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Let's break down the problem step by step to find the mean, variance, and standard deviation for the given probability distribution.

### Given:
The number of defects per 1000 machine parts inspected:
- Defects: 0, 1, 2, 3, 4, 5
- Probability: 0.265, 0.292, 0.246, 0.141, 0.043, 0.013

### Step 1: Calculate the Mean (Expected Value)
The mean (or expected value) \( \mu \) is calculated by the formula:

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

Where \( x \) is the number of defects, and \( P(x) \) is the corresponding probability.

So for this, we have:

\[
\mu = (0 \times 0.265) + (1 \times 0.292) + (2 \times 0.246) + (3 \times 0.141) + (4 \times 0.043) + (5 \times 0.013)
\]

### Step 2: Calculate the Variance
The variance \( \sigma^2 \) is calculated by the formula:

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

### Step 3: Calculate the Standard Deviation
The standard deviation \( \sigma \) is simply the square root of the variance:

\[
\sigma = \sqrt{\sigma^2}
\]

I will now compute these values. Given data
defects = [0, 1, 2, 3, 4, 5]
probabilities = [0.265, 0.292, 0.246, 0.141, 0.043, 0.013]

# Step 1: Calculate the mean (expected value)
mean = sum(x * p for x, p in zip(defects, probabilities))

# Step 2: Calculate the variance
variance = sum((x - mean)**2 * p for x, p in zip(defects, probabilities))

# Step 3: Calculate the standard deviation
std_deviation = variance**0.5

mean, variance, std_deviationThe results for the probability distribution are as follows:

- **Mean** (Expected Value): 1.4 (rounded to one decimal place)
- **Variance**: 1.47 (rounded to two decimal places)
- **Standard Deviation**: 1.21 (rounded to two decimal places)

Would you like further details on these calculations, or do you have any questions? 

Here are some related questions to deepen your understanding:

1. How does the mean relate to the center of a probability distribution?
2. How do you interpret the variance and standard deviation in the context of this problem?
3. What is the significance of rounding in probability distributions?
4. How can you calculate the expected value for continuous distributions?
5. How would the mean and standard deviation change if the probabilities were altered?

**Tip**: When working with probability distributions, always double-check that the sum of the probabilities equals 1, as this ensures the distribution is valid!

Use the probability distribution to complete parts (a) and (b) below. The number of defects per 1000 machine parts inspected. Defects: 0, 1, 2, 3, 4, 5. Probability: 0.265, 0.292, 0.246, 0.141, 0.043, 0.013. (a) Find the mean, variance, and standard deviation of the probability distribution.

Learn how to calculate the mean, variance, and standard deviation for a given probability distribution involving defects per 1000 machine parts inspected. This detailed solution is ideal for students in Grades 9-12.

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