2.5 Measures of the Center of the Data
SLO 2
Identify appropriate graphs and summary statistics for variables and relationships between them and correctly interpret information from graphs and summary statistics.
The mean, median, and mode are the summary statistics for the center. Knowing all three lets you pick the honest one — the median when one huge value drags the mean off, the mode when a repeated value is what actually describes the group — and estimate a mean from grouped intervals.
Learning Objectives
By the end of this section, you will be able to:
- calculate the mean of a data set from a raw list of values and from a table of values and frequencies;
- locate and identify the median of an ordered data set using the location formula;
- identify the mode of a data set, including a data set with two modes;
- recommend the mean, the median, or the mode as the better measure of the center for a given data set and defend the choice;
- estimate the mean of a data set that is only reported as grouped intervals and frequencies.
The last two sections were about position — where a single value sits relative to everything else, and how to draw that picture. The "center" of a data set is one more way of describing location, and it is the one people reach for first. Ask anyone how a class did on a test, or what houses cost in a neighborhood, and the answer they want is a single number that stands in for the whole pile.
There are three such numbers in common use, and this section is about all three: the mean, the median, and the mode. They do not always agree, and the interesting part is learning to say why they disagree and which one you should quote.
2.5.1 The Mean and the Median
The mean of a data set is the sum of all the data values divided by the number of data values. The mean of a sample is written \(\overline{x}\) (read "x-bar"), and the mean of a population is written with the Greek letter \(\mu\) (pronounced "mew"). One of the requirements for a sample mean to be a good estimate of the population mean is that the sample was taken truly at random.
There are two ways to do the arithmetic, and they always agree. You can add up every value one at a time, or — when values repeat — you can multiply each distinct value by how many times it occurs and divide that sum by the total count. Take the sample
1; 1; 1; 2; 2; 3; 4; 4; 4; 4; 4
Adding the eleven values one at a time gives
$$\overline{x} = \frac{1 + 1 + 1 + 2 + 2 + 3 + 4 + 4 + 4 + 4 + 4}{11} = 2.7$$Now do it the other way. The distinct values are 1, 2, 3, and 4, and their frequencies are 3, 2, 1, and 5:
$$\overline{x} = \frac{(1)(3) + (2)(2) + (3)(1) + (4)(5)}{11} = \frac{3 + 4 + 3 + 20}{11} = \frac{30}{11} = 2.7$$Same number, less writing. That second form is worth holding onto — it is the seed of the grouped-data formula at the end of this section.
The median is the other half of the story, and finding it is a two-step job: first you find where it is, then you read off what it is.
The median of a data set is the value that splits the ordered data into two halves, so that approximately half the values fall below it and half fall above it. Its location in the ordered list is given by
$$\frac{n + 1}{2}$$where \(n\) is the total number of data values. If \(n\) is odd, the median is the single middle value. If \(n\) is even, the median is the two middle values added together and divided by two. The median is often written with a capital \(M\).
Asking "where is the median?" is like asking which seat in a sold-out row is the middle seat. Seat 21 is the address; the person sitting in it is the answer. Confusing the two is the single most common mistake in this section.
Definition 2.5.1 — The mean is the level every bar reaches once the total is shared out evenly.
Two quick examples of the location formula at work. If a data set has 97 values, then \(\frac{n+1}{2} = \frac{97+1}{2} = 49\), so the median is the 49th value in the ordered list. If it has 100 values, then \(\frac{n+1}{2} = \frac{100+1}{2} = 50.5\), and the median sits midway between the 50th and 51st values — you average those two. Notice again that 49 and 50.5 are addresses: they tell you where to look, not what you will find there.
The median is generally the better measure of the center when the data contains extreme values or outliers, because it does not care about the precise size of those extremes — only about how many values sit on each side. The mean, meanwhile, remains the most commonly reported measure of the center. The next example computes both for the same data so you can see them land in almost the same place.
The two most widely used measures of the center are the mean (the average) and the median. Both are easy to describe. To find the mean weight of 50 people, add the 50 weights together and divide by 50. To find the median weight of those same 50 people, put the weights in order and find the number that splits them into two equal halves.
