Question

A random sample of 100 foxes was examined by a team of veterinarians to determine the prevalence of a particular type of parasite. Counting the number of parasites per fox, the veterinarians found that 69 foxes had no parasites, 17 had one parasite, and so on. A frequency tabulation of the data is given here: $$ \begin{array}{l|rrrrrrrrr} \text { Number of Parasites, } x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline \text { Number of Foxes, } f & 69 & 17 & 6 & 3 & 1 & 2 & 1 & 0 & 1 \end{array} $$ a. Construct a relative frequency histogram for \(x\), the number of parasites per fox. b. Calculate \(\bar{x}\) and \(s\) for the sample. c. What fraction of the parasite counts fall within two standard deviations of the mean? Within three standard deviations? Do these results agree with Tchebysheff's Theorem? With the Empirical Rule?

Short answer

Based on the given frequency table and the constructed histogram, the sample mean (𝑥̅) was found to be 0.49 parasites per fox, and the sample standard deviation (s) was approximately 1.29. When comparing fractions within two and three standard deviations of the mean, our results were 95% and 96%, respectively. These results agree with Tchebysheff's Theorem, which states that at least 75% of data should be within two standard deviations, and 89% within three standard deviations of the mean. However, our results cannot be directly compared to the Empirical Rule because our data does not appear to follow a normal distribution, as indicated by the histogram.

Step-by-step solution

Construct a relative frequency histogram

To construct a relative frequency histogram, first, calculate the relative frequency for each number of parasites (x) by dividing the number of foxes with that number of parasites (f) by the total number of foxes (100). Relative frequency for \(x=0\) is $\frac{69}{100}=0.69\\$ Relative frequency for \(x=1\) is $\frac{17}{100}=0.17\\$ ... Continue this for all values of x. Now, plot these values of x on the x-axis and the corresponding relative frequencies on the y-axis as the heights of the bars.

Calculate the sample mean (𝑥̅) and sample standard deviation (s)

Sample mean can be calculated using the following formula: $$\bar{x}=\frac{\sum_{i=1}^{n} x_if_i}{N_n}$$ Where \(x_i\) is the number of parasites, \(f_i\) is the number of foxes having that number of parasites, and \(N_n\) is the total number of foxes/sample size (100). $$\bar{x}=\frac{(0)(69)+(1)(17)+(2)(6)+(3)(3)+(4)(1)+(5)(2)+(6)(1)+(8)(1)}{100}$$ $$\bar{x}= \frac{49}{100}$$ $$\bar{x}= 0.49$$ To calculate the sample standard deviation (s), we need to compute the variance first: $$s^2=\frac{\sum_{i=1}^n(x_i-\bar{x})^2f_i}{N_n-1}$$ $$s^2=\frac{(0-0.49)^2(69)+(1-0.49)^2(17)+...+(8-0.49)^2(1)}{99}$$ \(s^2 \approx 1.66\) Sample Standard Deviation(s) is the square root of the variance: $$s=\sqrt{1.66}$$ \(s \approx 1.29\)

Fractions and comparing with Theorems

Calculate the fractions within two standard deviations (\(2s\)) and three standard deviations (\(3s\)) of the mean: For \(2s \approx 2.58\), the range will be \([-2.09, 3.07]\). Since the count can't be negative, we consider only [0, 3], which includes 69+17+6+3 = 95 foxes. The fraction is \(\frac{95}{100}=0.95\). For \(3s \approx 3.87\), the range will be \([-3.38, 4.36]\). Again, we consider only [0, 4], which includes 69+17+6+3+1 = 96 foxes. The fraction is \(\frac{96}{100}=0.96\). Now, let's compare these results with Tchebysheff's Theorem and the Empirical Rule: Tchebysheff's Theorem states that at least \(1-\frac{1}{k^2}\) of the data falls within \(k\) standard deviations of the mean. For \(k=2\), at least \(1-\frac{1}{2^2}=1-\frac{1}{4}=0.75\) or 75% of the data should be within two standard deviations. For \(k=3\), at least \(1-\frac{1}{3^2}=1-\frac{1}{9}\approx 0.89\) or 89% of the data should be within three standard deviations. Our results (\(95\%\) and \(96\%\)) agree with Tchebysheff's Theorem. The Empirical Rule states that approximately \(68\%\) of data falls within one standard deviation, \(95\%\) within two standard deviations, and \(99.7\%\) within three standard deviations of the mean for a normally distributed dataset. Our results (\(95\%\) and \(96\%\)) are very close to the Empirical Rule; however, since our data does not appear to be normally distributed (based on the histogram), we cannot say our results agree with the Empirical Rule.

