2.3 Measures of the Location 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.
Here the summary statistics are quartiles, percentiles, and the interquartile range, plus the 1.5 × IQR rule for flagging outliers. Half the section is interpretation alone — writing the sentence that says what a percentile means in context, and whether landing high in it is good news or bad.
Learning Objectives
By the end of this section, you will be able to:
- order a data set and locate its median, first quartile, and third quartile;
- calculate the interquartile range and use the 1.5 × IQR rule to flag potential outliers;
- find the kth percentile of a data set, and find the percentile of a given data value;
- read quartiles and percentiles off the cumulative relative frequency column of a frequency table;
- write a plain-English interpretation of a percentile that names the context, the value, and the percent falling above and below it.
The previous two sections asked what shape does this data have? This one asks a narrower and more practical question: where does one particular value sit inside the pile? A score of 1220 on the SAT means nothing on its own. A score of 1220 that beats three out of every four test-takers means something immediately. Measures of location are the tools that turn a bare number into a position.
The two common measures of location are quartiles and percentiles.
2.3.1 Quartiles, Percentiles, and the Median
The median is the number that separates ordered data into halves. Half of the values are the same number or smaller than the median, and half are the same number or larger. The median measures the "center" of the data, but it does not have to be one of the observed values.
You can think of the median as the "middle value," as long as you remember that with an even number of observations there is no single middle value — there are two, and the median is their average.
Here is the idea on a small data set. Consider the following fourteen values:
1; 11.5; 6; 7.2; 4; 8; 9; 10; 6.8; 8.3; 2; 2; 10; 1
Ordered from smallest to largest:
1; 1; 2; 2; 4; 6; 6.8; 7.2; 8; 8.3; 9; 10; 10; 11.5
Since there are 14 observations, the median sits between the seventh value, 6.8, and the eighth value, 7.2. To find it, add the two values and divide by two.
$$ \frac{6.8 + 7.2}{2} = \frac{14}{2} = 7 $$The median is seven. Half of the values are smaller than seven and half are larger — and notice that seven is not itself in the data set. That is allowed. The median marks a place on the number line, not a member of the list.
Definition 2.3.1 — The median is the place where an ordered list splits into two equal halves, not necessarily a value in the list.
Quartiles are the numbers that separate ordered data into quarters. There are three of them, labeled \(Q_1\), \(Q_2\), and \(Q_3\):
- about one-fourth of the data falls on or below the first quartile, \(Q_1\);
- about one-half of the data falls on or below the second quartile, \(Q_2\) — which is just another name for the median;
- about three-fourths of the data falls on or below the third quartile, \(Q_3\).
Quartiles may or may not be part of the data.
To find the quartiles, find the median first. The first quartile, \(Q_1\), is then the middle value of the lower half of the data, and the third quartile, \(Q_3\), is the middle value — the median — of the upper half. In other words, you take the median three times: once on the whole list, then once on each half.
Take the same ordered data set again:
1; 1; 2; 2; 4; 6; 6.8; | 7.2; 8; 8.3; 9; 10; 10; 11.5
The median, or second quartile, is seven. The lower half of the data is 1, 1, 2, 2, 4, 6, 6.8, and the middle value of that half is two.
$$ 1;\ 1;\ 2;\ \mathbf{2};\ 4;\ 6;\ 6.8 $$So \(Q_1 = 2\). One-fourth of the values are two or less, and three-fourths are more than two. Here the quartile is one of the data values — sometimes it is, sometimes it isn't.
The upper half of the data is 7.2, 8, 8.3, 9, 10, 10, 11.5, and its middle value is nine. So the third quartile, \(Q_3\), is nine. Three-fourths (75%) of the ordered data set is less than nine, and one-fourth (25%) is greater than nine.
Definition 2.3.2 — Quartiles come from taking the median three times: once on the whole list, then once on each half.
Percentiles divide ordered data into 100 equal parts. The \(k\)th percentile is the value at or below which \(k\) percent of the data falls. Percentiles may or may not be part of the data.
Scoring in the 90th percentile does not mean you got 90% on the test. It means 90% of scores were the same as or lower than yours. You could have answered half the questions wrong and still land there, if everyone else did worse.
Quartiles are just the percentiles you use most often: \(Q_1\) is the 25th percentile, the median is the 50th percentile, and \(Q_3\) is the 75th percentile.
That distinction is exactly why colleges lean on percentiles so heavily. A raw SAT score changes meaning every year as tests change; a percentile does not. When a university says it accepts scores at or above the 75th percentile, it is saying "we take the top quarter of test-takers," and that sentence stays true no matter how the test is scaled. For that particular admissions rule, the 75th percentile translates into a score of at least 1220.
Percentiles are mostly used with very large populations. That is why it is acceptable to say "90% of the test scores are less than your score" rather than the more careful "the same or less" — with tens of thousands of test-takers, removing one particular data value does not move the answer.
Every measure in this section starts the same way: put the data in order from smallest to largest. Nothing here works on an unsorted list, because "position in the pile" is meaningless until there is a pile to be positioned in.
Definition 2.3.3 — A percentile is a position in the ranked crowd: 90 of 100 at or below, 10 at or above.
Quinn asked eleven classmates how many minutes they spent commuting to campus this morning, and wrote the answers in their notebook:
24; 8; 40; 15; 8; 32; 12; 55; 20; 18; 30
Find the median, the first quartile, and the third quartile.
Solution
Step 1 — Order the data.
8; 8; 12; 15; 18; 20; 24; 30; 32; 40; 55
Step 2 — Find the median. There are 11 values, an odd count, so the median is the single middle value — the 6th one.
$$ Q_2 = 20 $$Step 3 — Find \(Q_1\). The lower half is the five values below the median: 8, 8, 12, 15, 18. Its middle value is the 3rd.
$$ Q_1 = 12 $$Step 4 — Find \(Q_3\). The upper half is the five values above the median: 24, 30, 32, 40, 55. Its middle value is the 3rd.
$$ Q_3 = 32 $$Answer: median = 20 minutes, \(Q_1 = 12\) minutes, \(Q_3 = 32\) minutes. A quarter of the class commutes 12 minutes or less; a quarter commutes 32 minutes or more.
