Problem 16 - Calculus and its applications

Question

Let \(x\) be a continuous random variable with a standard normal distribution. Using Table A, find each of the following. $$ P(-1.89 \leq x \leq 0.45) $$

Step-by-Step Solution

Answer

The probability \(P(-1.89 \leq x \leq 0.45)\) is approximately 0.6442.

Step 1: Understand the Standard Normal Distribution

A standard normal distribution is a special normal distribution where the mean is 0 and the standard deviation is 1. It is often represented by the variable \(x\) and is denoted by \(Z\). In this problem, we need to find the probability \(P(-1.89 \leq Z \leq 0.45)\).

Step 2: Use the Z-table for Lower Bound

To find \(P(-1.89 \leq Z \leq 0.45)\), we first find \(P(Z \leq -1.89)\) using the Z-table. According to the table, \(P(Z \leq -1.89)\) is approximately 0.0294.

Step 3: Use the Z-table for Upper Bound

Next, we find \(P(Z \leq 0.45)\) using the Z-table. The table shows that \(P(Z \leq 0.45)\) is approximately 0.6736.

Step 4: Calculate the Probability for the Range

Finally, we calculate \(P(-1.89 \leq Z \leq 0.45)\) by subtracting the probability of the lower bound from the probability of the upper bound: \(P(-1.89 \leq Z \leq 0.45) = P(Z \leq 0.45) - P(Z \leq -1.89) = 0.6736 - 0.0294 = 0.6442\).

Key Concepts

Z-table

A Z-table is an essential tool in statistics, especially when working with the standard normal distribution. It provides the probabilities that a standard normal random variable is less than or equal to a particular value of Z. In simpler terms, it helps determine the area under the curve to the left of a given Z-score.
In our exercise, we needed to find the probability that Z is between -1.89 and 0.45. To do this, we first used the Z-table to find the cumulative probability associated with the Z-score of -1.89, which was approximately 0.0294. We then did the same for the Z-score of 0.45, obtaining a cumulative probability of about 0.6736.
By using the Z-table, we could efficiently calculate these probabilities without complicated computations. It translates the Z-scores into meaningful probabilities through simple look-ups, allowing us to understand the likelihood of certain outcomes for a standard normal distribution.

Continuous Random Variable

Continuous random variables are variables that can take on an infinite number of values within a given range. Unlike discrete random variables, which have separate, distinct values, continuous random variables have outcomes that are not isolated but instead form a continuum.
The value of a continuous random variable is often represented by an interval within which the variable can lie. In the context of our exercise, the variable \( x \) is a continuous random variable with a standard normal distribution. This means \( x \) can take any real value between negative infinity and positive infinity, but it most often falls between -3 and 3, where the bulk of the distribution's probability is concentrated.
Understanding continuous random variables is crucial as it assists in modeling and interpreting real-life phenomena, such as heights, weights, and other metrics that can vary smoothly across a range.

Probability Calculation

Probability calculation involves determining the likelihood of a particular event occurring. In the context of the standard normal distribution, this often means finding the probability that a random variable falls within a certain range.
For the exercise given, to determine \( P(-1.89 \leq x \leq 0.45) \), we first found individual cumulative probabilities from the Z-table, specifically \( P(Z \leq -1.89) \) and \( P(Z \leq 0.45) \).
The final probability calculation was simple: subtract the smaller cumulative probability from the larger one. This calculation represented the probability that \( x \), our random variable, would fall between -1.89 and 0.45, yielding a result of 0.6442.
Probability calculations like this one give insight into the expected frequency of events, which is why they are a cornerstone in fields such as statistics, economics, and engineering.