five male 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. The data of the table is shown here : x = 0,1,2,3,4,5 and P(x) = 0.033,0,153,0,314,0.314,0.153,0.033. Solve questions here : A) does the probability show a probability distribution? B) find the mean of the random variable x. Select the correct choice below, if necessary fill in the answer box to complete your choice C) find the standard deviation of the random variable x. Select the correct choice below, if necessary fill in the answer box to complete your choice

### A) Does the data represent a valid probability distribution?

To determine whether the given data forms a valid probability distribution, the following two conditions must be met:

1. **The probabilities \( P(x) \) must all be between 0 and 1, inclusive.**
   - The given values of \( P(x) \) are:  
     \( P(0) = 0.033, P(1) = 0.153, P(2) = 0.314, P(3) = 0.314, P(4) = 0.153, P(5) = 0.033 \).  
     All of these probabilities are between 0 and 1, so this condition is satisfied.

2. **The sum of all probabilities must equal 1.**  
   - Let's check the sum of the probabilities:
     \[
     0.033 + 0.153 + 0.314 + 0.314 + 0.153 + 0.033 = 1.000
     \]
   Since the sum equals 1, this condition is also satisfied.

Since both conditions are met, **yes**, the given data represents a probability distribution.

---

### B) Find the mean of the random variable \( x \).

The mean (or expected value \( \mu \)) of a discrete random variable is given by the formula:
\[
\mu = \sum [x \cdot P(x)]
\]

Let's calculate the mean:

\[
\mu = (0 \cdot 0.033) + (1 \cdot 0.153) + (2 \cdot 0.314) + (3 \cdot 0.314) + (4 \cdot 0.153) + (5 \cdot 0.033)
\]
\[
\mu = 0 + 0.153 + 0.628 + 0.942 + 0.612 + 0.165 = 2.5
\]

Thus, the mean of the random variable \( x \) is **2.5**.

---

### C) Find the standard deviation of the random variable \( x \).

The standard deviation \( \sigma \) is the square root of the variance. The variance \( \sigma^2 \) is given by:
\[
\sigma^2 = \sum [x^2 \cdot P(x)] - \mu^2
\]

First, calculate \( \sum [x^2 \cdot P(x)] \):
\[
\sum [x^2 \cdot P(x)] = (0^2 \cdot 0.033) + (1^2 \cdot 0.153) + (2^2 \cdot 0.314) + (3^2 \cdot 0.314) + (4^2 \cdot 0.153) + (5^2 \cdot 0.033)
\]
\[
= 0 + 0.153 + 1.256 + 2.826 + 2.448 + 0.825 = 7.508
\]

Now, calculate the variance:
\[
\sigma^2 = 7.508 - (2.5)^2 = 7.508 - 6.25 = 1.258
\]

Finally, calculate the standard deviation:
\[
\sigma = \sqrt{1.258} \approx 1.122
\]

Thus, the standard deviation is approximately **1.122**.

---

### Summary:
A) The data represents a valid probability distribution.  
B) The mean is \( 2.5 \).  
C) The standard deviation is approximately \( 1.122 \).

---

Do you want details on any of the steps or have additional questions?

Here are 5 related questions to expand on this:

1. How does the concept of expected value relate to real-life decision-making scenarios?
2. What other methods can be used to validate a probability distribution?
3. Can we use similar methods to calculate expected values for continuous random variables?
4. How does changing probabilities affect the mean and standard deviation of a distribution?
5. What role does variance play in understanding the spread of a distribution?

**Tip:** Variance measures how spread out the values of a random variable are from the mean. The higher the variance, the more spread out the values.

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. The data of the table is shown here: X = 0,1,2,3,4,5 and P(X) = 0.033, 0.153, 0.314, 0.314, 0.153, 0.033. Solve questions: A) Does the probability show a probability distribution? B) Find the mean of the random variable X. C) Find the standard deviation of the random variable X.

Analyze a probability distribution for children inheriting an X-linked genetic disorder. Calculate the mean and standard deviation of the random variable X using the given probability data. Learn key concepts like expected value and variance, suitable for students in Grades 10-12.

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