2.3.2 The Interquartile Range and Potential Outliers
The interquartile range (IQR) is a number that indicates the spread of the middle half — the middle 50% — of the data. It is the difference between the third quartile and the first quartile:
$$ IQR = Q_3 - Q_1 $$Because the IQR ignores everything outside the middle 50%, a single wild value at either end cannot inflate it. That is its whole point, and it is what makes the next definition possible.
Definition 2.3.4 — The interquartile range measures only the middle half, so the extremes cannot inflate it.
A value is suspected to be a potential outlier if it is more than \(1.5 \times IQR\) below the first quartile, or more than \(1.5 \times IQR\) above the third quartile. In symbols, a value is a potential outlier when it falls outside the interval
$$ \left[\, Q_1 - 1.5(IQR),\ \ Q_3 + 1.5(IQR) \,\right] $$It beeps at anything unusual. Sometimes that is a nail in your shoe, sometimes it is buried treasure — but you always dig.
A potential outlier is a data point that is significantly different from the other data points. It may be an error, it may be some kind of abnormality, or it may be the single most important thing in the data set. Potential outliers always require further investigation.
The two boundary numbers in that interval are often called the fences. Anything inside the fences is ordinary; anything outside gets a second look. Notice the word potential — the rule flags candidates, it does not convict them. A typo where someone entered 5,500,000 instead of 550,000 is an outlier you delete. A genuine mansion in a neighborhood of starter homes is an outlier you keep and then talk about. The arithmetic cannot tell you which one you have; only knowing the situation can.
The two worked examples in this subsection run the rule end to end. In the first, the flagged value is so far outside the upper fence that it is obvious even before the arithmetic; in the second, two values sneak past a fence that is much closer in. Both examples follow the same four moves: order the data, find the three quartiles, compute the IQR, then build the fences and see what falls outside them. Once you have done it twice, the procedure stops feeling like a formula and starts feeling like a checklist.
Knowing where the middle of the data sits is useful. Knowing how wide that middle is tells you something the median alone cannot.
Definition 2.3.5 — Fences step 1.5 × IQR out from each quartile; anything beyond them is a potential outlier.
For the following 11 salaries, calculate the \(IQR\) and determine if any salaries are outliers. The salaries are in dollars.
$33,000; $64,500; $28,000; $54,000; $72,000; $68,500; $69,000; $42,000; $54,000; $120,000; $40,500
Solution
Step 1 — Order the data.
28,000; 33,000; 40,500; 42,000; 54,000; 54,000; 64,500; 68,500; 69,000; 72,000; 120,000
Step 2 — Find the median. There are 11 values, so the median is the 6th.
$$ M = \$54{,}000 $$Step 3 — Find the quartiles. The lower half is the five values below the median (28,000; 33,000; 40,500; 42,000; 54,000), so \(Q_1\) is its 3rd value. The upper half is the five values above it (64,500; 68,500; 69,000; 72,000; 120,000).
$$ Q_1 = \$40{,}500 \qquad Q_3 = \$69{,}000 $$Step 4 — Compute the IQR and the fences.
$$ IQR = 69{,}000 - 40{,}500 = \$28{,}500 $$ $$ 1.5 \times IQR = \$42{,}750 $$ $$ Q_1 - 1.5(IQR) = 40{,}500 - 42{,}750 = -\$2{,}250 $$ $$ Q_3 + 1.5(IQR) = 69{,}000 + 42{,}750 = \$111{,}750 $$Answer: \(IQR = \$28{,}500\). The salary of \(\$120{,}000\) exceeds the upper fence of \(\$111{,}750\), so it is a potential outlier. No salary falls below the lower fence.
For the following 13 real estate prices, calculate the \(IQR\) and determine if any prices are potential outliers. Prices are in dollars.
389,950; 230,500; 158,000; 479,000; 639,000; 114,950; 5,500,000; 387,000; 659,000; 529,000; 575,000; 488,800; 1,095,000
Solution
Step 1 — Order the data from smallest to largest.
114,950; 158,000; 230,500; 387,000; 389,950; 479,000; 488,800; 529,000; 575,000; 639,000; 659,000; 1,095,000; 5,500,000
Step 2 — Find the median. There are 13 values, so the median is the 7th.
$$ M = \$488{,}800 $$Step 3 — Find \(Q_1\). The lower half is the six values below the median. Its middle is the average of the 3rd and 4th of those.
$$ Q_1 = \frac{230{,}500 + 387{,}000}{2} = \$308{,}750 $$Step 4 — Find \(Q_3\). The upper half is the six values above the median.
$$ Q_3 = \frac{639{,}000 + 659{,}000}{2} = \$649{,}000 $$Step 5 — Compute the IQR.
$$ IQR = 649{,}000 - 308{,}750 = \$340{,}250 $$Step 6 — Build the fences.
$$ 1.5 \times IQR = 1.5 \times 340{,}250 = 510{,}375 $$ $$ Q_1 - 1.5(IQR) = 308{,}750 - 510{,}375 = -\$201{,}625 $$ $$ Q_3 + 1.5(IQR) = 649{,}000 + 510{,}375 = \$1{,}159{,}375 $$Step 7 — Compare. No house price is less than \(-\$201{,}625\) — no house has a negative price, so the lower fence catches nothing. But \(\$5{,}500{,}000\) is well above \(\$1{,}159{,}375\).
Answer: \(IQR = \$340{,}250\), and \(\$5{,}500{,}000\) is a potential outlier. Note that \(\$1{,}095{,}000\) — the second-most expensive house — sits just under the fence and is not flagged.