In everyday speech people swap the two freely. The technical term for what we are computing is the arithmetic mean, and "average" is really a loose word for any center location — but among non-statisticians "average" is universally accepted for the arithmetic mean, and we will use both words.
Definition 2.5.2 — The location formula gives an address; the median is the value standing at it.
Rowan Whitfield coordinates transplant cases at a regional hospital, and they keep a record of how many months their patients typically wait on the list before getting surgery. The data are ordered from smallest to largest. Calculate the mean and the median.
3; 4; 5; 7; 7; 7; 7; 8; 8; 9; 9; 10; 10; 10; 10; 10; 11; 12; 12; 13; 14; 14; 15; 15; 17; 17; 18; 19; 19; 19; 21; 21; 22; 22; 23; 24; 24; 24; 24
Solution
Step 1 — Count the values. There are 39 waiting times in the list.
Step 2 — Compute the mean. The 39 values add to 544, so
$$\overline{x} = \frac{544}{39} \approx 13.9$$Step 3 — Locate the median. With \(n = 39\),
$$\frac{n + 1}{2} = \frac{39 + 1}{2} = 20$$so the median is the 20th value in the ordered list — not an average of two, because 39 is odd.
Step 4 — Read it off. Counting in to the 20th value gives 13.
Answer: \(\overline{x} \approx 13.9\) months and \(M = 13\) months.
A hospital administrator keeps track of the ages (in years) of patients visiting the emergency room over a one-week period. The data are sorted from smallest to largest:
3; 4; 8; 8; 10; 11; 12; 13; 14; 15; 15; 16; 16; 17; 17; 18; 21; 22; 22; 24; 24; 25; 26; 26; 27; 27; 29; 29; 31; 32; 33; 33; 34; 34; 35; 37; 40; 44; 44; 47
Calculate the mean and the median.
Solution
Step 1 — Add up the values and divide by how many there are. There are 40 ages. Grouping the repeats to save writing:
$$\overline{x} = \frac{3 + 4 + (8)(2) + 10 + 11 + 12 + 13 + 14 + (15)(2) + (16)(2) + \ldots + 35 + 37 + 40 + (44)(2) + 47}{40} = \frac{943}{40} = 23.6$$Step 2 — Find the location of the median. With \(n = 40\),
$$\frac{n + 1}{2} = \frac{40 + 1}{2} = 20.5$$Step 3 — Read the median off the ordered list. A location of 20.5 means the median sits midway between the 20th and 21st values. Counting in from the smallest value, both of those are 24:
3; 4; 8; 8; 10; 11; 12; 13; 14; 15; 15; 16; 16; 17; 17; 18; 21; 22; 22; 24; 24; 25; 26; 26; 27; 27; 29; 29; 31; 32; 33; 33; 34; 34; 35; 37; 40; 44; 44; 47
$$M = \frac{24 + 24}{2} = 24$$Answer: \(\overline{x} = 23.6\) and \(M = 24\). The two agree closely here, which is what you expect when no single value is wildly out of line with the rest.
Try it in rāSHio
Open rāSHio, paste the forty emergency-room ages (3; 4; 8; 8; 10; …) into File → Delimited List… — the semicolons parse as-is — then choose Stats → Summary Statistics. The mean 23.6 and the median 24 you just worked out by hand come back in one panel, side by side, which is the comparison the rest of this section is about.
Figure 2.5.1 — Reading the mean and the median in rāSHio: Stats → Summary Statistics.
2.5.2 Choosing Between the Mean and the Median
So far the mean and the median have stayed close together. That is the normal case, and when it happens either number is a fair summary. The reason we keep both around is what happens when one value is very far from the rest.
Think about what each calculation actually does with an extreme value. The mean multiplies it in at full strength — a single enormous number drags the whole average toward it. The median only counts it as one more value on the high side, no matter how high. Swap a $5,000,000 salary for a $500,000,000 salary and the mean lurches; the median does not move at all.
When a news report quotes an average income, salary, or house price, ask which center they used. A handful of very large incomes can push a mean far above what a typical household actually earns, and the median is usually the honest answer to "what does a normal person make?"
