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Chapter 3: Problem 17

In this problem, we explore the effect on the mean, median, and mode of multiplying each data value by the same number. Consider the data set 2,2,3,6,10. (a) Compute the mode, median, and mean. (b) Multiply each data value by \(5 .\) Compute the mode, median, and mean. (c) Compare the results of parts (a) and (b). In general, how do you think the mode, median, and mean are affected when each data value in a set is multiplied by the same constant? (d) Suppose you have information about average heights of a random sample of airplane passengers. The mode is 70 inches, the median is 68 inches, and the mean is 71 inches. To convert the data into centimeters, multiply each data value by \(2.54 .\) What are the values of the mode, median, and mean in centimeters?

Short Answer

Expert verified
The mode, median, and mean are multiplied by the constant: 2, 3, 4.6 becomes 10, 15, 23, respectively. Converting heights: mode 177.8 cm, median 172.72 cm, mean 180.34 cm.

Step by step solution

01

Calculate Mode, Median, and Mean of Original Data

Given the data set is 2, 2, 3, 6, 10.- **Mode:** The number that appears most frequently is 2.- **Median:** Sort the data to find the middle value, which is 3.- **Mean:** Compute by summing all values and dividing by the number of values: \[ \text{Mean} = \frac{2 + 2 + 3 + 6 + 10}{5} = \frac{23}{5} = 4.6 \]
02

Multiply Each Data Value by 5

Multiply each number in the data set by 5: 10, 10, 15, 30, 50.
03

Calculate Mode, Median, and Mean of New Data

The new data set is 10, 10, 15, 30, 50.- **Mode:** The most frequent number remains 10.- **Median:** The middle value remains 15.- **Mean:** Calculate by summing all values and dividing by the number of values: \[ \text{Mean} = \frac{10 + 10 + 15 + 30 + 50}{5} = \frac{115}{5} = 23 \]
04

Compare Results from Steps 1 and 3

Comparing the original and multiplied data results: - **Mode:** Multiplied by 5, from 2 to 10. - **Median:** Multiplied by 5, from 3 to 15. - **Mean:** Multiplied by 5, from 4.6 to 23. When each data value is multiplied by the same constant, the mode, median, and mean are also multiplied by that constant.
05

Convert Heights from Inches to Centimeters

Given mode, median, and mean in inches are 70, 68, and 71 respectively.- Multiply each by 2.54 to convert to centimeters: - **Mode:** \(70 \times 2.54 = 177.8\) cm - **Median:** \(68 \times 2.54 = 172.72\) cm - **Mean:** \(71 \times 2.54 = 180.34\) cmEach statistical measure is multiplied by the conversion factor 2.54, converting from inches to centimeters.

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Key Concepts

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

Mode
The mode of a data set is the value that appears most frequently. It is a measure of central tendency that shows the most common value. In the data set 2, 2, 3, 6, 10, the mode is 2 because it appears more times than any other number. Understanding the mode is particularly useful in datasets where we want to know the most typical or popular item.
The mode may not always be unique; a set can be bimodal or multimodal if it has two or more values that appear with the highest frequency.
  • Calculation: Look for the most frequently occurring number.
  • Multiplied Effect: When each data value is multiplied by the same number, the mode is also multiplied by that number.
Median
To find the median of a data set, you first need to arrange the data in numerical order and identify the middle value. This measure of central tendency can often give a better sense of the "center" of a dataset than the mean, especially in skewed distributions.
For the original data set 2, 2, 3, 6, 10, when ordered, the median is 3, as it's the middle number in the list.
If the data set has an even number of observations, the median will be the average of the two middle numbers.
  • Calculation: Sort the list and find the middle number.
  • Multiplied Effect: When each data value is multiplied by the same constant, the median is also multiplied by that constant.
Mean
The mean, often referred to as the average, is calculated by adding all the values in a data set and then dividing by the number of values. It provides a measure of central tendency that balances all data points. In the given data set 2, 2, 3, 6, 10, the mean is calculated as \[\text{Mean} = \frac{2+2+3+6+10}{5} = 4.6.\]The mean gives us a general idea of the size of the data points, but it can be influenced by extreme values, or outliers.
  • Calculation: Sum all values and divide by the count of values.
  • Multiplied Effect: Multiplying each value by a constant results in multiplying the mean by that constant.
Data Transformation
Data transformation involves modifying each data value by the same operation. In this exercise, each data point was multiplied by a constant. Uniform transformations such as these are common when converting between units, like inches to centimeters.
In our example, when each data value in the original set 2, 2, 3, 6, 10 is multiplied by 5, they transform into 10, 10, 15, 30, 50. This transformation scales the dataset but maintains the relative differences and ranks.
  • Mode, Median, and Mean: Each is multiplied by the transformation constant, here 5. This keeps the measures directly proportional to the original dataset.
  • Practical Application: Converting measurements from one unit to another without altering the inherent meaning of the data, such as converting inches to centimeters using multiplication by 2.54.