Try it in rāSHio
Open rāSHio, paste the thirteen house prices (389950; 230500; 158000; …) into File → Delimited List… — the semicolons parse as-is — then choose Stats → Summary Statistics. The quartiles and the \(IQR\) you just worked out by hand (\(\$308{,}750\), \(\$649{,}000\), \(\$340{,}250\)) come back in one panel, so you can check the arithmetic and then spend your effort on the part the tool will not do for you — deciding what to make of that \(\$5{,}500{,}000\).
Figure 2.3.1 — Reading quartiles and the IQR in rāSHio: Stats → Summary Statistics.
Marisol tutors two study groups and wants to know which one is more spread out. Find the interquartile range for her two data sets and compare them.
Test scores for Class A: 69; 96; 81; 79; 65; 76; 83; 99; 89; 67; 90; 77; 85; 98; 66; 91; 77; 69; 80; 94
Test scores for Class B: 90; 72; 80; 92; 90; 97; 92; 75; 79; 68; 70; 80; 99; 95; 78; 73; 71; 68; 95; 100
Solution
Step 1 — Order Class A.
65; 66; 67; 69; 69; 76; 77; 77; 79; 80; 81; 83; 85; 89; 90; 91; 94; 96; 98; 99
With 20 values, the lower half is the first ten and the upper half is the last ten.
$$ Q_1 = \frac{69 + 76}{2} = 72.5 \qquad Q_3 = \frac{90 + 91}{2} = 90.5 $$ $$ IQR_A = 90.5 - 72.5 = 18 $$Step 2 — Order Class B.
68; 68; 70; 71; 72; 73; 75; 78; 79; 80; 80; 90; 90; 92; 92; 95; 95; 97; 99; 100
$$ Q_1 = \frac{72 + 73}{2} = 72.5 \qquad Q_3 = \frac{92 + 95}{2} = 93.5 $$ $$ IQR_B = 93.5 - 72.5 = 21 $$Answer: \(IQR_A = 18\) and \(IQR_B = 21\). The two classes start their middle half at exactly the same place (\(Q_1 = 72.5\) for both), but Class B's middle half reaches three points higher, so Class B's scores are slightly more spread out through the middle.
Test scores for a college statistics class held during the day are:
99; 56; 78; 55.5; 32; 90; 80; 81; 56; 59; 45; 77; 84.5; 84; 70; 72; 68; 32; 79; 90
Test scores for a college statistics class held during the evening are:
98; 78; 68; 83; 81; 89; 88; 76; 65; 45; 98; 90; 80; 84.5; 85; 79; 78; 98; 90; 79; 81; 25.5
For the two data sets, find the following:
a) The interquartile range. Compare the two interquartile ranges.
b) Any outliers in either set.
Solution
Step 1 — Order each set and read off its five-number summary.
| Class | Minimum | Q1 | Median | Q3 | Maximum |
|---|---|---|---|---|---|
| Day | 32 | 56 | 74.5 | 82.5 | 99 |
| Night | 25.5 | 78 | 81 | 89 | 98 |
Step 2 (part a) — Compute each IQR.
$$ IQR_{\text{day}} = Q_3 - Q_1 = 82.5 - 56 = 26.5 $$ $$ IQR_{\text{night}} = Q_3 - Q_1 = 89 - 78 = 11 $$The interquartile range — the spread, or variability — for the day class is more than twice that of the night class. This suggests more variation in the day class's test scores: its middle half is stretched across 26.5 points, while the night class's middle half is packed into 11.
Step 3 (part b) — Build the fences for the day class.
$$ Q_1 - 1.5(IQR) = 56 - 1.5(26.5) = 16.25 $$ $$ Q_3 + 1.5(IQR) = 82.5 + 1.5(26.5) = 122.25 $$Since the minimum (32) is greater than 16.25 and the maximum (99) is less than 122.25, the day class has no outliers.
Step 4 — Build the fences for the night class.
$$ Q_1 - 1.5(IQR) = 78 - 1.5(11) = 61.5 $$ $$ Q_3 + 1.5(IQR) = 89 + 1.5(11) = 105.5 $$Any night-class test score less than 61.5 is an outlier, so the scores of 45 and 25.5 are outliers. No score is greater than 105.5, so there is no upper-end outlier.
Answer: The day class has the larger IQR (26.5 versus 11) but no outliers; the tightly clustered night class has two low outliers, 45 and 25.5. This is worth sitting with — the more consistent class is the one that produced outliers, precisely because its narrow IQR makes narrow fences.
Try it in rāSHio
Open rāSHio, put the day scores in one column and the evening scores in another, then choose Graph → Box Plot. Each box is the five-number summary drawn to scale — the box spans \(Q_1\) to \(Q_3\), so the day class’s wider box is the larger \(IQR\), and the night class’s two low outliers are plotted as their own points beyond the whisker. This is the picture §2.4 builds by hand.
Figure 2.3.2 — Comparing two five-number summaries in rāSHio: Graph → Box Plot.
2.3.3 Reading Quartiles and Percentiles from a Frequency Table
You will not always be handed a raw list. Often the data arrives already summarized in a frequency table — and the cumulative relative frequency column is a percentile lookup table hiding in plain sight. Each entry in that column tells you what fraction of the data sits at or below that row, which is exactly the question a percentile asks.
Cody dispatches for a transit agency. He asked forty bus drivers how many hours they spend each day running their routes (rounded to the nearest hour). Find the 65th percentile.
| Amount of time spent on route (hours) | Frequency | Relative frequency | Cumulative relative frequency |
|---|---|---|---|
| 2 | 12 | 0.30 | 0.30 |
| 3 | 14 | 0.35 | 0.65 |
| 4 | 10 | 0.25 | 0.90 |
| 5 | 4 | 0.10 | 1.00 |
Solution
Step 1 — Find 0.65 in the cumulative relative frequency column. It sits on the "3 hours" row, and it lands there exactly.
Step 2 — Convert to a count. Sixty-five percent of 40 data values is 26 values, and there are exactly 26 values at or below three hours (twelve 2s plus fourteen 3s).