In a sample of 60 households, one house is worth $2,500,000. Twenty-nine houses are worth $280,000, and all the others are worth $315,000. Which is the better measure of the "center": the mean or the median?
Solution
Step 1 — Account for every house. There are 60 houses: 1 at $2,500,000, 29 at $280,000, and the remaining \(60 - 1 - 29 = 30\) at $315,000.
Step 2 — Compute the mean.
$$\overline{x} = \frac{2{,}500{,}000 + (280{,}000)(29) + (315{,}000)(30)}{60} = \frac{20{,}070{,}000}{60} = 334{,}500$$Step 3 — Find the median. In order, locations 1 through 29 are the $280,000 houses, locations 30 through 59 are the $315,000 houses, and location 60 is the $2,500,000 house. The location is \(\frac{60+1}{2} = 30.5\), so the median averages the 30th and 31st values — both $315,000. So \(M = \$315{,}000\).
Answer: The median, $315,000, is the better measure of the center. The single $2,500,000 house pulls the mean up to $334,500, which is higher than 59 of the 60 houses in the sample.
Marisol Vega and her wife publish the weekly paper in a small town of 50 people, and she is writing up what a typical household there earns. One person in town earns $5,000,000 per year and the other 49 each earn $30,000. Which is the better measure of the "center" for her article: the mean or the median?
Solution
Step 1 — Compute the mean. Forty-nine people earn $30,000 and one earns $5,000,000:
$$\overline{x} = \frac{(30{,}000)(49) + 5{,}000{,}000}{50} = \frac{1{,}470{,}000 + 5{,}000{,}000}{50} = \frac{6{,}470{,}000}{50} = 129{,}400$$Step 2 — Find the median. With \(n = 50\), the location is \(\frac{50+1}{2} = 25.5\), so the median is the average of the 25th and 26th values in the ordered list. Both of those are $30,000, so \(M = \$30{,}000\).
Step 3 — Compare them to the data. Not one person in this town earns $129,400. Forty-nine of the fifty earn exactly $30,000.
Answer: The median is the better measure of the center. The $5,000,000 is an outlier, and the $30,000 gives us a much better sense of the middle of this data.
2.5.3 The Mode
The mode of a data set is the value that occurs most frequently. A data set can have more than one mode, as long as those values share the same frequency and that frequency is the highest one in the set.
Definition 2.5.3 — The mode is whichever score's stack finishes tallest.
A data set with exactly two modes is called bimodal.
Unlike the mean and the median, the mode is always an actual member of the data set — it is a value someone really recorded, not a computed balancing point. That is both its strength and its limitation.
There is a third measure of the center, and it answers a different question than the other two. Instead of "what is the balancing point?" or "what splits the data in half?", it asks "what happens most often?"
Definition 2.5.4 — Bimodal: two scores tie for the highest frequency, so the mode label lands twice.
The number of books checked out from the library by 25 students are as follows:
0; 0; 0; 1; 2; 3; 3; 4; 4; 5; 5; 7; 7; 7; 7; 8; 8; 8; 9; 10; 10; 11; 11; 12; 12
Find the mode.
Solution
Step 1 — Tally the repeats. 0 appears three times; 3, 4, 5, 10, 11, and 12 each appear twice; 7 appears four times; 8 appears three times.
Step 2 — Take the largest tally. Four is the highest frequency, and only 7 achieves it.
Answer: Mode = 7 books.
Statistics exam scores for 20 students are as follows:
50; 53; 59; 59; 63; 63; 72; 72; 72; 72; 72; 76; 78; 81; 83; 84; 84; 84; 90; 93
Find the mode.
Solution
Step 1 — Count how many times each score appears. Working left to right: 59 appears twice, 63 appears twice, 72 appears five times, 84 appears three times, and every other score appears once.
Step 2 — Pick the highest count. Five is the largest frequency, and it belongs to 72.
Answer: Mode = 72.
Try it in rāSHio
Open rāSHio, paste the twenty exam scores into File → Delimited List…, then choose Graph → Frequency Table with Discrete values checked. Every distinct score gets its own row with its count beside it, so the mode stops being something you hunt for by hand and becomes the row with the largest number in it — 72, at five.
