A variable, \(x\), has a normal distribution with mean 6 and standard deviation 1.5. Calculate the probability that (a) \(x<7.5\) (b) \(x<5\) (c) \(x>7\) (d) \(x>5.5\) (e) \(5

Understanding the mean and standard deviation is fundamental in statistics, as they are measures that describe the distribution of a data set. The mean, denoted by the symbol \( \mu \), is the average of all data points and serves as the center of the distribution. \( \mu \) represents where the bulk of the values lie. In this exercise, the mean is given as 6.

On the other hand, the standard deviation, symbolized by \( \sigma \), reflects the amount of variability or spread in a set of data. A low standard deviation indicates that the data points are close to the mean, whereas a high standard deviation suggests the data is spread out over a broader range of values. In our case, the standard deviation is 1.5, meaning that most of the data points are expected to fall within 1.5 units of the mean.

Together, these two statistical measures are used to describe the normal distribution of data, also known as a bell curve due to its shape. Recognizing that the mean and standard deviation define this curve helps us in predicting the probability of finding a particular value within this distribution, which is exactly what we do when solving exercises like the one given.

The z-score is a mathematical tool used to determine the position of a data point in relation to the mean of a group of values, measured in terms of standard deviations. It is calculated by taking the difference between a value and the mean, then dividing by the standard deviation. The formula is given by \( z = \frac{x - \mu}{\sigma} \).

In the exercise, calculating a z-score for a given value allows us to standardize different data points and thus compare them directly to the standard normal distribution. This distribution has a mean of 0 and a standard deviation of 1, which simplifies the calculation of probabilities.

Example:

For part (a) of the exercise, the z-score calculation for \( x=7.5 \) would be \( z_a = \frac{7.5 - 6}{1.5} = 1 \). This result tells us that 7.5 is one standard deviation above the mean. Z-scores are particularly helpful because they transform our normal distribution into a standard normal distribution, where we can easily find the relevant probabilities, as demonstrated in the subsequent steps of the solution.

Once we have the z-scores, we use the z-table to determine the probabilities associated with different values from our normal distribution. A z-table, also known as the standard normal table, represents the probability of a z-score falling within a particular interval from the mean in a standard normal distribution. The table provides the area to the left of a z-score, which translates to the probability of obtaining a value less than or equal to that particular z-score in a standard normal distribution.

For example, in part (a) of the exercise, the z-table tells us the area to the left of \( z_a = 1 \) is 0.8413, which means there's a 84.13% chance that a randomly chosen data point from our original normal distribution would be less than 7.5. When the z-score is negative, such as in part (b), we use the symmetry of the normal distribution to find the probability of \( x \) being less than the given value. Importantly, the total area under the normal curve is always 1 (or 100%), so to find the probability of \( x \) being greater than a value, we subtract the table value from 1.

For exercises that ask for the probability of \( x \) lying between two values, like in part (e), we simply calculate the z-scores for both values and subtract the smaller area from the larger one. This use of the z-table is a powerful tool in statistics, as it allows us to calculate the probability of any range of values within a normal distribution, once we have their corresponding z-scores.