Step 3 — The percentile falls between the 26th and 27th values. The 26th value is a three (the last of the 3s) and the 27th is a four (the first of the 4s).
$$ P_{65} = \frac{3 + 4}{2} = 3.5 \text{ hours} $$Answer: the 65th percentile is 3.5 hours. This is the same situation as the 28th percentile in the sleep-hours example in this subsection — when a cumulative relative frequency lands exactly on your target percent, the percentile falls on the boundary between two rows, so you average them.
Yasmin asked the fifty students in her statistics class how much sleep they get per school night (rounded to the nearest hour). The results were:
| Amount of sleep per school night (hours) | Frequency | Relative frequency | Cumulative relative frequency |
|---|---|---|---|
| 4 | 2 | 0.04 | 0.04 |
| 5 | 5 | 0.10 | 0.14 |
| 6 | 7 | 0.14 | 0.28 |
| 7 | 12 | 0.24 | 0.52 |
| 8 | 14 | 0.28 | 0.80 |
| 9 | 7 | 0.14 | 0.94 |
| 10 | 3 | 0.06 | 1.00 |
Find the 28th percentile, the median, and the third quartile.
Solution
Step 1 — Find the 28th percentile. Notice the 0.28 sitting in the cumulative relative frequency column, on the "6 hours" row. Twenty-eight percent of 50 data values is 14 values, and there are exactly 14 values at or below six hours: the two 4s, the five 5s, and the seven 6s. So the 28th percentile falls between the last six and the first seven.
$$ P_{28} = \frac{6 + 7}{2} = 6.5 \text{ hours} $$Step 2 — Find the median. Look again at the cumulative relative frequency column and find 0.52. The median is the 50th percentile, or the second quartile, and 50% of 50 is 25. There are 25 values at or below the median: the two 4s, the five 5s, the seven 6s, and eleven of the 7s. The median is therefore between the 25th value (a seven) and the 26th value (also a seven).
$$ M = 7 \text{ hours} $$Step 3 — Find the third quartile. The third quartile is the same as the 75th percentile, and you can "eyeball" this one. The cumulative column jumps from 0.52 to 0.80: once you have all the fours, fives, sixes, and sevens you have 52% of the data, and once you include all the eights you have 80%. The 75th percentile must therefore be an eight.
Another way to see it: find 75% of 50, which is 37.5, and round up to 38. The third quartile is the 38th value, which is an eight.
$$ Q_3 = 8 \text{ hours} $$You can check this by counting: there are 37 values below the third quartile and 12 above it.
Answer: the 28th percentile is 6.5 hours, the median is 7 hours, and \(Q_3\) is 8 hours.
Try it in rāSHio
Open rāSHio, enter the fifty sleep answers, then choose Graph → Frequency Table with Discrete values checked. The cumulative relative frequency column you read 0.28, 0.52 and 0.80 off is built for you — which is the whole trick of this subsection, since that column is a percentile lookup table.
Figure 2.3.3 — Building the cumulative relative frequency column in rāSHio: Graph → Frequency Table with Discrete values checked.
2.3.4 A Formula for Finding the kth Percentile
Let \(k\) be the percentile you want, \(i\) be the index (the ranking or position of a data value), and \(n\) be the total number of data values. To find the \(k\)th percentile:
- order the data from smallest to largest;
- calculate the index
- if \(i\) is an integer, the \(k\)th percentile is the data value in the \(i\)th position of the ordered set;
- if \(i\) is not an integer, round \(i\) both down and up to the nearest integers, and average the two data values in those two positions.
Read the formula as a set of instructions for counting, not as algebra to be solved. It hands you a position in the line, and then you walk down the ordered list until you reach that position. The \((n+1)\) is there because you are counting the gaps between values, not the values themselves — with 29 values there are 30 gaps to spread the 100 percentage points across. The worked example in this subsection works both cases: one where the index lands on a whole number, and one where it lands between two.
Eyeballing a cumulative column works when the percent you want happens to line up with a row. When it doesn't, you need arithmetic. If you were to do a little research, you would find several formulas for calculating the \(k\)th percentile; here is one of them.
Definition 2.3.6 — The index formula returns a position: a whole index lands on one value, a fractional index between two.
Listed are 29 ages for Academy Award winning best actors:
18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77
Calculate the 20th percentile and the 55th percentile.
Solution
The data is ordered and \(n = 29\), so \(n + 1 = 30\).
The 20th percentile.
$$ i = \frac{20}{100}(30) = 6 $$Six is an integer, so take the 6th value in the ordered list.
$$ P_{20} = 27 \text{ years} $$The 55th percentile.
$$ i = \frac{55}{100}(30) = 16.5 $$Not an integer, so round down to 16 and up to 17. The 16th age is 52 and the 17th is 55.
$$ P_{55} = \frac{52 + 55}{2} = 53.5 \text{ years} $$Answer: the 20th percentile is 27 years and the 55th percentile is 53.5 years.
The same index formula works when the data arrives as a frequency table instead of a list — you just have to count your way down the frequency column to find which value occupies the \(i\)th position. That is the only extra step, and the sleep-table example in this subsection does it three times over.
Percentiles are one of the first things any statistics calculator, spreadsheet, or online tool will compute for you. Learn the index formula anyway — knowing what the tool is doing is what lets you catch it when it disagrees with you.
Harriet is writing an article about Academy Award winning best actors, and she has collected 29 of their ages:
18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77
a) Find the 70th percentile.
b) Find the 83rd percentile.
Solution
The data is already ordered, and \(n = 29\).
Part (a) — the 70th percentile. Here \(k = 70\):
$$ i = \frac{k}{100}(n + 1) = \frac{70}{100}(29 + 1) = 21 $$Twenty-one is an integer, so the 70th percentile is simply the data value in the 21st position. Counting down the ordered list, the 21st age is 64.
$$ P_{70} = 64 \text{ years} $$Part (b) — the 83rd percentile. Here \(k = 83\):
$$ i = \frac{k}{100}(n + 1) = \frac{83}{100}(29 + 1) = 24.9 $$This is not an integer, so round it down to 24 and up to 25. The age in the 24th position is 71 and the age in the 25th position is 72. Average them.
$$ P_{83} = \frac{71 + 72}{2} = 71.5 \text{ years} $$Answer: the 70th percentile is 64 years, and the 83rd percentile is 71.5 years.