Figure 2.5.2 — Counting each distinct value in rāSHio: Graph → Frequency Table.
2.5.4 When the Mode Is the Best Choice
The mode earns its keep whenever the typical value matters more than the balancing point — and especially when the data has two clusters instead of one, because that is a fact neither the mean nor the median will report to you.
Five credit scores are 680, 680, 700, 720, 720. Explain why this data set is bimodal. Then consider the annual earnings of workers at a factory: the mode is $25,000 and occurs 150 times out of 301, the median is $50,000, and the mean is $47,500. What would be the best measure of the "center" for the factory earnings?
Solution
Step 1 — Check the credit scores. 680 occurs twice and 720 occurs twice, and no value occurs more often than that. Two values tie for the highest frequency, so the data set is bimodal with modes 680 and 720.
Step 2 — Look at what the factory mode actually covers. 150 out of 301 workers earn $25,000. That is very nearly half of everyone at the factory earning exactly the same amount — a single wage that describes an enormous share of the workforce.
Step 3 — Compare against the other two centers. The median ($50,000) and the mean ($47,500) both sit at roughly twice the wage that half the workers actually receive. Neither one describes a typical worker's paycheck.
Answer: The mode, $25,000, is the best measure of the center for the factory earnings, because it is the wage that the largest group of workers actually earns.
Five real estate exam scores are 430, 430, 480, 480, 495. Explain why this data set is bimodal.
Solution
Step 1 — Count the frequencies. 430 occurs twice, 480 occurs twice, and 495 occurs once.
Step 2 — Check the rule. Two different values tie for the highest frequency, so both are modes.
Answer: The data set is bimodal because the scores 430 and 480 each occur twice.
So when is the mode the best measure of the center? Consider a weight loss program that advertises a mean weight loss of six pounds in the first week. That mean could easily be carried by a few people who lost a great deal while most participants lost very little. If the mode says that most people lose two pounds in the first week, the program suddenly looks far less appealing — and the mode is the number that told you so.
The mean and the median need numbers, but the mode only needs counting. If the data set is red, red, red, green, green, yellow, purple, black, blue, the mode is red — and that is a perfectly good answer to "what color are these?" This makes the mode the only one of the three measures available for qualitative (categorical) data.
In practice you will rarely compute any of these three by hand outside of a homework set. Statistical software calculates the mean, median, and mode instantly, and many graphing calculators will do it too. In the real world people use software; the reason to learn the arithmetic is so you can tell when the software is answering a different question than the one you asked.
2.5.5 The Law of Large Numbers and the Mean
The Law of Large Numbers states that as the sample size drawn from a population increases, the sample mean \(\overline{x}\) becomes very likely to be closer and closer to the population mean \(\mu\).
The law says "very likely," not "guaranteed." A large sample can still miss — it is just far less likely to miss by much. Think of it as the noise averaging out rather than the truth being revealed.
This idea is developed in much more detail later in the text, but it is worth meeting now because it is the reason a single sample mean is treated as an estimate of something real rather than as an accident of whoever happened to get surveyed.
Here is a promise that makes the sample mean worth computing in the first place. The Law of Large Numbers says that if you take samples of larger and larger size from any population, then the mean of the sample is very likely to get closer and closer to \(\mu\), the population mean.
Definition 2.5.5 — As the sample size grows, the sample mean stops wandering and settles toward the population mean (illustrative trace).
Dale Brennan runs a polling company, and his budget covers surveying either 30 voters or 3,000 voters from the same population. Which sample's mean is more likely to land close to the population mean \(\mu\), and what is the name of the result that says so?
Solution
Step 1 — Compare the sample sizes. 3,000 is far larger than 30, and both samples come from the same population.
Step 2 — Apply the result. The Law of Large Numbers says the sample mean gets closer and closer to the population mean as the sample size increases.
Answer: The mean of the 3,000-voter sample is more likely to be close to \(\mu\), by the Law of Large Numbers. Note that this is a statement about likelihood — the small sample could get lucky, it is just much less dependable.
