91Ó°ÊÓ

When dealing with probability, the goal is to figure out how likely a particular outcome is. In our exercise on refrigerators, we are interested in the probability of selecting a certain number of defective items out of a group. This type of problem is perfect for the hypergeometric distribution because it allows us to calculate the probability without replacement from a finite population.
To calculate the probability that exactly 4 out of the first 6 refrigerators examined have a defective compressor, we use the hypergeometric probability formula:\( P(X=k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \).
For this problem, \( N = 12 \), \( K = 7 \), \( n = 6 \), and \( k = 4 \). So, \( P(X=4) \) is computed by evaluating the combinatorial choices for picking defective and non-defective items, which results in approximately 0.378. This formula shows:

• \( \binom{K}{k} \): Ways to choose defective items.
• \( \binom{N-K}{n-k} \): Ways to choose non-defective items.
• \( \binom{N}{n} \): Total possible ways to choose any items.

The expected value, often referred to as the mean, provides a measure of the center of a probability distribution. In the context of a hypergeometric distribution as in our refrigerator problem, the mean tells us on average how many defective refrigerators we might expect to find among the first 6 examined.
For the hypergeometric distribution, the expected value \( \mu \) is given by the formula:\[ \mu = n \cdot \frac{K}{N} \] where \( n \) is the sample size (refrigerators examined), \( K \) is the number of successes in the population (defective refrigerators), and \( N \) is the population size (total refrigerators).
Plugging in our values, we have: \( \mu = 6 \cdot \frac{7}{12} \approx 3.5 \). This means, on average, we would expect approximately 3.5 defective refrigerators in our sample of 6.

Standard deviation provides insight into the variability or spread of a distribution. A smaller standard deviation indicates that the data points are closer to the mean, while a larger standard deviation indicates more spread out data.
For a hypergeometric distribution, the standard deviation \( \sigma \) is calculated using the variance formula, and then taking its square root:\[ \sigma^2 = n \cdot \frac{K}{N} \cdot \frac{N-K}{N} \cdot \frac{N-n}{N-1} \] Substitute the problem's values to find:\[ \sigma \approx \sqrt{6 \cdot \frac{7}{12} \cdot \frac{5}{12} \cdot \frac{6}{11}} \approx 1.14 \] This calculation shows us how much variation we can expect in the number of defective refrigerators each time a sample is checked. It's used to understand how much a typical observation might deviate from the mean.

Sometimes dealing with hypergeometric calculations can be tedious, especially with large numbers. In such cases, we use the binomial approximation, which simplifies our calculations while still providing a fairly accurate result.
A hypergeometric distribution closely resembles a binomial distribution when the sample size is much smaller than the population. This is the case in part c of our problem, where we have a large shipment of 400 refrigerators.
Here, instead of calculating probabilities through the hypergeometric setup, we can use the binomial distribution with the formula:\[ X \sim \text{Binomial}(n, p) \] where \( n \) is the number of trials (selected refrigerators), and \( p \) represents the probability of success (probability of picking a defective refrigerator), calculated as:\( p = \frac{K}{N} = \frac{40}{400} = 0.1 \).
Using this approximation allows us to easily compute \( P(X \leq 5) \) and understand probabilities for large populations efficiently.