Chayton is reviewing the same route data in Table 2.3.2. He needs the third quartile. Find it, and give another name for the third quartile.
Solution
Step 1 — Translate. The third quartile is another name for the 75th percentile.
Step 2 — Compute the index. There are 40 data values, so \(n = 40\) and \(n + 1 = 41\).
$$ i = \frac{75}{100}(41) = 30.75 $$Step 3 — Locate the 30th and 31st values. Not an integer, so average those two positions. Counting down the frequency column: values 1–12 are 2 hours, values 13–26 are 3 hours, values 27–36 are 4 hours. Both the 30th and the 31st values fall in the 4-hour block.
$$ Q_3 = \frac{4 + 4}{2} = 4 \text{ hours} $$Answer: the third quartile is 4 hours, also called the 75th percentile.
Your instructor or a member of the class will ask everyone in class how many sweaters they own. Answer the following questions:
1. How many students were surveyed?
2. What kind of sampling did you do?
3. Construct two different histograms. For each, record the starting value and the ending value.
4. Find the median, first quartile, and third quartile.
5. Construct a table of the data to find the following:
a. the 10th percentile
b. the 70th percentile
c. the percent of students who own less than four sweaters
Try it in rāSHio
Open rāSHio and use File → Delimited List… to paste the class’s sweater counts straight in as you collect them, rather than retyping them cell by cell. Once the column is loaded, the median, \(Q_1\) and \(Q_3\) that question 4 asks for are one click away under Stats → Summary Statistics.
Figure 2.3.4 — Getting collected data into rāSHio: File → Delimited List…
Using Table 2.3.3:
a) Find the 80th percentile.
b) Find the 90th percentile.
c) Find the first quartile. What is another name for the first quartile?
Solution
Using the data from the frequency table, notice there are 50 data values, so \(n = 50\) and \(n + 1 = 51\).
Part (a) — the 80th percentile.
$$ i = \frac{80}{100}(50 + 1) = 40.8 $$Since \(i = 40.8\) is not an integer, calculate the mean of the 40th and 41st data values. Counting down the frequency column, the 40th data value is 8 and the 41st is 9.
$$ P_{80} = \frac{8 + 9}{2} = 8.5 \text{ hours} $$Part (b) — the 90th percentile.
$$ i = \frac{90}{100}(50 + 1) = 45.9 $$Again not an integer, so average the 45th and 46th data values. The 45th data value is 9 and the 46th is also 9.
$$ P_{90} = \frac{9 + 9}{2} = 9 \text{ hours} $$Part (c) — the first quartile. Another name for the first quartile is the 25th percentile, so calculate that:
$$ i = \frac{25}{100}(50 + 1) = 12.75 $$Since \(i = 12.75\) is not an integer, average the 12th and 13th data values. Both are 6.
$$ Q_1 = \frac{6 + 6}{2} = 6 \text{ hours} $$Answer: the 80th percentile is 8.5 hours, the 90th percentile is 9 hours, and the first quartile is 6 hours — also called the 25th percentile.
2.3.5 A Formula for Finding the Percentile of a Value in a Data Set
To find the percentile of a given value in a data set:
- order the data from smallest to largest;
- let \(x\) be the number of data values counting from the bottom of the list up to but not including the value whose percentile you want;
- let \(y\) be the number of data values equal to that value;
- let \(n\) be the total number of data values;
- calculate
then round to the nearest integer.
The \(0.5y\) term is the piece worth pausing on. If three people tie at your exact score, it would be unfair to count all three as "below you," and equally unfair to count none of them. Splitting the tie down the middle — counting half of them — is the compromise the formula makes.
So far you have started with a percent and asked "which value sits there?" Now run the question backwards: you are handed a value, and you want the percent. This is the direction that matters when someone tells you their score and you want to know how they did.
Definition 2.3.7 — Values below the target count in full and tied values count half, which is what the 0.5y term does.
Listed are 30 ages for Academy Award winning best actors, in order from smallest to largest:
18; 21; 22; 25; 26; 27; 29; 30; 31; 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77
Find the percentiles for 47 and 31.
Solution
Here \(n = 30\). Note this list contains two 31s — that is the detail the \(0.5y\) term exists for.
The percentile for 47. Counting from the bottom, 15 values are less than 47 (18, 21, 22, 25, 26, 27, 29, 30, 31, 31, 33, 36, 37, 41, 42), and one value equals 47. So \(x = 15\) and \(y = 1\).
$$ \frac{15 + 0.5(1)}{30}(100) = \frac{15.5}{30}(100) = 51.67 $$Rounded, 47 is the 52nd percentile.
The percentile for 31. Counting from the bottom, 8 values are less than 31 (18, 21, 22, 25, 26, 27, 29, 30), and two values equal 31. So \(x = 8\) and \(y = 2\).
$$ \frac{8 + 0.5(2)}{30}(100) = \frac{9}{30}(100) = 30 $$So 31 is the 30th percentile.
Answer: 47 is the 52nd percentile and 31 is the 30th percentile.
Listed are 29 ages for Academy Award winning best actors:
18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77
a) Find the percentile for 58.
b) Find the percentile for 25.
Solution
Here \(n = 29\).
Part (a) — the percentile for 58. Counting from the bottom of the list, there are 18 data values less than 58, and there is one value equal to 58. So \(x = 18\) and \(y = 1\).
$$ \frac{x + 0.5y}{n}(100) = \frac{18 + 0.5(1)}{29}(100) = \frac{18.5}{29}(100) = 63.8 $$Rounded to the nearest integer, 58 is the 64th percentile.