2.5.6 Sampling Distributions and Statistics
A statistic is a number calculated from a sample. The mean, the median, and the mode are all statistics, and there are many others. The sample mean \(\overline{x}\) is the statistic used to estimate the population mean \(\mu\).
That last sentence is the whole logic of inferential statistics in miniature: you cannot measure \(\mu\), so you compute \(\overline{x}\) from a sample and use it as your best available stand-in.
You can think of a sampling distribution as a relative frequency distribution built from a great many samples. (See Sampling and Data for a review of relative frequency.)
Suppose thirty randomly selected students were asked the number of movies they watched the previous week. The results are in the relative frequency table below.
| # of movies | Relative Frequency |
|---|---|
| 0 | 5/30 |
| 1 | 15/30 |
| 2 | 6/30 |
| 3 | 3/30 |
| 4 | 1/30 |
If you let the number of samples get very large — say, 300 million or more — the relative frequency table becomes a relative frequency distribution.
Try it in rāSHio
Open rāSHio, enter the thirty students' movie counts, then choose Graph → Frequency Table with Discrete values checked. The relative frequency column Table 2.5.1 prints as fifths and halves of thirty is built for you — and that column is exactly what turns into a sampling distribution once the number of samples gets large.
Definition 2.5.6 — A statistic is a number computed from a sample, standing in for a population number nobody can reach.
Elliot Hayes is reviewing the movie-watching survey above with his husband, and he points out that the value \(\overline{x}\) computed from those thirty students is a statistic. What is the corresponding population quantity it is being used to estimate, and why can we not simply compute that quantity directly?
Solution
Step 1 — Name the population quantity. The population mean, \(\mu\) — the mean number of movies watched per week by every student in the population, not just the thirty surveyed.
Step 2 — Explain why it is out of reach. Computing \(\mu\) would require surveying every member of the population, which is normally impossible, too expensive, or too slow to be worth doing.
Answer: \(\overline{x}\) estimates the population mean \(\mu\). We use a statistic from a sample precisely because the population parameter cannot practically be measured directly — and the Law of Large Numbers is what makes that substitution trustworthy.
2.5.7 Calculating the Mean of Grouped Frequency Tables
The midpoint of an interval is
$$m = \frac{\text{lower} + \text{upper}}{2}$$the average of the interval's lower and upper boundaries. It is the single value used to represent every data point falling inside that interval.
Using the midpoint assumes that within each interval, the data sits about as often above the middle as below it. Over several intervals those small errors tend to cancel, which is why the estimate usually lands close — and why it is still only an estimate.
With midpoints in hand, the mean of a grouped frequency table is
$$\mu = \frac{\sum fm}{\sum f}$$where \(f\) is the frequency of an interval and \(m\) is that interval's midpoint. Read it as "multiply each midpoint by how many values sit in its interval, add all of those up, and divide by the total number of values" — which is exactly the frequency-weighted mean from the start of this section.
Sometimes data arrives already summarized: you are told that 4 students scored somewhere between 62.5 and 68.5, but not what any one of them actually scored. When only grouped data is available you do not know the individual values — only the intervals and their frequencies — so you cannot compute an exact mean. What you can do is estimate it.
The plan is to keep the basic definition of the mean,
$$\text{mean} = \frac{\text{data sum}}{\text{number of data values}}$$and modify it to fit what a frequency table actually gives you. Since the individual values are unknown, we stand in the midpoint of each interval for every value inside it.
Definition 2.5.7 — The midpoint is the single value that stands in for every unknown value inside an interval.
Mai Xiong conducted a study on the effect that playing video games has on memory recall. As part of their study, they compiled the following data. What is the best estimate for the mean number of hours spent playing video games?
| Hours Teenagers Spend on Video Games | Number of Teenagers |
|---|---|
| 0-3.5 | 3 |
| 3.5-7.5 | 7 |
| 7.5-11.5 | 12 |
| 11.5-15.5 | 7 |
| 15.5-19.5 | 9 |
Solution
Step 1 — Compute the midpoints. \(\frac{0+3.5}{2} = 1.75\), \(\frac{3.5+7.5}{2} = 5.5\), \(\frac{7.5+11.5}{2} = 9.5\), \(\frac{11.5+15.5}{2} = 13.5\), and \(\frac{15.5+19.5}{2} = 17.5\).
