Problem 28 - Calculus and its applications

Question

Let \(x\) be a continuous random variable that is normally distributed with mean \(\mu=22\) and standard deviation \(\sigma=5 .\) Using Table A, find the following. $$ P(19 \leq x \leq 25) $$

Step-by-Step Solution

Answer

The probability \( P(19 \leq x \leq 25) \) is approximately 0.4514.

Step 1: Standardize the Normal Distribution

To find the probability for a specific range in a normal distribution, we need to convert the variable to a standard normal distribution using the z-score formula. The z-score is given by \( z = \frac{x - \mu}{\sigma} \). Here, \( x \) is the value of the variable, \( \mu = 22 \), and \( \sigma = 5 \).

Step 2: Calculate Z-Scores for Boundaries

Compute the z-scores for the boundaries 19 and 25. - For \( x = 19 \): \[ z = \frac{19 - 22}{5} = \frac{-3}{5} = -0.6 \] - For \( x = 25 \): \[ z = \frac{25 - 22}{5} = \frac{3}{5} = 0.6 \]

Step 3: Use Z-Table to Find Probabilities

Using a standard normal distribution table (Table A), find the probabilities corresponding to the z-scores.- For \( z = -0.6 \), the probability \( P(Z < -0.6) \) is approximately 0.2743.- For \( z = 0.6 \), the probability \( P(Z < 0.6) \) is approximately 0.7257.

Step 4: Calculate the Desired Probability

The probability that \( x \) falls between 19 and 25 is the difference between the probabilities found:\[ P(19 \leq x \leq 25) = P(Z < 0.6) - P(Z < -0.6) = 0.7257 - 0.2743 = 0.4514 \]

Key Concepts

Understanding the Z-Score

The concept of a z-score is central when working with normal distribution problems. A z-score represents how many standard deviations a data point is from the mean. It offers a way to standardize scores on different scales into a common frame of reference, allowing you to compare different data points. This is done by using the formula: \[ z = \frac{x - \mu}{\sigma} \] Where:

• \( x \) is the data point in question.
• \( \mu \) is the mean of the distribution.
• \( \sigma \) is the standard deviation of the distribution.

For instance, if a score is exactly on the mean, its z-score is 0. If a score is one standard deviation above or below the mean, it gets a z-score of 1 or -1, respectively. This transformation to z-scores is crucial for comparing scores within any normal distribution and is the basis for using the standard normal distribution table, commonly known as the z-table.

Exploring the Standard Normal Distribution

Once you calculate z-scores, they relate directly to the standard normal distribution, which is a special case of a normal distribution. The standard normal distribution has a mean of 0 and a standard deviation of 1. By converting any normal distribution into z-scores, the distribution becomes a standard normal distribution. Thus, you can use a standard normal distribution table to find probabilities. This table lists the probabilities that a standard normal distributed variable will be less than a given z-score. Every point on the z-table represents the area under the curve to the left of the z-score. This property allows you to determine how likely a value is to occur within a standard normal distribution.

Performing Probability Calculations

Probability calculations in a normal distribution context may seem complex, but breaking them down into steps simplifies the process. Once you have determined the z-scores from your desired range, you use the z-table to find the probabilities. First, find the probability that the random variable is less than the higher boundary using the z-score. Next, do the same for the lower boundary. To find the probability that the variable lies between the two boundaries, subtract the probability of the lower boundary from the probability of the higher boundary:\[ P(a \leq x \leq b) = P(Z < b) - P(Z < a) \] In our example, we calculated that \( P(19 \leq x \leq 25) \) is \( 0.4514 \). This calculation tells us that there is a 45.14% chance that a value of \( x \) drawn from this normal distribution will lie between 19 and 25.