) children.**
3. **Standard deviation \( \sigma \approx 1.1 \) children.**

---

Do you have any questions or would you like further clarification?

Here are 5 follow-up questions:
1. Can you explain the significance of the expected value in probability?
2. How does the standard deviation help in interpreting the spread of data?
3. What would happen to the distribution if one of the probabilities were larger than 1?
4. Why is it important that the sum of probabilities equals 1?
5. Can you calculate the variance based on the standard deviation?

**Tip**: The mean represents the "average" outcome in a probability distribution, and the standard deviation tells us how much the outcomes deviate from the mean on average.

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 not, identify the requirements not satisfied.

Determine if the table represents a valid probability distribution for an X-linked genetic disorder case, and calculate the mean (expected value) and standard deviation. Concepts include probability distributions, expected value, and standard deviation.

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