Step 2 — Multiply each midpoint by its frequency and add.
$$\sum fm = 1.75(3) + 5.5(7) + 9.5(12) + 13.5(7) + 17.5(9) = 5.25 + 38.5 + 114 + 94.5 + 157.5 = 409.75$$Step 3 — Total the frequencies. \(\sum f = 3 + 7 + 12 + 7 + 9 = 38\) teenagers.
Step 4 — Divide.
$$\mu = \frac{409.75}{38} \approx 10.78$$Answer: The best estimate for the mean number of hours spent playing video games is about 10.78 hours.
A frequency table displaying professor Blount's last statistics test is shown. Find the best estimate of the class mean.
| Grade Interval | Number of Students |
|---|---|
| 50-56.5 | 1 |
| 56.5-62.5 | 0 |
| 62.5-68.5 | 4 |
| 68.5-74.5 | 4 |
| 74.5-80.5 | 2 |
| 80.5-86.5 | 3 |
| 86.5-92.5 | 4 |
| 92.5-98.5 | 1 |
Solution
Step 1 — Find the midpoint of every interval. Average each interval's two boundaries.
| Grade Interval | Midpoint |
|---|---|
| 50-56.5 | 53.25 |
| 56.5-62.5 | 59.5 |
| 62.5-68.5 | 65.5 |
| 68.5-74.5 | 71.5 |
| 74.5-80.5 | 77.5 |
| 80.5-86.5 | 83.5 |
| 86.5-92.5 | 89.5 |
| 92.5-98.5 | 95.5 |
Step 2 — Multiply each midpoint by its frequency and add.
$$\sum fm = 53.25(1) + 59.5(0) + 65.5(4) + 71.5(4) + 77.5(2) + 83.5(3) + 89.5(4) + 95.5(1) = 1460.25$$Step 3 — Add up the frequencies. \(\sum f = 1 + 0 + 4 + 4 + 2 + 3 + 4 + 1 = 19\) students.
Step 4 — Divide.
$$\mu = \frac{\sum fm}{\sum f} = \frac{1460.25}{19} = 76.86$$Answer: The best estimate of the class mean is 76.86.
Problem Set 2.5
Problem 1. Find the mean for the following frequency tables.
a)
| Grade | Frequency |
|---|---|
| 49.5-59.5 | 2 |
| 59.5-69.5 | 3 |
| 69.5-79.5 | 8 |
| 79.5-89.5 | 12 |
| 89.5-99.5 | 5 |
b)
| Daily Low Temperature | Frequency |
|---|---|
| 49.5-59.5 | 53 |
| 59.5-69.5 | 32 |
| 69.5-79.5 | 15 |
| 79.5-89.5 | 1 |
| 89.5-99.5 | 0 |
c)
| Points per Game | Frequency |
|---|---|
| 49.5-59.5 | 14 |
| 59.5-69.5 | 32 |
| 69.5-79.5 | 15 |
| 79.5-89.5 | 23 |
| 89.5-99.5 | 2 |
Solution
Step 1 — Find each interval's midpoint: every interval in all three tables is 10 units wide with the same boundaries, so the midpoints are the same throughout: \(\frac{49.5+59.5}{2} = 54.5\), then 64.5, 74.5, 84.5, and 94.5.
