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Chapter 3: Problem 6

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

Short Answer

Expert verified
Mean: 15.4, Median: 15, Mode: 20.

Step by step solution

01

Arrange the Data Set

First, arrange the data set in ascending order. The given data set is: 10, 12, 15, 20, 20.
02

Calculate the Mean

To find the mean, add all the numbers in the data set and divide by the total number of values. \[ \text{Mean} = \frac{10 + 12 + 15 + 20 + 20}{5} = \frac{77}{5} = 15.4 \]
03

Find the Median

The median is the middle value in a set of ordered numbers. Since there are 5 numbers, the median is the third number in the ordered set. Thus, the median is 15.
04

Determine the Mode

The mode is the number that appears most frequently in the dataset. In this data set, the number 20 appears twice, more than any other numbers. Therefore, the mode is 20.

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Key Concepts

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

Mean
The mean is a fundamental concept in descriptive statistics and represents the average of a data set. To calculate the mean, you need to sum all the values in the set and then divide by the number of values. This helps give a single number that summarizes the data:
\[ \text{Mean} = \frac{\text{Sum of all values}}{\text{Total number of values}} \]
For example, in the data set \( \{10, 12, 15, 20, 20\} \), you can calculate the mean by adding all the numbers together, which is 77, and dividing by 5 (since there are 5 numbers). This gives you a mean of 15.4.

Understanding the mean can help to identify the "central" or "average" tendency of the data, though bear in mind it can be skewed by very large or small values.
Median
The median is another measure of central tendency, offering a way to find the middle value in a data set. Unlike the mean, the median is not affected by extremely high or low values in the data set. To find the median, the data set must be arranged in order from smallest to largest.
When you have an odd number of values, like in the given data set \( \{10, 12, 15, 20, 20\} \), the median is the middle number. In this case, it is the third number, which is 15.

For an even number of values, you'd calculate the median by taking the average of the two central numbers. Using the median helps to better understand the central point of a data set, providing insights that the mean might not always display clearly.
Mode
The mode is a helpful statistical measure that identifies the most frequently occurring value in a data set. It shows you which value appears most often, providing insights into common elements within the data.

In the data set \( \{10, 12, 15, 20, 20\} \), the mode is 20 because it appears twice, more frequently than any other value.
Here are some key points about the mode:
  • A data set can have no mode if all numbers appear with the same frequency.
  • If two or more numbers share the highest frequency, the set can be bimodal or multimodal.
The mode is particularly useful when you want to understand the prevalence of data points within the set, helping to highlight repeated patterns.

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Most popular questions from this chapter

Consider the mode, median, and mean. Which average represents the middle value of a data distribution? Which average represents the most frequent value of a distribution? Which average takes all the specific values into account?

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 \(\mu\).

Critical Thinking Consider a data set of 15 distinct measurements with mean \(A\) and median \(B\) (a) If the highest number were increased, what would be the effect on the median and mean? Explain. (b) If the highest number were decreased to a value still larger than \(B\), what would be the effect on the median and mean? (c) If the highest number were decreased to a value smaller than \(B\), what would be the effect on the median and mean?

Do bonds reduce the overall risk of an investment portfolio? Let \(x\) be a random variable representing annual percent return for Vanguard Total Stock Index (all stocks). Let \(y\) be a random variable representing annual return for Vanguard Balanced Index \((60 \% \text { stock and } 40 \%\) bond). For the past several years, we have the following data (Reference: Morning star Research Group, Chicago). $$\begin{array}{ccccccccccc} x: & 11 & 0 & 36 & 21 & 31 & 23 & 24 & -11 & -11 & -21 \\ y: & 10 & -2 & 29 & 14 & 22 & 18 & 14 & -2 & -3 & -10 \end{array}$$ (a) Compute \(\Sigma x, \Sigma x^{2}, \Sigma y,\) and \(\Sigma y^{2}\). (b) Use the results of part (a) to compute the sample mean, variance, and standard deviation for \(x\) and for \(y\). (c) Compute a \(75 \%\) Chebyshev interval around the mean for \(x\) values and also for \(y\) values. Use the intervals to compare the two funds. (d) Compute the coefficient of variation for each fund. Use the coefficients of variation to compare the two funds. If \(s\) represents risks and \(\bar{x}\) represents expected return, then \(s / \bar{x}\) can be thought of as a measure of risk per unit of expected return. In this case, why is a smaller CV better? Explain.

Consider sample data with \(\bar{x}=15\) and \(s=3\) (a) Compute the coefficient of variation. (b) Compute a \(75 \%\) Chebyshev interval around the sample mean.
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