Calculating Fruit Fly Abdomen Length Standard Deviation

This problem requires calculating the standard deviation (SD) for a given set of measurements of male fruit fly abdomen lengths. The standard deviation measures the dispersion or spread of the data points around the mean.

Data Provided

• Abdomen Lengths (mm): 1.9, 2.4, 2.1, 2.0, 2.2, 2.4, 1.7, 1.8, 2.0, 2.0, 2.3, 2.1, 1.6, 2.3, 2.2
• Sample Size ($n$): 15

Standard Deviation Formula

For a sample dataset, the formula for sample standard deviation ($s$) is:
$s = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{n-1}}$ where $x_i$ represents each individual data point, $\bar{x}$ is the sample mean, and $n$ is the total number of data points.

Calculation Steps

1. Calculate the Sample Mean ($\bar{x}$):

   First, sum all the abdomen length measurements:
   $\sum x_i = 1.9 + 2.4 + 2.1 + 2.0 + 2.2 + 2.4 + 1.7 + 1.8 + 2.0 + 2.0 + 2.3 + 2.1 + 1.6 + 2.3 + 2.2 = 31.0$ mm.
   Next, divide the sum by the sample size ($n$) to find the mean:
   $\bar{x} = \frac{\sum x_i}{n} = \frac{31.0}{15} \approx 2.0667$ mm.

2. Calculate the Standard Deviation ($s$):

   Using the sample data and the calculated mean, the standard deviation is determined by applying the formula. The calculation yields a standard deviation of approximately $0.25$ mm.

Result

The computed standard deviation for the abdomen lengths of the male fruit flies is approximately $0.25$ mm, which corresponds to Option B.