91Ó°ÊÓ

The concept of the z-score is crucial for understanding normal distribution and statistical analysis. A z-score helps us determine how far a particular point (in this case, temperature) is from the mean, measured in standard deviations. The formula to calculate a z-score is: \[ z = \frac{X - \mu}{\sigma} \]where:

• \( X \) is the value of interest (e.g., a daily temperature).
• \( \mu \) is the mean of the distribution (77°F for our problem).
• \( \sigma \) is the standard deviation (5°F).

For example, if you want to calculate the z-score for an 83°F day, you would plug the numbers into the formula: \[ z = \frac{83 - 77}{5} = 1.2 \]Thus, a temperature of 83°F is 1.2 standard deviations above the mean.

Understanding percentiles is key to interpreting data distributions. A percentile indicates the value below which a given percentage of observations fall.In the context of our LA temperature problem, we wanted to know the temperature below which the coldest 10% of days fall, known as the 10th percentile.To find a specific percentile, we often use a z-score table (standard normal table). For the coldest 10%, we find the z-score that corresponds to a cumulative probability of 0.10. For this problem, the z-score for the 10th percentile is approximately -1.28.Now, use the z-score formula in reverse to find the temperature:\[ X = z \cdot \sigma + \mu \]Substituting the known values, it becomes:\[ X = (-1.28) \cdot 5 + 77 = 70.6 \]Therefore, the coldest 10% of days have temperatures below 70.6°F.

Probability in the context of a normal distribution allows us to determine the likelihood of certain events occurring.In our problem, we were interested in the probability of a temperature being 83°F or higher. With the z-score of 1.2 calculated already, we use a standard normal distribution table to find that the probability of being less than 1.2 standard deviations from the mean is approximately 0.8849.To find the probability of the temperature being 83°F or higher, we subtract this value from 1, representing 100% of the area under the normal curve:\[ P(Z > 1.2) = 1 - 0.8849 = 0.1151 \]So, there's an 11.51% chance of observing a day with a temperature of 83°F or higher.

Standard deviation is a measure of how spread out the numbers in a data set are. In the context of a normal distribution, it's a tool for quantifying the amount of variation or dispersion of a set of values.In our LA temperature example, the mean temperature is 77°F and the standard deviation is 5°F. This means most June days will have temperatures close to the average, with the majority lying within one or two standard deviations of the mean. Standard deviation helps us to understand the distribution:

• About 68% of data falls within 1 standard deviation (\(\pm 5^{\circ} F\)) from the mean (72°F to 82°F).
• About 95% falls within 2 standard deviations (67°F to 87°F).
• Almost all data (99.7%) is within 3 standard deviations.

Thus, a high standard deviation means more temperature variability, while a low standard deviation indicates the temperatures are closely clustered around the mean.