Problem 3 - Calculus for Biology and Medicine

Question

The following data represent the age of patients in a clinical trial: $$ 28,45,34,36,30,42,35,45,38,27 $$ Find the median, the sample mean, and the sample variance.

Step-by-Step Solution

Answer

Median is 35.5, sample mean is 36, and sample variance is approximately 43.11.

Step 1: Sort the data

First, arrange the given ages in ascending order: 27, 28, 30, 34, 35, 36, 38, 42, 45, 45.

Step 2: Find the Median

The median is the middle value of a data set. If the number of observations is even, the median is the average of the two middle numbers. Since our data set has 10 numbers, which is even, the median is calculated by averaging the 5th and 6th values: \( \frac{35 + 36}{2} = 35.5 \).

Step 3: Calculate the Sample Mean

The sample mean is calculated by summing all the values and dividing by the count of the values. Sum: \( 27 + 28 + 30 + 34 + 35 + 36 + 38 + 42 + 45 + 45 = 360 \). Count: 10. Sample mean \( \bar{x} = \frac{360}{10} = 36 \).

Step 4: Compute Each Deviation from the Mean

Subtract the sample mean from each observation: \(27-36, 28-36, 30-36, 34-36, 35-36, 36-36, 38-36, 42-36, 45-36, 45-36\) resulting in deviations: \(-9, -8, -6, -2, -1, 0, 2, 6, 9, 9\).

Step 5: Square Each Deviation

Square all of the deviations to eliminate negatives: \(81, 64, 36, 4, 1, 0, 4, 36, 81, 81\).

Step 6: Mean of Squared Deviations for Sample Variance

Sum of squared deviations: \( 81 + 64 + 36 + 4 + 1 + 0 + 4 + 36 + 81 + 81 = 388 \). To find the sample variance, divide by \(n-1\) (since this is a sample, not a population), where \(n = 10\). So, sample variance \( s^2 = \frac{388}{9} \approx 43.11 \).

Key Concepts

Median Calculation

The median is a measure of central tendency that represents the midpoint of a data set. It is especially useful because it is not affected by extreme values or outliers. To calculate the median, you must first arrange your data in ascending order.
In our example, the dataset is already sorted as: 27, 28, 30, 34, 35, 36, 38, 42, 45, 45.

With 10 numbers, an even data set, the median will be the average of the two middle numbers.

• Identify the 5th and 6th numbers in the sequence, which are 35 and 36.
• Average these two numbers: \( \frac{35 + 36}{2} = 35.5 \).

Thus, the median of the ages in this clinical trial is 35.5. This indicates that half of the patients are younger than 35.5, and half are older.

Sample Mean

The sample mean is another measure of central tendency that provides the average of a data set. It is calculated by summing all the values and dividing by the number of values. This measure is sensitive to outliers, meaning that extremely high or low values can significantly affect it.
Here's how to calculate the sample mean for our data.

First, sum all of the ages:

• Sum: \( 27 + 28 + 30 + 34 + 35 + 36 + 38 + 42 + 45 + 45 = 360 \)
• Count of ages: 10

Then, divide the total sum by the number of observations:

• Sample mean \( \bar{x} = \frac{360}{10} = 36 \)

The sample mean of 36 indicates that, on average, the patients in this clinical trial are 36 years old.

Sample Variance

Sample variance is a measure of how much the data varies from the sample mean. It gives an idea of the amount of spread or dispersion within a set of data. Especially useful in understanding variability, sample variance helps in predicting the range of future variations in forms of data.

To find sample variance:

• First, calculate each deviation from the mean by subtracting the mean from each data point.
• Our calculated deviations were: \(-9, -8, -6, -2, -1, 0, 2, 6, 9, 9\).
• Next, square each deviation to eliminate negatives: \(81, 64, 36, 4, 1, 0, 4, 36, 81, 81\).

Now, perform the final calculation:

• Find the sum of these squared deviations: \(81 + 64 + 36 + 4 + 1 + 0 + 4 + 36 + 81 + 81 = 388\).
• Since this is a sample and not a full population, divide by \(n-1\), where \(n\) is the sample size: \(\frac{388}{9} \approx 43.11 \).

The sample variance of approximately 43.11 demonstrates how spread out the ages are in this clinical trial, showing a moderate level of variability around the mean age.