Problem 5 - Calculus for Biology and Medicine

Question

The following data represent the frequency distribution of seed numbers per flower head in a flowering plant: $$ \begin{array}{cc} \hline \text { Seed Number } & \text { Frequency } \\ \hline 9 & 37 \\ 10 & 48 \\ 11 & 53 \\ 12 & 49 \\ 13 & 61 \\ 14 & 42 \\ 15 & 31 \\ \hline \end{array} $$ Calculate the sample mean and the sample variance.

Step-by-Step Solution

Answer

The sample mean is approximately 11.33 and the sample variance is approximately 1.99.

Step 1: Calculate the Sample Mean

The sample mean is calculated using the formula \( \bar{x} = \frac{\sum{(f_i \times x_i)}}{n} \), where \( x_i \) is the seed number, \( f_i \) is the frequency, and \( n \) is the total number of data points. First, we calculate \( n = 37 + 48 + 53 + 49 + 61 + 42 + 31 = 321 \). Next, we compute \( \sum{(f_i \times x_i)} = 9 \times 37 + 10 \times 48 + 11 \times 53 + 12 \times 49 + 13 \times 61 + 14 \times 42 + 15 \times 31 = 3637 \). Now, calculate the mean: \( \bar{x} = \frac{3637}{321} \approx 11.33 \).

Step 2: Calculate the Sample Variance

The sample variance is given by \( s^2 = \frac{\sum{f_i(x_i - \bar{x})^2}}{n-1} \). We calculate \( (x_i - \bar{x})^2 \) for each seed number: \( (9 - 11.33)^2 = 5.4289 \), \( (10 - 11.33)^2 = 1.7689 \), \( (11 - 11.33)^2 = 0.1089 \), \( (12 - 11.33)^2 = 0.4489 \), \( (13 - 11.33)^2 = 2.7889 \), \( (14 - 11.33)^2 = 7.1289 \), and \( (15 - 11.33)^2 = 13.4689 \). Multiply these by their frequencies and sum the results: \( 37 \times 5.4289 + 48 \times 1.7689 + 53 \times 0.1089 + 49 \times 0.4489 + 61 \times 2.7889 + 42 \times 7.1289 + 31 \times 13.4689 = 636.4667 \). Finally, compute the variance: \( s^2 = \frac{636.4667}{320} \approx 1.99 \).

Key Concepts

Understanding Frequency Distribution

Frequency distribution is a way of summarizing data by showing the number of times each possible outcome occurs. Let's break down this concept with our example using seed numbers from a flowering plant.

The data you've encountered is laid out in a table with two columns:

• Seed Number, which shows different values that the seed counts take.
• Frequency, which indicates how many times each seed number appears in the sample.

This table makes it easier to see quickly how the data is distributed.

For instance, if you have a seed number of 9 with a frequency of 37, this means there are 37 instances where the plants had 9 seeds in a flower head. This pattern continues for each seed number, revealing a clear structure in the data.

When looking at frequency distributions, you can visually grasp the shape of your data, note any patterns, and identify which data points are most common across the sample.

Sample Mean Calculation

The sample mean is a measure of central tendency, which provides a single value representing the average of the dataset. We find this average by weighing each seed number by its occurrence frequency and then dividing by the total number of occurrences.

Here's a quick way to calculate the sample mean:

• First calculate the total data points, known as \( n \). Add all frequencies: \( 37 + 48 + 53 + 49 + 61 + 42 + 31 = 321 \).
• Then calculate the sum of all outcomes multiplied by their frequencies: \( 9 \times 37 + 10 \times 48 + 11 \times 53 + 12 \times 49 + 13 \times 61 + 14 \times 42 + 15 \times 31 = 3637 \).
• Finally, divide this sum by the total number of data points to find the mean: \( \bar{x} = \frac{3637}{321} \approx 11.33 \).

The calculated sample mean of approximately 11.33 tells us that, on average, a flower head contains about 11 seeds.

Sample Variance Calculation

Sample variance gives us an idea of how much the data varies from the mean. It measures the spread of the data showing the average degree to which each number is different from the sample mean.

Follow these steps to compute the sample variance:

• For each seed number, calculate the squared deviation from the mean. For example, for 9 seeds: \( (9 - 11.33)^2 = 5.4289 \).
• Multiply each squared deviation by its corresponding frequency: \( 37 \times 5.4289, 48 \times 1.7689, \) and so on.
• Sum all these values to get \( \sum{f_i(x_i - \bar{x})^2} = 636.4667 \).
• Divide this sum by \( n-1 \) (where \( n \) is 321): \( s^2 = \frac{636.4667}{320} \approx 1.99 \).

The sample variance calculated here is about 1.99, which implies a moderate level of variability among the seed numbers per flower head in this sample. This means while many seed heads are close to the mean, there is some variability in the data.