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The gamma distribution is a two-parameter family of continuous probability distributions. It is defined by a shape parameter \(\alpha\) and a rate parameter \(\beta\). The formula looks a little complicated, but its essence is in its flexibility in modeling distributions of various shapes. In the context of the exercise, the gamma distribution has \(\alpha = 2\) and \(\beta = 4\). This means that the distribution of our random sample with size 128 has:

• a mean (expected value) of \(\alpha \times \beta = 8\)
• a variance of \(\alpha \times \beta^2 = 32\)

The gamma distribution is often used to model waiting times or life durations. It can take on many shapes depending on the values of \(\alpha\) and \(\beta\). When \(\alpha\) is an integer, it reflects the time until the \(\alpha\)-th event in a series of events happening independently and continuously at the average rate \(\beta\). This connection to event timing makes the gamma distribution a versatile tool in statistical modeling. The Central Limit Theorem is applied here to approximate the problem's distribution as a normal distribution to find probability within a range.

The standard normal distribution is a special case of the normal distribution. It is a continuous probability distribution with a mean of 0 and a standard deviation of 1. To put it simply, it's the bell curve you might think of when picturing a normal distribution. In this exercise, the Central Limit Theorem allows the mean and variance from our gamma (non-normal) distribution to approximate a normal distribution due to the sample size of 128, which is quite large.
A standard normal distribution makes it easier to calculate probabilities because:

• it is symmetric around the center (mean value)
• and its total area (under the curve) sums to 1

This characteristic symmetrical shape means that the deviation from the mean can be easily calculated in terms of z-scores, a measure of how many standard deviations an element is from the mean. In this problem, transitioning from the gamma distribution to a normal distribution simplifies solving for probabilities in the 7 to 9 range of our dataset, making it much simpler to use standardized tables or computational tools.

Z-scores are a statistical measurement that describe a value's relation to the mean of a group of values. They help to determine where a data point lies in relation to the rest of the data set using the standard deviation. In formula terms, a z-score is calculated using:

• \(Z = \frac{x - \mu}{\sigma}\)

where \(x\) is your data point, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. Z-scores tell us how many standard deviations an element \(x\) is from the mean \(\mu\).
In the given exercise, when we wanted to find the probability that \(7 < X < 9\), we first translated these values into z-scores:

• For \(x = 7: Z_1 = -1\)
• For \(x = 9: Z_2 = 1\)

Understanding z-scores lets you utilize the standard normal distribution table easily. You can then find probabilities of different elements occurring within a certain range. This way, even tricky non-normal distributions like gamma can be analyzed efficiently when they are approximated to normal through standardization.