Step 2 — Table (a), multiply midpoint by frequency and add: each midpoint stands in for every value inside its interval.
$$\sum fm = 54.5(2) + 64.5(3) + 74.5(8) + 84.5(12) + 94.5(5) = 109 + 193.5 + 596 + 1014 + 472.5 = 2385$$The frequencies total \(\sum f = 2 + 3 + 8 + 12 + 5 = 30\), so
$$\mu = \frac{2385}{30} = 79.5$$Step 3 — Table (b), same two calculations: the frequencies are much larger here, but nothing about the method changes.
$$\sum fm = 54.5(53) + 64.5(32) + 74.5(15) + 84.5(1) + 94.5(0) = 2888.5 + 2064 + 1117.5 + 84.5 + 0 = 6154.5$$With \(\sum f = 53 + 32 + 15 + 1 + 0 = 101\),
$$\mu = \frac{6154.5}{101} \approx 60.94$$Step 4 — Table (c), one more time:
$$\sum fm = 54.5(14) + 64.5(32) + 74.5(15) + 84.5(23) + 94.5(2) = 763 + 2064 + 1117.5 + 1943.5 + 189 = 6077$$With \(\sum f = 14 + 32 + 15 + 23 + 2 = 86\),
$$\mu = \frac{6077}{86} \approx 70.66$$Answer: (a) 79.5 (b) about 60.94 (c) about 70.66.
Camila Reyes keeps a slip at the marina with her wife, and she recorded the lengths of all the boats moored there. The data are ordered from smallest to largest: 16; 17; 19; 20; 20; 21; 23; 24; 25; 25; 25; 26; 26; 27; 27; 27; 28; 29; 30; 32; 33; 33; 34; 35; 37; 39; 40. Use this data for the next three problems.
Problem 2. Calculate the mean.
Solution
Step 1 — Count the values: there are 27 boat lengths in the ordered list.
Step 2 — Add them all up: working left to right through the list gives a total of 738 feet.
Step 3 — Divide by the number of values: that is the definition of the mean.
$$\overline{x} = \frac{738}{27} \approx 27.33$$Answer: The mean boat length is about 27.33 feet.
Problem 3. Identify the median.
Solution
Step 1 — Find the location of the median: with \(n = 27\) values,
$$\frac{n+1}{2} = \frac{27+1}{2} = 14$$Because 27 is odd the location is a whole number, so the median is a single value — no averaging needed.
Step 2 — Count in to the 14th value: the data is already ordered, so count from the smallest: 16, 17, 19, 20, 20, 21, 23, 24, 25, 25, 25, 26, 26, 27.
Answer: The median is 27 feet.
Problem 4. Identify the mode.
Solution
Step 1 — Tally how often each length appears: 20 appears twice, 25 appears three times, 26 appears twice, 27 appears three times, 33 appears twice, and every other length appears once.
Step 2 — Pick out the highest frequency: three is the largest count, and two different lengths reach it — 25 and 27.
Answer: The data set is bimodal, with modes 25 feet and 27 feet.
Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in one week. Fourteen people answered that they generally sell three cars; nineteen generally sell four cars; twelve generally sell five cars; nine generally sell six cars; eleven generally sell seven cars. Use this data for the next three problems.
Problem 5. Calculate the sample mean.
Solution
Step 1 — Write the data as values with frequencies: 14 salespeople sell 3 cars, 19 sell 4, 12 sell 5, 9 sell 6, and 11 sell 7.
Step 2 — Total the frequencies: this is the sample size.
$$n = 14 + 19 + 12 + 9 + 11 = 65$$Step 3 — Multiply each value by its frequency and add: this is the frequency-weighted form of the mean.
$$\sum fx = 3(14) + 4(19) + 5(12) + 6(9) + 7(11) = 42 + 76 + 60 + 54 + 77 = 309$$Step 4 — Divide:
$$\overline{x} = \frac{309}{65} \approx 4.75$$Answer: The sample mean is about 4.75 cars per week.
Problem 6. Identify the median.
Solution
Step 1 — Find the location of the median: with \(n = 65\),
$$\frac{n+1}{2} = \frac{65+1}{2} = 33$$so the median is the 33rd value in the ordered list.
Step 2 — Build a running count to find where the 33rd value falls: locations 1 through 14 are the salespeople who sell 3 cars; locations 15 through \(14 + 19 = 33\) are the ones who sell 4 cars.
Step 3 — Read off location 33: it lands inside that second block.
Answer: The median is 4 cars.
Problem 7. Identify the mode.
Solution
Step 1 — Compare the frequencies: the counts are 14, 19, 12, 9, and 11 for 3, 4, 5, 6, and 7 cars respectively.
