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 0.77% of the population has a weight exceeding 5000 g.

Step 1: Identify the Normal Distribution

The problem states that the animal's weight is normally distributed with a mean (\(\mu\)) of \(3720\, \mathrm{g}\) and a standard deviation (\(\sigma\)) of \(527\, \mathrm{g}\).

Step 2: Calculate the Z-score

To find the percentage of the population exceeding \(5000\, \mathrm{g}\), we need to calculate the Z-score using the formula \[ Z = \frac{X - \mu}{\sigma} \] where \(X\) is the given weight threshold of \(5000\, \mathrm{g}\). Plugging in the values: \[ Z = \frac{5000 - 3720}{527} \] \(\Rightarrow Z \approx 2.428\).

Step 3: Find the Probability from the Z-table

Using the Z-score of \(2.428\), find the corresponding probability from a standard normal distribution table. The Z-table gives the probability that a value is less than \(Z\). For \(Z=2.428\), this probability is approximately \(0.9923\).

Step 4: Calculate Percentage Greater Than the Threshold

Since we are interested in the percentage of the population with weights exceeding \(5000\, \mathrm{g}\), subtract the Z-table probability from \(1\): \[ 1 - 0.9923 = 0.0077 \] Convert this to a percentage by multiplying by \(100\): \[ 0.0077 \times 100 = 0.77\% \] Therefore, approximately \(0.77\%\) of the population exceeds \(5000\, \mathrm{g}\).

Key Concepts

Understanding the Mean

In any normal distribution, the mean represents the average value around which data points are spread. It is a measure of central tendency. In our exercise, the weight of the animal has a mean of \(3720 \, \mathrm{g}\). This means if we measured the weight of all animals in this population, the average weight would be \(3720 \, \mathrm{g}\). Understanding the mean is crucial because it gives us a point of reference to understand how much individual data points (like the weight of the animals) deviate or fluctuate from the average. In real-world applications, such as animal weight, having a clear understanding of the mean allows us to anticipate what a typical example should be.

Exploring Standard Deviation

Standard deviation is a measure that tells us how much the values in a distribution vary or "deviate" from the mean. In simpler terms, it provides an insight into the spread of the data. A smaller standard deviation indicates that the data points tend to be closer to the mean, whereas a larger standard deviation suggests more spread-out data.For the weights of our animals, the standard deviation is \(527 \, \mathrm{g}\). This standard deviation helps us determine the variability in the weight. It is essential in understanding how spread out the weights are around the mean weight of \(3720\, \mathrm{g}\). Knowing the standard deviation is indispensable in fields like zoology and other sciences, as it can indicate diversity or uniformity within a population.

Calculating the Z-score

The Z-score is a powerful tool in statistics that tells us how many standard deviations an element is from the mean. For any given value in a data set, you can calculate the Z-score using the formula: \[ Z = \frac{X - \mu}{\sigma} \]where:

• \(X\) is the value in question,
• \(\mu\) is the mean of the distribution,
• \(\sigma\) is the standard deviation.

In our exercise, we've calculated the Z-score for a weight of \(5000 \, \mathrm{g}\). The result, approximately \(2.428\), tells us that a weight of \(5000 \, \mathrm{g}\) is \(2.428\) standard deviations above the mean. Calculating a Z-score is invaluable in comparing different data points or assessing the rarity of an observation within a normal distribution.

Unraveling Probability

Probability in the context of a normal distribution quantifies the likelihood of an event occurring within a given range. After calculating the Z-score, we refer to the standard normal distribution table, often called the Z-table, to find the probability that a value is less than our calculated Z-score.In our problem, the Z-score of \(2.428\) corresponds to a probability of approximately \(0.9923\). This figure represents the probability of selecting an animal at random that weighs less than \(5000\, \mathrm{g}\). To find the percentage exceeding this weight, we subtract this probability from \(1\), resulting in \(0.0077\). Converting this to a percentage gives us about \(0.77\%\). Thus, only \(0.77\%\) of the animal population weighs more than \(5000\, \mathrm{g}\). Understanding probability is critical in statistics for making predictions and informed decisions based on data.