91Ó°ÊÓ

In probability and statistics, a **random variable** is a numerical outcome of a random process. It is essential in describing how probabilities are distributed across different possible outcomes. In the context of our PVC pipe exercise, we are dealing with three random variables: the length of the first pipe piece (\(X\)), the length of the second pipe piece (\(Y\)), and the amount of overlap between them (\(Z\)).

Each of these random variables follows a normal distribution with a specified mean and variance:

• The first pipe: \(X \sim N(20, 0.5^2)\)
• The second pipe: \(Y \sim N(15, 0.4^2)\)
• The overlap: \(Z \sim N(1, 0.1^2)\)

Understanding these distributions helps in calculating how these random variables interact, particularly concerning their roles in forming the total length after the pipes are inserted into each other with some overlap.

Calculating the probability of a **total length** falling within a specific range requires a combination of addition and subtraction of our random variables. First, we derive the total length by expressing it as the function \( T = X + Y - Z \), where each of those terms is normally distributed.

The probability distribution of \( T \) has a mean \( E(T)\) calculated as:\[ E(T) = E(X) + E(Y) - E(Z) = 20 + 15 - 1 = 34 \]
The variance \( Var(T) \) is derived from the individual variances:\[ Var(T) = Var(X) + Var(Y) + Var(Z) = 0.5^2 + 0.4^2 + 0.1^2 = 0.42 \]
Once we have \( N(34, 0.42) \) for the total length, the next task is standardizing this variable to effectively use tables or calculators available for probability estimation.
The goal is to find the probability that \( T \) lies between \( 34.5 \) and \( 35 \) by transforming those limits into the standard normal distribution.

The **standard normal distribution** is a useful statistical tool with a mean of 0 and a standard deviation of 1. To utilize it for our probability range of the total length, we transform (or "standardize") the interval \([34.5, 35]\) for \( T \sim N(34, 0.42)\) into a standard normal variable \( Z \).

The transformation is done by calculating:

• \( z_1 = \frac{34.5 - 34}{\sqrt{0.42}} \approx 0.77 \)
• \( z_2 = \frac{35 - 34}{\sqrt{0.42}} \approx 1.54 \)

These \( z \)-scores indicate how many standard deviations away from the mean 34 the values 34.5 and 35 are.

Finally, we calculate the probability that the standardized variable falls between these \( z \)-scores using tables or calculators. For our case:\[ P(0.77 < Z < 1.54) = P(Z < 1.54) - P(Z < 0.77) \]Finding these values from the standard normal distribution table gives:

• \( P(Z < 1.54) \approx 0.9382 \)
• \( P(Z < 0.77) \approx 0.7794 \)

Thus, the probability that the total length is between 34.5 and 35 is \( 0.9382 - 0.7794 = 0.1588 \). This transformation and calculation are crucial steps in interpreting normally distributed random variables in practical scenarios.