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When you're dealing with binomial distributions, like the probability of U.S. households having at least one credit card, using normal approximation can be a lifesaver. Calculating probabilities for each possible outcome in large samples can be tedious. Thankfully, with a sufficiently large sample size, we can approximate the binomial distribution to a normal distribution. This is known as the normal approximation to the binomial distribution.

This works because, as the sample size increases, the shape of the binomial distribution starts resembling the bell-shaped curve of a normal distribution. A rule of thumb is if the sample size is large enough such that both np and n(1-p) are greater than 5, a normal approximation can be applied. In our problem, the sample size is 500, and we're trying to predict the behavior between 375 to 385 successes, which fits the criteria.

However, when converting to a normal distribution, remember we need to adjust for something called continuity correction. This involves adjusting the discrete nature of the binomial into the continuous nature of the normal by subtracting 0.5 from your desired lower bound and adding 0.5 to your upper bound.

The process of probability calculation in statistics often involves identifying the desired range of outcomes, and then utilizing statistical techniques to find the likelihood of these events. Let's break this down for our given scenario.

In the exercise, you need to find the probability that between 375 and 385 households out of a sample of 500 have at least one credit card. First, you determine whether it's feasible to use normal approximation, which it is in this case due to the sample size and proportion criteria. Your next step is to express the lower bound (375) and upper bound (385) as part of the normally distributed model.

With the inclusion of the continuity correction, you adjust these limits to 374.5 and 385.5. The probabilities can then be calculated by finding the corresponding areas under the normal curve between these adjusted bounds using Z-scores.

In probability and statistics, standard deviation measures the amount of variation or dispersion of a set of values. It tells us how much the values in a data set are spread out around the mean.

For a binomial distribution with probability of success p, and number of trials n, the formula to determine the standard deviation is \( \sigma = \sqrt{np(1 - p)} \). In this problem, the probability of having at least one credit card (p) is 0.78 and the sample size (n) is 500. Therefore, the standard deviation can be calculated by plugging these values into the formula, resulting in a numerical value that helps us understand the spread of potential values around the mean (expected value).

By determining the standard deviation, we can understand the typical deviation of household counts with at least one credit card from the expected mean, which is crucial for applying a normal approximation and calculating probabilities accurately.

Z-scores are a powerful tool in statistics that represent the number of standard deviations a data point is from the mean. They allow us to determine how unusual or typical a particular outcome is within a normal distribution.

In our problem, once we have the mean \( \mu \) and standard deviation \( \sigma \), we convert the bounds (375 and 385) with the continuity correction to Z-scores using the formula: \[ Z = \frac{x - \mu}{\sigma} \]where \(x\) is the value for which you are finding the Z-score. This conversion is important because it transforms our initial problem from a binomial context to using the properties of the normal distribution.

After finding the Z-scores for 374.5 and 385.5, we can refer to a standard normal distribution table, or Z-table, to find the probabilities that correspond to these scores. The final step is to determine the probability of falling between these Z-scores by subtracting the smaller probability from the larger one, giving us the desired probability of our event.