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 that \(-3.5 \leq X \leq 0.5\) is approximately 0.6678.

Step 1: Identify the normal distribution parameters

The problem states that the random variable \(X\) follows a normal distribution with a mean \(\mu = -1\) and a standard deviation \(\sigma = 2\). Since the normal distribution is fully described by these two parameters, we denote this distribution as \(X \sim N(-1, 2^2)\).

Step 2: Standardize the variable

To find probabilities from a normal distribution, we convert the variable \(X\) to a standard normal variable \(Z\) which has a mean of 0 and a standard deviation of 1. The standardization formula is \(Z = \frac{X - \mu}{\sigma}\). We will compute \(Z\) for the boundary values \(-3.5\) and \(0.5\).

Step 3: Calculate the standardized value for -3.5

Apply the standardization formula: \[ Z_1 = \frac{-3.5 - (-1)}{2} = \frac{-3.5 + 1}{2} = \frac{-2.5}{2} = -1.25 \].

Step 4: Calculate the standardized value for 0.5

Similarly, apply the standardization formula for the upper boundary: \[ Z_2 = \frac{0.5 - (-1)}{2} = \frac{0.5 + 1}{2} = \frac{1.5}{2} = 0.75 \].

Step 5: Use the standard normal distribution table

Now that we have \(Z_1 = -1.25\) and \(Z_2 = 0.75\), use the standard normal distribution (Z) table to find the probabilities: \(P(Z < -1.25)\) and \(P(Z < 0.75)\). From the Z-table, \(P(Z < -1.25) \approx 0.1056\) and \(P(Z < 0.75) \approx 0.7734\).

Step 6: Calculate the probability

To find \(P(-3.5 \leq X \leq 0.5)\), we need \(P(Z_1 < Z < Z_2)\). This is equivalent to \(P(Z < Z_2) - P(Z < Z_1)\). Calculate: \[ P(-3.5 \leq X \leq 0.5) = P(Z < 0.75) - P(Z < -1.25) = 0.7734 - 0.1056 = 0.6678 \].

Key Concepts

Standardization

Standardization is an essential process when dealing with normal distributions. It allows us to transform any normal distribution into a standard normal distribution, making it easier to calculate probabilities.
Imagine you have a normal distribution with a mean (the average) and a standard deviation (which describes how spread out the distribution is). Usually, your data won't align perfectly with the standard normal distribution, which has a mean of 0 and a standard deviation of 1.

To standardize, we use the formula:\[Z = \frac{X - \mu}{\sigma}\]where:

• \(X\) is a value from your original distribution
• \(\mu\) is the mean of your original distribution
• \(\sigma\) is the standard deviation of your original distribution

This process converts the original data point to a Z-score, which tells us how many standard deviations the original value is from the mean.
For example, if we standardize the boundary value of our exercise, we find \(Z = -1.25\) for \(-3.5\) and \(Z = 0.75\) for \(0.5\). These Z-scores are now comparable within the standard normal distribution.

Probability Calculation

Once we've standardized our values, we can proceed to calculate probabilities using the Z-scores. Probability calculation involves determining the likelihood of a random variable falling within a particular range.
In this context, once we have our Z-scores, the next step is to find the probability that our variable falls between them. This is done using a standard normal distribution table, which provides the probability that a standard normal variable will be less than a given Z-score.

So, to find the probability that \(-3.5 \leq X \leq 0.5\), we first check the probability for the upper Z-score \(Z = 0.75\), and the lower Z-score \(Z = -1.25\):

• \(P(Z < 0.75) = 0.7734\)
• \(P(Z < -1.25) = 0.1056\)

Then, we subtract the probability of the lower bound from the upper bound:\[P(-3.5 \leq X \leq 0.5) = 0.7734 - 0.1056 = 0.6678\]This means the probability that the variable falls between \(-3.5\) and \(0.5\) is about 66.78%.

Standard Normal Distribution Table

The standard normal distribution table, or Z-table, is a valuable tool in statistics. It tells us the probabilities associated with Z-scores in a standard normal distribution.
Standard normal distribution, also known as the Z-distribution, has a bell-shaped curve centered at zero. It allows statisticians to find the probability of a variable being less than a particular Z-score.

Using the Z-table:

• Locate the Z-score on the table
  
• Find the corresponding probability
  

The leftmost column and the top row of a Z-table indicate the Z-score, while the cell where they intersect shows the probability.
You'd use a Z-table when you've standardized your data to determine how likely a given outcome is. This helps in various statistical assessments, including hypothesis testing, confidence intervals, and comparing different data sets.
For our example, referring to the Z-table allowed us to find that \(P(Z < 0.75)\) is approximately 0.7734 and \(P(Z < -1.25)\) is approximately 0.1056, leading to a final probability of about 66.78% for the original problem.