Part (b) — the percentile for 25. Counting from the bottom, there are three data values less than 25 (18, 21, 22), and there is one value equal to 25. So \(x = 3\) and \(y = 1\).
$$ \frac{x + 0.5y}{n}(100) = \frac{3 + 0.5(1)}{29}(100) = \frac{3.5}{29}(100) = 12.07 $$Rounded, 25 is the 12th percentile.
Answer: 58 is the 64th percentile and 25 is the 12th percentile of these ages.
2.3.6 Interpreting Percentiles, Quartiles, and the Median
A percentile indicates the relative standing of a data value when the data is sorted into numerical order from smallest to largest. Percentages of data values are less than or equal to the \(p\)th percentile: for example, 15% of data values are less than or equal to the 15th percentile.
- Low percentiles always correspond to lower data values.
- High percentiles always correspond to higher data values.
What a percentile does not carry is a value judgment. Whether a certain percentile is "good" or "bad" depends entirely on the context of the situation the data describes. In some situations a low percentile is good; in others a high percentile is good; in many situations no value judgment applies at all. A 90th percentile finishing time in a race means you were slower than nearly everyone. A 90th percentile score on that same day's chemistry exam means the opposite. The arithmetic is identical; only the context tells you whether to celebrate.
Understanding how to interpret percentiles properly matters not only when describing data, but also when calculating probabilities in later chapters of this text.
When writing the interpretation of a percentile in the context of the given data, the sentence should contain all of the following:
- information about the context of the situation being considered;
- the data value (value of the variable) that represents the percentile;
- the percent of individuals or items with data values below the percentile;
- the percent of individuals or items with data values above the percentile.
For the 100-meter dash, the third quartile for times for finishing the race was 11.5 seconds. Interpret the third quartile in the context of the situation.
Solution
Step 1 — Translate the quartile into a percent. The third quartile is the 75th percentile.
Step 2 — State both sides in context.
- Seventy-five percent of runners finished the 100-meter dash in 11.5 seconds or less.
- Twenty-five percent of runners finished in 11.5 seconds or more.
Step 3 — Judge it in context. For a race, a low time is a fast run, so a low percentile is the desirable place to be. A runner sitting exactly at the third quartile was slower than three-quarters of the field.
Answer: 11.5 seconds is the time that separates the slowest quarter of runners from the rest of the field.
On a timed math test, the first quartile for time it took to finish the exam was 35 minutes. Interpret the first quartile in the context of this situation.
Solution
The first quartile is the 25th percentile, so 25% of the times are at or below 35 minutes and 75% are at or above it. Writing that out with all four required pieces:
- Twenty-five percent of students finished the exam in 35 minutes or less.
- Seventy-five percent of students finished the exam in 35 minutes or more.
- A low percentile could be considered good here, as finishing more quickly on a timed exam is desirable. (If you take too long, you might not be able to finish.)
Answer: 35 minutes is the time that separates the quickest quarter of the class from the rest.
On a 60-point written assignment, the 80th percentile for the number of points earned was 49. Interpret the 80th percentile in the context of this situation.
Solution
Step 1 — State the two sides.
- Eighty percent of students earned 49 points or fewer on the assignment.
- Twenty percent of students earned 49 points or more.
Step 2 — Judge it in context. More points is better on a graded assignment, so a high percentile is desirable. Sitting at the 80th percentile means you scored at or above four out of every five classmates.
Answer: 49 points out of 60 is the score that separates the top fifth of the class from the rest.
On a 20-question math test, the 70th percentile for number of correct answers was 16. Interpret the 70th percentile in the context of this situation.
Solution
Step 1 — State the two sides. The 70th percentile means 70% of the data lies at or below 16 correct answers, and 30% lies at or above it.
- Seventy percent of students answered 16 or fewer questions correctly.
- Thirty percent of students answered 16 or more questions correctly.
Step 2 — Judge it in context. On a test, more correct answers is better, so a high percentile is the desirable place to be. A student who answered exactly 16 correctly did better than about 70% of the class.
Answer: 16 correct answers is the score that separates the top 30% of the class from the rest.
Kiran keeps stats for a basketball league. In their season records, the 40th percentile for points scored per player in a game is eight. Interpret the 40th percentile in the context of this situation.
Solution
Step 1 — State the two sides.
- Forty percent of players scored eight points or fewer in a game.
- Sixty percent of players scored eight points or more.
Step 2 — Judge it in context. In scoring, higher is generally seen as better, so a high percentile is the desirable place to be — though a player whose job is defense or setting up teammates may sit low here and still be excellent at their role.
Answer: eight points is the game total that separates the lowest-scoring 40% of players from the other 60%.
Jelani advises students at a community college. In their enrollment report, the 30th percentile of credit units that students are enrolled for is seven units. Interpret the 30th percentile in the context of this situation.
Solution
Step 1 — State the two sides.
- Thirty percent of students are enrolled in seven credit units or fewer.
- Seventy percent of students are enrolled in seven credit units or more.
Step 2 — Judge it in context. This is the case where no value judgment applies. Being enrolled in few units is not "bad" — a student working full time, or caring for family, may deliberately take a light load, and a student who is nearly finished with a degree may only need one class. The percentile describes enrollment behavior; it does not rank the students.
Answer: seven units is the course load that separates the lightest-enrolled 30% of students from the other 70%. It says something useful about how the college's students schedule their lives, and nothing at all about how well they are doing.
A college statistics instructor is investigating the amount of time students spend working on a final project in the course. The instructor would like students to spend approximately 3 to 4 hours as the typical amount of time to be spent on the project. The instructor collects data from a random sample of 10 students for the number of hours spent working on the final project. The results obtained are as follows:
2; 3; 5; 3; 4; 4; 3; 11; 3; 2 (hours)
a) Determine the five-number summary: minimum, \(Q_1\), median, \(Q_3\), and maximum.
b) Should the instructor modify the final project or leave it as is?
Solution
Part (a) — Step 1: order the data.
