Problem 30 - Calculus for Biology and Medicine

Question

Suppose that \(X\) is normally distributed with mean \(-1\) and standard deviation 2. Find \(P(-3.5 \leq X \leq 0.5)\).

Step-by-Step Solution

Answer

The probability is approximately 0.6678.

Step 1: Understand the Problem

We need to find the probability that the normally distributed random variable \(X\) with a mean of \(-1\) and a standard deviation of 2 falls within the interval \([-3.5, 0.5]\).

Step 2: Standardize the Given Range

To use the standard normal distribution table, we first convert the interval \([-3.5, 0.5]\) into \(Z\)-scores using the formula: \[ Z = \frac{{X - \mu}}{\sigma} \] where \(\mu = -1\) and \(\sigma = 2\).

Step 3: Calculate the Z-scores

First, calculate the \(Z\)-score for \(X = -3.5\):\[ Z_1 = \frac{-3.5 - (-1)}{2} = \frac{-2.5}{2} = -1.25 \]Next, calculate the \(Z\)-score for \(X = 0.5\):\[ Z_2 = \frac{0.5 - (-1)}{2} = \frac{1.5}{2} = 0.75 \]

Step 4: Use the Standard Normal Distribution Table

Look up the probabilities corresponding to the \(Z\)-scores \(-1.25\) and \(0.75\) using a standard normal distribution table, or a calculator:\[ P(Z \leq -1.25) \approx 0.1056 \]\[ P(Z \leq 0.75) \approx 0.7734 \]

Step 5: Calculate the Probability Within the Interval

The probability that \(X\) falls within the range \([-3.5, 0.5]\) is found by calculating \(P(-1.25 \leq Z \leq 0.75)\):\[ P(-1.25 \leq Z \leq 0.75) = P(Z \leq 0.75) - P(Z \leq -1.25) \]\[ = 0.7734 - 0.1056 = 0.6678 \]

Key Concepts

Understanding Z-scores

Z-scores are a key concept when working with normal distributions. A Z-score helps to determine how far away a point is from the mean, measured in terms of standard deviations. This allows the conversion of different values to a common scale. In the formula for calculating Z-scores, we use:\[ Z = \frac{X - \mu}{\sigma} \]where: - \(X\) is the value from the data set, - \(\mu\) is the mean of the data set, - \(\sigma\) is the standard deviation. Converting values into Z-scores makes it easy to compare data points from different normal distributions or datasets. It's like using a ruler that measures in standard deviations instead of centimeters or inches. By doing this, you can use Z-scores to find probabilities and percentiles in a standard normal distribution table.

Understanding Standard Deviation

Standard deviation is a measure of how spread out numbers are in a data set. It indicates the typical distance between the data points and the mean of the data set. A larger standard deviation means that the data points are spread out over a wider range, while a smaller standard deviation means they are clustered closely around the mean.This is significant in the formula:\[ Z = \frac{X - \mu}{\sigma} \]because if the standard deviation is large, the Z-score will be smaller, as the same difference (\(X - \mu\)) will not be as significant. Conversely, a smaller standard deviation will make the Z-score larger for the same difference. Understanding standard deviation is crucial because it reflects the variability in your data and affects the calculation of Z-scores.

Probability Calculation Using Z-scores

Probability calculations often involve finding the likelihood of a value falling within a certain range in a normal distribution. Here, Z-scores play a crucial role. Once you convert your data points to Z-scores, you can use these scores to find probabilities using a standard normal distribution table.For example, in the problem we studied, we converted the range \([-3.5, 0.5]\) into Z-scores: - For \(X = -3.5\), the Z-score was \(-1.25\), - For \(X = 0.5\), it was \(0.75\). Using the standard normal distribution table, we found the probability \(P(Z \leq -1.25)\), which was approximately 0.1056, and \(P(Z \leq 0.75)\), which was approximately 0.7734.The probability that \(X\) falls within the interval \([-3.5, 0.5]\) is given by the difference:\[ P(-1.25 \leq Z \leq 0.75) = P(Z \leq 0.75) - P(Z \leq -1.25) = 0.7734 - 0.1056 = 0.6678 \]Therefore, there is a 66.78% chance that the value will lie between \(-3.5\) and \(0.5\). This method emphasizes the importance of Z-scores in probability calculations within normal distributions.