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Chapter 3: Problem 10

Watch out for these little buggers. Each of these exercises involves some feature that is somewhat tricky. Find the (a) mean, (b) median, (c) mode, \((d)\) midrange, and then answer the given question. Biologists conducted experiments to determine whether a deficiency of carbon dioxide in the soil affects the phenotypes of peas. Listed below are the phenotype codes, where \(1=\) smooth-yellow, \(2=\) smooth-green, \(3=\) wrinkled- yellow, and \(4=\) wrinkled-green. Can the measures of center be obtained for these values? Do the results make sense? 2 \(\begin{array}{rrrrrrrrrrrrr}1 & 4 & 1 & 2 & 2 & 1 & 2 & 3 & 3 & 2 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 2 & 2\end{array}\)

Short Answer

Expert verified
Mean: 2.1, Median: 2, Mode: 2, Midrange: 2.5. These measures may not be very meaningful for categorical data.

Step by step solution

01

- List the Data

First, list the given phenotype codes in order to make the calculations clearer: 2, 1, 4, 1, 2, 2, 1, 2, 3, 3, 2, 3, 1, 3, 1, 3, 1, 3, 2, 2.
02

- Calculate the Mean

To find the mean, sum all the values and then divide by the number of values. Sum: \[2 + 1 + 4 + 1 + 2 + 2 + 1 + 2 + 3 + 3 + 2 + 3 + 1 + 3 + 1 + 3 + 1 + 3 + 2 + 2 = 42\] Number of values: 20 Mean: \[ \text{Mean} \ = \ \frac{42}{20} = 2.1\]
03

- Determine the Median

To find the median, first arrange the values in ascending order: 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4. The median is the middle value. Since there are 20 values, the median is the average of the 10th and 11th values. \[ \text{Median} \ = \ \frac{2 + 2}{2} = 2\]
04

- Identify the Mode

The mode is the value(s) that appear most frequently. From the ordered list: 1 appears 6 times, 2 appears 7 times, 3 appears 6 times, and 4 appears 1 time. Mode = 2 (since it appears the most frequently)
05

- Compute the Midrange

The midrange is the average of the maximum and minimum values in the data set. Minimum value = 1 Maximum value = 4 Midrange: \[ \text{Midrange} \ = \ \frac{1 + 4}{2} = 2.5\]
06

- Interpretation and Conclusion

The measures of center for the phenotype codes are obtained as follows: Mean = 2.1, Median = 2, Mode = 2, Midrange = 2.5.However, these measures of center may not be very meaningful for categorical data like phenotype codes, as these are more appropriately analyzed using distribution and frequency.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

mean
The mean is often referred to as the average. It is calculated by summing all the values in a data set and dividing by the number of values. For the provided phenotype data, we summed the values to get 42, and there were 20 values in total. Therefore, the mean is calculated as follows:\[\text{Mean} = \frac{42}{20} = 2.1\]The mean gives us a central value, but remember, it might not always represent the most typical value in the set, especially when dealing with categorical data like the phenotype codes here. The mean can be skewed by outliers, though it offers a useful way to compare different data sets.
median
The median represents the middle value in an ordered data set. To find the median, we first arrange the data in ascending order. For the given phenotype codes, the arranged data set is:1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4.If the number of observations is even, the median is the average of the two middle numbers. Since we have 20 values, we take the 10th and 11th values (both 2) and average them:\[\text{Median} = \frac{(2 + 2)}{2} = 2\]The median is particularly useful because it is not affected by outliers. This makes it a better measure of central tendency for skewed distributions or categorical data.
mode
The mode is the value that appears most frequently in a data set. In the provided phenotype codes, we count the occurrences of each value:
  • 1 appears 6 times
  • 2 appears 7 times
  • 3 appears 6 times
  • 4 appears 1 time
Here, the mode is 2, as it appears the most frequently (7 times). The mode can be especially useful in categorical data to identify the most common category. Unlike the mean and median, the mode can be non-numeric. It's possible to have no mode (if no number repeats) or multiple modes (if multiple numbers tie for the highest frequency).
midrange
The midrange is the average of the maximum and minimum values in a data set. It provides a measure of central tendency based on the range of the data. For the phenotype codes, the minimum value is 1 and the maximum value is 4. The midrange is calculated as:\[\text{Midrange} = \frac{(1 + 4)}{2} = 2.5\]The midrange can give us a quick sense of the central value, but it is particularly sensitive to outliers. In cases where the data set includes significantly high or low values, the midrange might not accurately represent the data set's center. This measure is less commonly used but can be valuable in understanding the dispersion in the data.

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Most popular questions from this chapter

Based on Data Set 1 "Body Data" in Appendix B, blood platelet counts of women have a bell-shaped distribution with a mean of 255.1 and a standard deviation of \(65.4 .\) (All units are 1000 cells \(/ \mu\) L.) Using Chebyshev's theorem, what do we know about the percentage of women with platelet counts that are within 3 standard deviations of the mean? What are the minimum and maximum platelet counts that are within 3 standard deviations of the mean?

Use z scores to compare the given values. Oscars In the 87 th Academy Awards, Eddie Redmayne won for best actor at the age of 33 and Julianne Moore won for best actress at the age of \(54 .\) For all best actors, the mean age is 44.1 years and the standard deviation is 8.9 years. For all best actresses, the mean age is 36.2 years and the standard deviation is 11.5 years. (All ages are determined at the time of the awards ceremony.) Relative to their genders, who had the more extreme age when winning the Oscar: Eddie Redmayne or Julianne Moore? Explain.

Here are four of the Verizon data speeds (Mbps) from Figure 3-1: \(13.5,10.2,21.1,15.1 .\) Find the mean and median of these four values. Then find the mean and median after including a fifth value of \(142,\) which is an outlier. (One of the Verizon data speeds is \(14.2 \mathrm{Mbps}\), but 142 is used here as an error resulting from an entry with a missing decimal point.) Compare the two sets of results. How much was the mean affected by the inclusion of the outlier? How much is the median affected by the inclusion of the outlier?

Find the mean of the data summarized in the frequency distribution. Also, compare the computed means to the actual means obtained by using the original list of data values, which are as follows: (Exercise 29) 36.2 years; (Exercise 30) 44. I years; (Exercise 31) 224.3; (Exercise 32) 255.I. $$\begin{array}{|c|c|}\hline \begin{array}{c}\text { Age (yr) of Best Actress } \\\\\text { When Oscar Was Won }\end{array} & \text { Frequency } \\\\\hline 20-29 & 29 \\\\\hline 30-39 & 34 \\ \hline 40-49 & 14 \\\\\hline 50-59 & 3 \\\\\hline 60-69 & 5 \\\\\hline 70-79 & 1 \\\\\hline 80-89 & 1 \\\\\hline\end{array}$$

Find the range, variance, and standard deviation for the given sample data. Include appropriate units (such as "minutes") in your results. (The same data were used in Section 3-I, where we found measures of center. Here we find measures of variation.) Then answer the given questions. In a study of speed dating conducted at Columbia University, female subjects were asked to rate the attractiveness of their male dates, and a sample of the results is listed below \((1=\text { not attractive; } 10=\) extremely attractive). Can the results be used to describe the variation among attractiveness ratings for the population of adult males? $$\begin{array}{rrrrrrrrrrrrrrrrrrr} 5 & 8 & 3 & 8 & 6 & 10 & 3 & 7 & 9 & 8 & 5 & 5 & 6 & 8 & 8 & 7 & 3 & 5 & 5 & 6 & 8 & 7 & 8 & 8 & 8 & 7 \end{array}$$
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