2; 2; 3; 3; 3; 3; 4; 4; 5; 11
Step 2 — Find the median. With 10 values, average the 5th and 6th.
$$ \text{Med} = \frac{3 + 3}{2} = 3 $$Step 3 — Find the quartiles. The lower half is 2, 2, 3, 3, 3, so \(Q_1\) is its 3rd value. The upper half is 3, 4, 4, 5, 11, so \(Q_3\) is its 3rd value.
$$ \text{Min} = 2 \qquad Q_1 = 3 \qquad \text{Med} = 3 \qquad Q_3 = 4 \qquad \text{Max} = 11 $$Part (b) — interpret it. The instructor wanted a typical time of 3 to 4 hours. The middle half of the class falls between \(Q_1 = 3\) and \(Q_3 = 4\) hours, which is exactly the target — so the project is doing what the instructor intended and does not need modifying.
One student reported 11 hours. Check whether that is an outlier:
$$ IQR = 4 - 3 = 1 \qquad Q_3 + 1.5(IQR) = 4 + 1.5 = 5.5 $$Since \(11 > 5.5\), that value is a potential outlier — a single student who struggled, or who went far beyond the assignment. Worth a conversation, but not a reason to redesign a project that is working for the other nine.
Answer: the five-number summary is 2, 3, 3, 4, 11; the instructor should leave the project as is and follow up with the student who spent 11 hours.
Sharpe Middle School is applying for a grant that will be used to add fitness equipment to the gym. The principal, Esperanza, surveyed 15 anonymous students to determine how many minutes a day the students spend exercising. The results from her 15 anonymous students are shown.
0; 40; 60; 30; 60; 10; 45; 30; 300; 90; 30; 120; 60; 0; 20 (minutes)
Determine the five-number summary. If you were the principal, would you be justified in purchasing new fitness equipment?
Solution
Step 1 — Order the data and read off the five values.
0; 0; 10; 20; 30; 30; 30; 40; 45; 60; 60; 60; 90; 120; 300
$$ \text{Min} = 0 \qquad Q_1 = 20 \qquad \text{Med} = 40 \qquad Q_3 = 60 \qquad \text{Max} = 300 $$Step 2 — Read the middle of the data. Since 75% of the students exercise for 60 minutes or less daily, and since the \(IQR\) is 40 minutes (\(60 - 20 = 40\)), we know that half of the students surveyed exercise between 20 minutes and 60 minutes daily. That seems a reasonable amount of time spent exercising, so the principal would be justified in purchasing the new equipment.
Step 3 — But check the fences first. Esperanza needs to be careful. The value 300 looks suspicious.
$$ Q_3 + 1.5(IQR) = 60 + 1.5(40) = 120 $$The value 300 is greater than 120, so it is a potential outlier.
Step 4 — Recompute without it. Delete the 300 and the remaining 14 values are 0; 0; 10; 20; 30; 30; 30; 40; 45; 60; 60; 60; 90; 120.
$$ \text{Min} = 0 \qquad Q_1 = 20 \qquad \text{Med} = \frac{30 + 40}{2} = 35 \qquad Q_3 = 60 \qquad \text{Max} = 120 $$Answer: we still have 75% of the students exercising for 60 minutes or less daily and half of the students exercising between 20 and 60 minutes a day — the conclusion survives dropping the outlier, which is exactly the reassurance the principal wanted. However, 15 students is a small sample, and the principal should survey more students to be sure of the survey results.
Mateo scored at the 72nd percentile on the SAT. What was his SAT score?
Solution
This is a trick question, and noticing that is the point of it.
A percentile tells you a position, not a value. Knowing that Mateo sits at the 72nd percentile tells you that 72% of test-takers scored the same as or lower than he did, and 28% scored the same or higher. To convert that position back into an actual score, you would need one more thing: the distribution of scores for everyone who took the test — a table, a curve, or the full ordered list.
Without it, the percentile alone cannot be turned into a number.
Answer: the score cannot be determined from the percentile alone. What we can say is that Mateo outperformed roughly 72% of the students who took the exam, and that about 28% of them scored at or above him.
Problem Set 2.3
Problem 1. Listed are 29 ages for Academy Award winning best actors:
18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77
a) Find the 40th percentile.
b) Find the 78th percentile.
Solution
The data is already ordered and \(n = 29\), so \(n + 1 = 30\).
a) The 40th percentile.
$$ i = \frac{40}{100}(30) = 12 $$Twelve is an integer, so take the 12th value in the ordered list.
$$ P_{40} = 37 \text{ years} $$b) The 78th percentile.
$$ i = \frac{78}{100}(30) = 23.4 $$Not an integer, so round down to 23 and up to 24, then average those two values. The 23rd age is 69 and the 24th is 71.
$$ P_{78} = \frac{69 + 71}{2} = 70 \text{ years} $$ <!-- Solutions -->Problem 2. Listed are 32 ages for Academy Award winning best actors:
18; 18; 21; 22; 25; 26; 27; 29; 30; 31; 31; 33; 36; 37; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77
a) Find the percentile of 37.
b) Find the percentile of 72.
Solution
Here you are given a value and asked for its percentile, so use
$$ \frac{x + 0.5y}{n}(100) $$with \(n = 32\).
a) The percentile of 37. Counting from the bottom, 13 ages are less than 37 (18, 18, 21, 22, 25, 26, 27, 29, 30, 31, 31, 33, 36), and two ages equal 37. So \(x = 13\) and \(y = 2\).
$$ \frac{13 + 0.5(2)}{32}(100) = \frac{14}{32}(100) = 43.75 $$Rounded, 37 is the 44th percentile.
b) The percentile of 72. Counting from the bottom, 27 ages are less than 72, and one age equals 72. So \(x = 27\) and \(y = 1\).
$$ \frac{27 + 0.5(1)}{32}(100) = \frac{27.5}{32}(100) = 85.94 $$Rounded, 72 is the 86th percentile.