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Most popular questions from this chapter

Each of the following data sets has a mean of \(\bar{x}=10\). \(\begin{array}{llllllllllll}\text { (i) } 8 & 9 & 10 & 11 & 12 & & \text { (ii) } 7 & 9 & 10 & 11 & 13 & & \text { (iii) } & 7 & 8 & 10 & 12 & 13\end{array}\) (a) Without doing any computations, order the data sets according to increasing value of standard deviations. (b) Why do you expect the difference in standard deviations between data sets (i) and (ii) to be greater than the difference in standard deviations between data sets (ii) and (iii)? Hint: Consider how much the data in the respective sets differ from the mean.

Find the mean, median, and mode of the data set 10 12 20 15 20

For mallard ducks and Canada geese, what percentage of nests are successful (at least one offspring survives)? Studies in Montana, Illinois, Wyoming, Utah, and California gave the following percentages of successful nests (Reference: The Wildlife Society Press, Washington, D.C.). \(x:\) Percentage success for mallard duck nests $$56 \quad 85 \quad 52 \quad 13 \quad 39$$ \(y:\) Percentage success for Canada goose nests $$24 \quad 53 \quad 60 \quad 69 \quad 18$$ (a) Use a calculator to verify that \(\Sigma x=245 ; \Sigma x^{2}=14,755 ; \Sigma y=224 ;\) and \(\Sigma y^{2}=12,070\). (b) Use the results of part (a) to compute the sample mean, variance, and standard deviation for \(x,\) the percent of successful mallard nests. (c) Use the results of part (a) to compute the sample mean, variance, and standard deviation for \(y,\) the percent of successful Canada goose nests. (d) Use the results of parts (b) and (c) to compute the coefficient of variation for successful mallard nests and Canada goose nests. Write a brief explanation of the meaning of these numbers. What do these results say about the nesting success rates for mallards compared to those of Canada geese? Would you say one group of data is more or less consistent than the other? Explain.