Key concepts

Relative Frequency Histogram

A Relative Frequency Histogram is a fantastic way to visualize data. It shows how often certain values appear in a dataset, relative to the total number of observations. In this exercise, we had 100 foxes categorized by the number of parasites each carried. To create the histogram, we divided the number of foxes for each category by the total number of foxes, finding the relative frequency. For example, 69 foxes had no parasites, so the relative frequency was \(0.69\) since \(\frac{69}{100} = 0.69\).

This approach helps us see which counts are most common. It's important to plot these frequencies as bars on a graph with the parasite counts on the x-axis and relative frequencies on the y-axis. This visual tool makes it easier to spot trends and patterns at a glance.

Sample Mean

The Sample Mean gives us an average of the observed data, providing a central value. It’s calculated by adding up all the data points and dividing by the total number of points. Here, the mean number of parasites is calculated using the formula:

\[ \bar{x} = \frac{\sum_{i=1}^{n} x_i f_i}{N_n} \]

Substituting the values, we get:
\( \bar{x} = \frac{(0)(69)+(1)(17)+(2)(6)+(3)(3)+(4)(1)+(5)(2)+(6)(1)+(8)(1)}{100} = 0.49 \).
This result tells us that, on average, there were about 0.49 parasites per fox. The sample mean is crucial as it helps us understand the general tendency or the expected number of parasites based on the sample.

Standard Deviation

Standard Deviation measures the amount of variation or spread in a set of data values. A lower standard deviation means data points tend to be closer to the mean, while a higher value indicates more spread out data.

To find the standard deviation \(s\), we first calculate the variance \(s^2\) using:
\[ s^2 = \frac{\sum_{i=1}^n(x_i-\bar{x})^2f_i}{N_n-1} \]
After computing it for our data, we find \(s^2 \approx 1.66\). Taking the square root gives us \(s \approx 1.29\).

This value shows us that the number of parasites per fox is not too dispersed, and a majority of the counts are close to the mean (0.49 parasites). Understanding standard deviation is essential as it gives context to the average and helps identify how typical or unusual particular values are.

Tchebysheff's Theorem

Tchebysheff's Theorem provides a way to understand data regardless of distribution shape. This theorem guarantees that at least \(1 - \frac{1}{k^2}\) of values fall within \(k\) standard deviations of the mean for any dataset.

For two standard deviations (\(k=2\)), we expect at least 75% of data points within this range. In our study, 95% fell within two standard deviations, well above the expected minimum. For three standard deviations (\(k=3\)), the theorem states at least 89% should fall within, and our data showed 96% do.

This indicates that our data aligns well with Tchebysheff's Theorem, providing a useful way to reason about datasets, even when they aren't normally distributed.

Empirical Rule

The Empirical Rule, or 68-95-99.7 rule, applies to datasets that are approximately normally distributed. It suggests:

• 68% of data fall within one standard deviation of the mean.
• 95% within two standard deviations.
• 99.7% within three standard deviations.

In our data, 95% of parasite counts fall within two standard deviations, closely aligning with this rule. However, our dataset doesn’t appear to be normally distributed, evidenced by the relative frequency histogram.

Although the percentages closely match, it’s crucial to interpret such results properly and consider the data distribution when applying the Empirical Rule. Consider it as a guiding principle for normally distributed data rather than a strict rule for all datasets.