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Normal distribution is a fundamental concept in statistics, describing how a set of values are spread around a central mean. It resembles a bell-shaped curve which is symmetrical, meaning the left and right sides of the curve are mirror images of each other. This symmetry indicates that values further from the mean are less frequent, relative to the values close to the mean.

The normal distribution is defined by two parameters:

• Mean (\( \mu \)): This is the average value of the distribution. In our exercise, the mean tire life is 57,500 miles.
• Standard Deviation (\( \sigma \)): This shows how much variation or dispersion there is from the mean. A smaller standard deviation means values are closely clustered around the mean. In this example, the standard deviation is 950 miles.

The probability of an event within a normal distribution sets the foundation for calculating other statistical concepts, like Z-scores.

A Z-score is a way of describing a data point's position relative to the mean of a group of values. It tells us how many standard deviations a data point is from the mean. The formula to calculate a Z-score is:\[ Z = \frac{x - \mu}{\sigma} \]where \( x \)is the value in question, \( \mu \)is the mean, and \( \sigma \)is the standard deviation. The Z-score provides a way to understand where a value lies on a normal distribution by transforming any normal distribution into the standard normal distribution.

For example, in finding the Z-scores of tire lives between 57,000 and 58,000 miles:

• For 57,000 miles: \( Z_1 = \frac{57,000 - 57,500}{950} = -0.5263 \)
• For 58,000 miles: \( Z_2 = \frac{58,000 - 57,500}{950} = 0.5263 \)

These Z-scores help in estimating probabilities using a Z-table.

Standard deviation is crucial in understanding how data is spread out from the mean in a dataset. It measures the average distance of each data point from the mean, providing insights into the variability of the data. In the context of normal distribution, a smaller standard deviation means the data points are closer to the mean. Why is standard deviation important?

• Predictability: In applications like tire life, knowing the standard deviation helps predict outcomes more accurately.
• Comparison: It enables the comparison of variability across different datasets.

In the exercise given, the standard deviation is 950 miles. This reflects how much the tire life data is expected to vary from the average of 57,500 miles. Calculating the standard deviation is key to determining the usual range that most tire lifespans will fall within.

The concept of independence of events implies that the occurrence of one event does not affect the probability of another. In probability, two events A and B are independent if:\[ P(A \cap B) = P(A) \times P(B) \]This formula shows that the probability of both events occurring together is the product of their individual probabilities. In our exercise, when considering the lifespan of four tires, each tire's lifespan does not influence the others.

Therefore, the probability that all four tires last between certain mileages is calculated by multiplying the probability for one tire four times:

• Probability of one tire lasting between 57,000 and 58,000 miles: \( 0.4008 \)
• Probability for all four tires: \( (0.4008)^4 \approx 0.0258 \)

This independence is critical when assessing products where the failure of one does not impact the reliability of others in a group.