This is a technique to break down the variation of a random variable into useful components (called stratum) in order to decrease experimental variation and increase accuracy of results. It has been found that a more accurate estimate of population mean \(\mu\) can often be obtained by taking measurements from naturally occurring subpopulations and combining the results using weighted averages. For example, suppose an accurate estimate of the mean weight of sixth grade students is desired for a large school system. Suppose (for cost reasons) we can only take a random sample of \(m=100\) students, Instead of taking a simple random sample of 100 students from the entire population of all sixth grade students, we use stratified sampling as follows. The school system under study consists three large schools. School A has \(N_{1}=310\) sixth grade students, School B has \(N_{2}=420\) sixth grade students, and School C has \(N_{3}=516\) sixth grade students. This is a total population of 1246 sixth grade students in our study and we have strata consisting of the 3 schools. A preliminary study in each school with relatively small sample size has given estimates for the sample standard deviation \(s\) of sixth grade student weights in each school. These are shown in the following table. $$\begin{array}{lll} \text { School A } & \text { School B } & \text { School C } \\ N_{1}=310 & N_{2}=420 & N_{3}=516 \\ s_{1}=3 \mathrm{lb} & s_{2}=12 \mathrm{lb} & s_{3}=6 \mathrm{lb} \end{array}$$ How many students should we randomly choose from each school for a best estimate \(\mu\) for the population mean weight? A lot of mathematics goes into the answer. Fortunately, Bill Williams of Bell Laboratories wrote a book called A Sampler on Sampling (John Wiley and Sons, publisher), which provides an answer. Let \(n_{1}\) be the number of students randomly chosen from School \(\mathrm{A}\), \(n_{2}\) be the number chosen from School \(\mathrm{B},\) and \(n_{3}\) be the number chosen from School C. This means our total sample size will be \(m=n_{1}+n_{2}+n_{3} .\) What is the formula for \(n_{i} ?\) A popular and widely used technique is the following. $$n_{i}=\left[\frac{N_{i} s_{i}}{N_{1} s_{1}+N_{2} s_{2}+N_{3} s_{3}}\right] m$$ The \(n_{i}\) are usually not whole numbers, so we need to round to the nearest whole number. This formula allocates more students to schools that have a larger population of sixth graders and/or have larger sample standard deviations. Remember, this is a popular and widely used technique for stratified sampling. It is not an absolute rule. There are other methods of stratified sampling also in use. In general practice, according to Bill Williams, the use of naturally occurring strata seems to reduce overall variability in measurements by about \(20 \%\) compared to simple random samples taken from the entire (unstratified) population. Now suppose you have taken a random sample size \(n_{i}\) from each appropriate school and you got a sample mean weight \(\bar{x}_{i}\) from each school. How do you get the best estimate for population mean weight \(\mu\) of the all 1246 students? The answer is, we use a weighted average. $$\mu \approx \frac{n_{1}}{m} \overline{x_{1}}+\frac{n_{2}}{m} \overline{x_{2}}+\frac{n_{3}}{m} \overline{x_{3}}$$ This is an example with three strata. Applications with any number of strata can be solved in a similar way with obvious extensions of formulas. (a) Compute the size of the random samples \(n_{1}, n_{2}, n_{3}\) to be taken from each school. Round each sample size to the nearest whole number and make sure they add up to \(m=100\). (b) Suppose you took the appropriate random sample from each school and you got the following average student weights: \(\overline{x_{1}}=82 l b, \overline{x_{2}}=115 l b, \overline{x_{3}}=90 l b\). Compute your best estimate for the population mean weight \(\mu\).

Some data sets include values so high or so low that they seem to stand apart from the rest of the data. These data are called outliers. Outliers may represent data collection errors, data entry errors, or simply valid but unusual data values. It is important to identify outliers in the data set and examine the outliers carefully to determine if they are in error. One way to detect outliers is to use a box-and-whisker plot. Data values that fall beyond the limits, $$\begin{aligned} &\text { Lower limit: } Q_{1}-1.5 \times(I Q R)\\\ &\text { Upper limit: } Q_{3}+1.5 \times(I Q R) \end{aligned}$$ where \(I Q R\) is the interquartile range, are suspected outliers. In the computer software package Minitab, values beyond these limits are plotted with asterisks (*). Students from a statistics class were asked to record their heights in inches. The heights (as recorded) were $$\begin{array}{cccccccccccc} 65 & 72 & 68 & 64 & 60 & 55 & 73 & 71 & 52 & 63 & 61 & 74 \\ 69 & 67 & 74 & 50 & 4 & 75 & 67 & 62 & 66 & 80 & 64 & 65 \end{array}$$ (a) Make a box-and-whisker plot of the data. (b) Find the value of the interquartile range \((I Q R)\) (c) Multiply the IQR by 1.5 and find the lower and upper limits. (d) Are there any data values below the lower limit? above the upper limit? List any suspected outliers. What might be some explanations for the outliers?
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