Step 2 — Take the largest one: 19 is the highest frequency, and it belongs to the value 4. No other value ties it, so the data set has a single mode.
Answer: The mode is 4 cars.
Problem 8. The countries with the highest rates of obesity in the world have obesity rates that range from 11.4% to 74.6%. This data is summarized in the following table.
| Percent of Population | Number of Countries |
|---|---|
| 11.4-20.45 | 29 |
| 20.45-29.45 | 13 |
| 29.45-38.45 | 4 |
| 38.45-47.45 | 0 |
| 47.45-56.45 | 2 |
| 56.45-65.45 | 1 |
| 65.45-74.45 | 0 |
| 74.45-83.45 | 1 |
a) What is the best estimate of the average obesity percentage for these countries?
b) The United States has an average obesity rate of 33.9%. Is this rate above average or below?
c) How does the United States compare to other countries?
Solution
Step 1 — Compute the midpoint of each interval: average the two boundaries of each row.
15.925, 24.95, 33.95, 42.95, 51.95, 60.95, 69.95, 78.95
Step 2 — Multiply each midpoint by its frequency and add: intervals with a frequency of 0 contribute nothing.
$$\sum fm = 15.925(29) + 24.95(13) + 33.95(4) + 51.95(2) + 60.95(1) + 78.95(1) = 1165.775$$Step 3 — Total the frequencies:
$$\sum f = 29 + 13 + 4 + 0 + 2 + 1 + 0 + 1 = 50$$Step 4 — Divide to estimate the mean (part a):
$$\mu = \frac{1165.775}{50} \approx 23.32$$Step 5 — Place the United States against that estimate (part b): the U.S. rate of 33.9% is larger than the estimated average of 23.32%, so it sits above average.
Step 6 — Say where the U.S. lands in the whole distribution (part c): 42 of the 50 countries fall in the two lowest intervals, which top out at 29.45%. The U.S. rate of 33.9% falls in the 29.45-38.45 interval, so only a handful of countries in the table have a higher obesity rate than the United States.
Answer: (a) about 23.32% (b) above average (c) the United States has a higher obesity rate than most of the countries in this table — the great majority of them sit below 29.45%.
Problem 9. The table below gives the percent of children under five considered to be underweight. What is the best estimate for the mean percentage of underweight children?
| Percent of Underweight Children | Number of Countries |
|---|---|
| 16-21.45 | 23 |
| 21.45-26.9 | 4 |
| 26.9-32.35 | 9 |
| 32.35-37.8 | 7 |
| 37.8-43.25 | 6 |
| 43.25-48.7 | 1 |
Solution
Step 1 — Compute the midpoint of each interval:
18.725, 24.175, 29.625, 35.075, 40.525, 45.975
Step 2 — Multiply each midpoint by its frequency and add:
$$\sum fm = 18.725(23) + 24.175(4) + 29.625(9) + 35.075(7) + 40.525(6) + 45.975(1) = 1328.65$$Step 3 — Total the frequencies:
$$\sum f = 23 + 4 + 9 + 7 + 6 + 1 = 50$$Step 4 — Divide:
$$\mu = \frac{1328.65}{50} \approx 26.57$$Answer: The best estimate for the mean percentage of underweight children is about 26.57%.
Key Terms
mean (arithmetic mean) — the sum of all data values divided by the number of values; written \(\overline{x}\) for a sample and \(\mu\) for a population.
median — the value that splits the ordered data into two halves, whose location in the ordered list is \(\frac{n+1}{2}\).
mode — the value that occurs most frequently in a data set; the only center measure that also works on qualitative data.
bimodal — describing a data set with exactly two modes.
Law of Large Numbers — the result that a sample mean grows very likely to approach the population mean \(\mu\) as the sample size increases.
statistic — a number calculated from a sample, such as \(\overline{x}\), used to estimate a population parameter.
sampling distribution — a relative frequency distribution built from a great many samples.
frequency table — a data representation in which grouped data is displayed alongside the corresponding frequencies.
midpoint — the average of an interval's lower and upper boundaries, used to represent every value inside that interval.