Problem 26 - 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. $$ P(24 \leq x \leq 30) $$

Step-by-Step Solution

Answer

The probability that \(24 \leq x \leq 30\) is approximately 0.2898.

Step 1: Understand the Problem

We need to find the probability that a normally distributed random variable \( x \) with a mean \( \mu = 22 \) and standard deviation \( \sigma = 5 \) takes on values between 24 and 30.

Step 2: Convert to Standard Normal Variable

We convert the values 24 and 30 to standard normal variables using the formula \( z = \frac{x - \mu}{\sigma} \). For 24, \( z = \frac{24 - 22}{5} = 0.4 \). For 30, \( z = \frac{30 - 22}{5} = 1.6 \).

Step 3: Use the Standard Normal Table

Using Table A, we find the probabilities for \( z = 0.4 \) and \( z = 1.6 \). The table gives \( P(Z \leq 0.4) \approx 0.6554 \) and \( P(Z \leq 1.6) \approx 0.9452 \).

Step 4: Find the Probability of Interest

To find \( P(24 \leq x \leq 30) \), we calculate \( P(0.4 \leq Z \leq 1.6) = P(Z \leq 1.6) - P(Z \leq 0.4) \). Substituting the probabilities: \( 0.9452 - 0.6554 = 0.2898 \).

Key Concepts

Continuous Random Variable

A continuous random variable is a type of random variable that can take on an infinite number of possible values. Unlike discrete random variables, which have specific points of values, continuous variables can represent any value within a given range. For example, variables like height, weight, and temperature are classified as continuous because they can be measured to any level of precision. In the context of a normal distribution, a continuous random variable like our variable \( x \), characterized by a mean \( \mu \) and a standard deviation \( \sigma \), can take on any value on the real number line, often in a specific interval. Here are some features of continuous random variables:

• They are described by a probability density function (PDF) instead of a probability mass function (PMF) used for discrete variables.
• The area under the curve of the PDF represents probabilities, with the total area totaling to 1.
• Probabilities for specific values are actually zero because there are infinitely many possible values; instead, we find probabilities over intervals.

Understanding this concept helps explain why we need to look at an interval, such as \( P(24 \leq x \leq 30) \), to determine probabilities for \( x \).

Standard Normal Distribution

The standard normal distribution is a special case of the normal distribution. It has a mean of 0 and a standard deviation of 1. In this system, every normal distribution, no matter its mean or standard deviation, can be transformed into a standard normal distribution using a process called standardization. This transformation, which relies on converting a random variable \( x \) to a \( z \)-score, helps us compare different normal distributions universally.To convert any normal distribution into a standard normal distribution, follow these steps:

• Subtract the mean \( \mu \) from the value \( x \).
• Divide the result by the standard deviation \( \sigma \).

This conversion results in a \( z \)-score, which tells us how many standard deviations away from the mean a particular value of \( x \) is. With standard normal distributions, we can use \( z \) tables, sometimes called standard normal tables, to find probabilities of different values or intervals directly, which simplifies the calculation process.

Z-Score

A \( z \)-score is a statistical measurement that describes a value's position relative to the mean of a group of values. In the context of a normal distribution, the \( z \)-score is crucial for standardizing different normal distributions to a common scale — the standard normal distribution.Here's how it works:

• A \( z \)-score is calculated using the formula: \( z = \frac{x - \mu}{\sigma} \).
• This formula indicates how many standard deviations \( x \) is from the mean \( \mu \). Positive \( z \)-scores indicate values above the mean, while negative \( z \)-scores indicate values below the mean.
• For example, in the original problem, converting 24 and 30 using \( z \)-score transformations gives us \( z = 0.4 \) and \( z = 1.6 \), respectively.

The \( z \)-score is essential for finding probabilities within a normal distribution because it allows us to use pre-calculated values in \( z \)-tables to find precise probabilities of occurrence between values. By standardizing, we eliminate the need to integrate the normal distribution function for each unique set of mean and standard deviation.