Problem 29 - Calculus for Biology and Medicine

Question

Suppose that \(X\) is normally distributed with mean 2 and standard deviation \(1 .\) Find \(P(0 \leq X \leq 3)\).

Step-by-Step Solution

Answer

The probability that \(0 \leq X \leq 3\) is approximately 0.8185.

Step 1: Identify the Parameters

The problem states that the random variable \( X \) is normally distributed with mean \( \mu = 2 \) and standard deviation \( \sigma = 1 \). This can be written as \( X \sim N(2, 1^2) \).

Step 2: Standardize the Variable

To find probabilities for a normal distribution, we first standardize the variable. This involves converting \( X \) into a standard normal variable \( Z \) using the formula: \[ Z = \frac{X - \mu}{\sigma} \]For the lower limit \( X = 0 \):\[ Z_0 = \frac{0 - 2}{1} = -2 \]For the upper limit \( X = 3 \):\[ Z_3 = \frac{3 - 2}{1} = 1 \]

Step 3: Find Standard Normal Probabilities

To find \( P(0 \leq X \leq 3) \), we need \( P(-2 \leq Z \leq 1) \).Using the standard normal distribution table or calculator, find:- \( P(Z \leq 1) \): approximately 0.8413- \( P(Z \leq -2) \): approximately 0.0228Then calculate the probability as \( P(-2 \leq Z \leq 1) = P(Z \leq 1) - P(Z \leq -2) \).

Step 4: Calculate the Probability

Using the probabilities found:\[ P(-2 \leq Z \leq 1) = 0.8413 - 0.0228 = 0.8185 \]

Step 5: State the Result

The probability that \( X \) takes a value between 0 and 3 is \( 0.8185 \).

Key Concepts

Standard Deviation

In statistics, the standard deviation is a measure that indicates the amount of variation or dispersion in a set of values. It tells us how much the values in a data set deviate from the mean, giving us insight into the spread of data. When you're dealing with a normal distribution, the standard deviation plays a vital role in determining the shape and spread of the curve.

The larger the standard deviation, the more spread out the data points are around the mean, resulting in a flatter, wider curve. Conversely, a smaller standard deviation indicates that the data points are closer to the mean, leading to a steeper, narrower curve.

• The formula for standard deviation is: \[ \sigma = \sqrt{\frac{1}{N} \sum_{i=1}^{N} (x_i - \mu)^2} \]
• Here, \( \mu \) is the mean, \( N \) is the number of observations, and \( x_i \) represents each individual observation.

In our exercise, the standard deviation is 1, which simplifies calculations and directly influences the transformation of the normal random variable into the standard normal variable.

Standard Normal Variable

The standard normal variable is a special type of random variable that has a mean of 0 and a standard deviation of 1. This transformation makes it easier to calculate probabilities because we can use a standard normal distribution table or calculator. The process of converting a normal variable into a standard normal variable is called standardization.

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

• \( X \) is the original normal variable.
• \( \mu \) is the mean of \( X \).
• \( \sigma \) is the standard deviation of \( X \).

For example, in our exercise:

• For \( X = 0 \): \( Z_0 = \frac{0 - 2}{1} = -2 \)
• For \( X = 3 \): \( Z_3 = \frac{3 - 2}{1} = 1 \)

These transformations allow us to find probabilities using the standard normal distribution. This is essential, as all tables or calculators for normal distributions are based on this standardized form.

Probability Calculation

Calculating probability in a normal distribution involves determining the likelihood that a random variable falls within a certain range. Once we have a standard normal variable, we can compute these probabilities more straightforwardly. For the given problem, after standardization, we need to find the probability that the standardized variable \( Z \) falls between -2 and 1.

• Find \( P(Z \leq 1) \), which refers to the cumulative probability up to \( Z = 1 \). This value is approximately 0.8413.
• Find \( P(Z \leq -2) \), which is the cumulative probability up to \( Z = -2 \). This value is approximately 0.0228.

To find \( P(-2 \leq Z \leq 1) \), subtract \( P(Z \leq -2) \) from \( P(Z \leq 1) \): \[ P(-2 \leq Z \leq 1) = P(Z \leq 1) - P(Z \leq -2) \]This calculation gives: \[ 0.8413 - 0.0228 = 0.8185 \]This result means there is an 81.85% probability that \( X \) will be within the range of 0 to 3. Understanding this process involves using a standard normal distribution table or calculator to determine these cumulative probabilities efficiently.