Question

To study the spatial distribution of Japanese beetle larvae in the soil, researchers divided a \(12-\times 12\) -foot section of a cornfield into 144 one-foot squares. They counted the number of larvae \(Y\) in each square, with the results shown in the following table. \(^{57}\) $$ \begin{array}{|cc|} \hline & \text { Frequency (Number } \\ \text { Number of larvae } & \text { of squares) } \\ \hline 0 & 13 \\ 1 & 34 \\ 2 & 50 \\ 3 & 18 \\ 4 & 16 \\ 5 & 10 \\ 6 & 2 \\ 7 & 1 \\ \hline \text { Total } & 144 \\ \hline \end{array} $$ (a) The mean and SD of \(Y\) are \(\bar{y}=2.23\) and \(s=1.47\). What percentage of the observations are within (i) \(1 \mathrm{SD}\) of the mean? (ii) 2 SDs of the mean? (b) Determine the total number of larvae in all 144 squares. How is this number related to \(\bar{y} ?\) (c) Determine the median value of the distribution.

Short answer

Part (a) 102 squares (70.83%) are within 1 SD, and 141 squares (97.92%) are within 2 SDs. Part (b) The total number of larvae is 322, which relates to the mean (\(\bar{y}\)) as 322 equals 144 times \(\bar{y}\). Part (c) With a cumulative approach, the median falls within the range for 2 larvae since the 72nd and 73rd observations lie within this category.

Step-by-step solution

- Understanding the Standard Deviation Range

To calculate the percentage of observations within 1 SD of the mean, we consider all observations that fall between \(\bar{y} - s\) and \(\bar{y} + s\). Given that the mean (\(\bar{y}\)) is 2.23 and the standard deviation (\(s\)) is 1.47, this range is from 2.23 - 1.47 to 2.23 + 1.47 (0.76 to 3.70). Since the number of larvae (\(Y\)) is discrete, we consider the range 1 to 3 for practical purposes.

- Calculating the Percentage within 1 SD

We add the frequencies of the squares with 1, 2, and 3 larvae (since these are the whole numbers within our practical range). The frequencies are 34, 50, and 18 respectively. Therefore, the number of squares within 1 SD is 34 + 50 + 18 = 102.

- Understanding the Range for 2 SDs

To calculate the percentage of observations within 2 SDs of the mean, we consider all observations that fall between \(\bar{y} - 2s\) and \(\bar{y} + 2s\). This range is from 2.23 - 2(1.47) to 2.23 + 2(1.47) (-0.71 to 5.17). Since larvae counts are non-negative and discrete, we consider the range 0 to 5.

- Calculating the Percentage within 2 SDs

We add the frequencies of squares with 0 to 5 larvae to find the total number of squares within 2 SDs. The frequencies are 13, 34, 50, 18, 16, and 10 for 0, 1, 2, 3, 4, and 5 larvae respectively, giving a total of 13 + 34 + 50 + 18 + 16 + 10 = 141.

- Finding the Total Number of Larvae

The total number of larvae in all 144 squares can be found by summing up the products of the frequency of each number of larvae with the number itself. This is calculated as follows: \(0(13) + 1(34) + 2(50) + 3(18) + 4(16) + 5(10) + 6(2) + 7(1)\).

- Relating Total Number of Larvae to Mean

The total number calculated in the previous step should equal the product of the mean number of larvae per square (\(\bar{y}\)) and the total number of squares (144). Therefore, \(\text{Total larvae} = \bar{y} \times 144\).

- Determining the Median Value

The median value is the middle value when all observations are ordered from the least to greatest. With 144 observations (an even number), the median will be the average of the 72nd and 73rd observations. We need to find the number of larvae that corresponds to these two positions in the cumulative frequency distribution.

Key concepts

Understanding Standard Deviation

Standard deviation (SD) is a widely used measure of variability or diversity used in statistics and probability theory. It shows how much variation there is from the average (mean), or expected value. A low standard deviation indicates that the data points tend to be close to the mean, whereas a high standard deviation indicates that the data points are spread out over a wider range.

In the context of the Japanese beetle larvae exercise, the standard deviation helps researchers understand how larvae are distributed within the cornfield. If the number of larvae widely varies from square to square, this would be reflected in a high standard deviation. Conversely, if every square had a similar number of larvae to the others, the standard deviation would be low. Understanding this concept is critical when analyzing the spread of organisms in an environment or any other form of data distribution.

Mean and Median: Measures of Central Tendency

The mean and median are both measures of central tendency, which is a way to describe the center of a data set. The mean is calculated by adding all the numbers in a data set and then dividing by the count of numbers. It's sometimes referred to as the 'average.' However, the mean can be heavily influenced by outliers or extreme values.

The median, on the other hand, is the middle value in a list of numbers that has been arranged in ascending or descending order. If there’s an even number of observations, as in the beetle larvae case, the median will be the average of the two central numbers. The median is less affected by outliers and skewed data, which makes it a more accurate reflection of a typical value when the distribution is not symmetrical. Understanding the difference between these two measures can provide better insights when interpreting data.

Frequency Distribution: The Big Picture

Frequency distribution is a summary of how often each different value in a set of data occurs. It is commonly represented in a table or as a histogram. Seeing data in this format allows for a quick understanding of the distribution—whether most observations are clustered around certain values (indicative of a peak in the histogram), or if they're more spread out (creating a flatter histogram).

In the beetle larvae example, each square represents an observation and the count of larvae is the value. When these are tallied up into a frequency distribution table, it becomes easier to visualize and analyze the patterns. Researchers could quickly see, for instance, whether there are more squares without larvae than with, which could indicate possible issues with soil conditions or the presence of predators. Frequency distribution is an invaluable tool for making these kinds of assessments in many fields of study.