Problem 27 - Calculus for Biology and Medicine

Question

Suppose the weight of an animal is normally distributed with mean \(3720 \mathrm{~g}\) and standard deviation \(527 \mathrm{~g}\). What percentage of the population has a weight that exceeds \(5000 \mathrm{~g}\) ?

Step-by-Step Solution

Answer

Approximately 0.75% of the population has a weight exceeding 5000 grams.

Step 1: Understand the Problem

The problem is asking for the percentage of animals that have a weight greater than 5000 grams, given a normal distribution with a mean (\(\mu\)) of 3720 grams and a standard deviation (\(\sigma\)) of 527 grams.

Step 2: Transform to a Standard Normal Variable

Convert the weight of 5000 grams to a standard normal variable using the z-score formula: \(z = \frac{x - \mu}{\sigma}\). Here, \(x = 5000\), \(\mu = 3720\), and \(\sigma = 527\).

Step 3: Calculate the Z-Score

Substitute the values into the formula: \[z = \frac{5000 - 3720}{527} = \frac{1280}{527} \approx 2.43\]. The z-score is approximately 2.43.

Step 4: Find the Probability from the Z-Table

Look up the value of the z-score 2.43 in the standard normal distribution table (z-table). This gives the probability that a value is less than 5000 grams.

Step 5: Calculate the Complement Probability

The z-table gives the probability that the weight is less than 5000 grams. Since we need the probability of exceeding 5000 grams, subtract the z-table value from 1. For \(z = 2.43\), the table shows approximately 0.9925. Thus, \(P(X > 5000) = 1 - 0.9925 = 0.0075\).

Step 6: Convert Probability to Percentage

Convert the probability to a percentage by multiplying by 100. Thus, \(0.0075 \times 100 = 0.75\).

Key Concepts

Understanding Z-Score

In statistics, a z-score, also known as a standard score, is a key concept when working with normal distributions. It measures how many standard deviations a particular data point is away from the mean of the dataset.

To calculate a z-score, you use the formula:

• \( z = \frac{x - \mu}{\sigma} \)

Here:

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

A positive z-score indicates that the data point is above the mean, while a negative z-score indicates it's below the mean. A z-score close to 0 suggests that the data point is near the mean.

In our example, we calculated the z-score to find how much the weight of 5000 grams deviates from the average weight of 3720 grams, given a standard deviation of 527 grams. This distance and direction help us understand the relative position of 5000 grams within the normal distribution.

The Role of Standard Deviation

Standard deviation is a vital statistical measure that quantifies the amount of variation or dispersion in a dataset. It shows how spread out the values are around the mean. A small standard deviation means the data points are close to the mean, while a large standard deviation signifies that the data points are spread out over a broader range of values.

Standard deviation is calculated using the formula:

• \( \sigma = \sqrt{\frac{\sum (x_i - \mu)^2}{N}} \)

Where:

• \( x_i \) represents every data point in the dataset
• \( \mu \) is the mean of the data points
• \( N \) is the total number of data points

In our problem, the given standard deviation of 527 grams tells us how much the animal weights vary around the average weight of 3720 grams. This information is crucial to understand how typical or atypical the weight of 5000 grams is within this distribution.

Probability Calculation with the Z-Score

With a calculated z-score, we can determine the probability that a data point falls within a specific range of a normal distribution by using standard normal distribution tables (commonly called z-tables). These tables provide the probability that a standard normal random variable is less than a given z-score value.

In our exercise, the z-score for 5000 grams was calculated to be approximately 2.43. Using a z-table, we find the probability associated with this z-score, which indicates the likelihood of a value being less than 5000 grams in our distribution. For a z-score of 2.43, the probability is around 0.9925, meaning there's a 99.25% chance a weight will be below 5000 grams.

This step is essential for assessing how rare or common a specific weight is when compared to the normal distribution of weights.

Exploring Complement Probability

Complement probability is a concept used to determine the likelihood of the opposite of an event occurring. In statistical terms, if there's a probability \( P(A) \) of an event occurring, then there's \( 1 - P(A) \) probability of the event not occurring.

In the given exercise, we looked at the probability of the weights being less than 5000 grams. However, the problem asked us to find the probability of weights exceeding 5000 grams. Here, we applied complement probability:

• \( P(X > 5000) = 1 - P(X < 5000) \)
• \( P(X > 5000) = 1 - 0.9925 = 0.0075 \)

This calculation shows that only 0.75% of the population is expected to have weights greater than 5000 grams. Complement probability is a simple yet powerful tool to understand possibilities outside the initially considered range.