Problem 3. Leilani was ranked 37th in her graduating class of 180 students. At what percentile is her ranking?
Solution
Leilani graduated 37th out of a class of 180 students, so there are \(180 - 37 = 143\) students ranked below her, and there is one rank of 37. So \(x = 143\), \(y = 1\), and \(n = 180\).
$$ \frac{x + 0.5y}{n}(100) = \frac{143 + 0.5(1)}{180}(100) = \frac{143.5}{180}(100) = 79.72 $$Answer: Leilani's rank of 37 puts her at the 80th percentile. About 80% of her class ranked at or below her.
Problem 4. a) For runners in a race, a low time means a faster run. The winners in a race have the shortest running times. Is it more desirable to have a finish time with a high or a low percentile when running a race?
b) The 20th percentile of run times in a particular race is 5.2 minutes. Write a sentence interpreting the 20th percentile in the context of the situation.
c) A bicyclist in the 90th percentile of a bicycle race completed the race in 1 hour and 12 minutes. Is he among the fastest or slowest cyclists in the race? Write a sentence interpreting the 90th percentile in the context of the situation.
Solution
a) A low percentile is more desirable. In a race the clock is the measurement, and a low time is a fast run — so being near the bottom of the ordered list of times is where the winners are.
b) Twenty percent of the runners in this race finished in 5.2 minutes or less, and 80% finished in 5.2 minutes or more.
c) He is among the slowest cyclists in the race. Ninety percent of the cyclists completed the race in 1 hour and 12 minutes or less, and only 10% took that long or longer.
Problem 5. a) For runners in a race, a higher speed means a faster run. Is it more desirable to have a speed with a high or a low percentile when running a race?
b) The 40th percentile of speeds in a particular race is 7.5 miles per hour. Write a sentence interpreting the 40th percentile in the context of the situation.
Solution
a) A high percentile is more desirable. Here the measurement is speed rather than time, so the ordering flips: a high percentile means a higher speed, which is faster.
b) Forty percent of the runners ran at speeds of 7.5 miles per hour or less (slower), and 60% ran at speeds of 7.5 miles per hour or more (faster).
Problem 6. On an exam, would it be more desirable to earn a grade with a high or low percentile? Explain.
Solution
A high percentile is more desirable on an exam. The measurement is your score, and more correct work means a higher score, so a high percentile means you scored at or above a large share of the other test-takers. A student at the 90th percentile outperformed about 90% of the class; a student at the 10th percentile outperformed only about 10%.
Problem 7. Mina is waiting in line at the Department of Motor Vehicles (DMV). Her wait time of 32 minutes is the 85th percentile of wait times. Is that good or bad? Write a sentence interpreting the 85th percentile in the context of this situation.
Solution
For a wait time, that is bad. When waiting in line at the DMV, the 85th percentile is a long wait compared to the other people waiting — 85% of people had shorter wait times than Mina. In this context she would prefer a wait time corresponding to a lower percentile.
Interpretation: 85% of people at the DMV waited 32 minutes or less, and 15% of people at the DMV waited 32 minutes or longer.
Problem 8. In a survey collecting data about the salaries earned by recent college graduates, Li found that her salary was in the 78th percentile. Should Li be pleased or upset by this result? Explain.
Solution
Li should be pleased. Salary is a measurement where more is generally better, so a high percentile is the desirable place to be.
Interpretation: 78% of the recent college graduates surveyed earned a salary the same as or less than Li's, and only 22% earned the same or more.
Problem 9. In a study collecting data about the repair costs of damage to automobiles in a certain type of crash test, a certain model of car had $1,700 in damage and was in the 90th percentile. Should the manufacturer and the consumer be pleased or upset by this result? Explain, and write a sentence that interprets the 90th percentile in the context of this problem.
Solution
Both the manufacturer and the consumer would be upset. This is a large repair cost for the damage, compared to the other cars in the sample — the measurement is cost, and for repair costs low is good.
Interpretation: 90% of the crash-tested cars had damage repair costs of $1,700 or less; only 10% had damage repair costs of $1,700 or more.
Problem 10. The University of Wisconsin has two criteria used to set admission standards for students to be admitted to a college in the UW system:
a) Students' GPAs and scores on standardized tests (SATs and ACTs) are entered into a formula that calculates an "admissions index" score. The admissions index score is used to set eligibility standards intended to meet the goal of admitting the top 12% of high school students in the state. In this context, what percentile does the top 12% represent?
b) Students whose GPAs are at or above the 96th percentile of all students at their high school are eligible (called eligible in the local context), even if they are not in the top 12% of all students in the state. What percentage of students from each high school are "eligible in the local context"?
Solution
a) Admitting the top 12% means setting the cutoff so that 88% of students fall at or below it, since \(100 - 12 = 88\). The top 12% therefore represents the 88th percentile and above.
b) Students at or above the 96th percentile of their own high school are the top \(100 - 96 = 4\%\). So 4% of students from each high school are "eligible in the local context."
Problem 11. Suppose that you are buying a house. You and your realtor have determined that the most expensive house you can afford is at the 34th percentile. The 34th percentile of housing prices is $240,000 in the town you want to move to. In this town, can you afford 34% of the houses or 66% of the houses?
Solution
You can afford 34% of the houses; 66% of the houses are too expensive for your budget. The percentile counts what falls at or below the value, and your ceiling is the value.
Interpretation: 34% of houses in the town cost $240,000 or less, and 66% of houses cost $240,000 or more.
Use the following information to answer the next six exercises. 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.
Problem 12. First quartile =
Solution
First build the ordered list from the counts. With 14 salespersons selling 3 cars, 19 selling 4, 12 selling 5, 9 selling 6, and 11 selling 7, the total is
$$ n = 14 + 19 + 12 + 9 + 11 = 65 $$and the ordered data occupies these positions:
- positions 1–14: 3 cars
- positions 15–33: 4 cars
- positions 34–45: 5 cars
- positions 46–54: 6 cars
- positions 55–65: 7 cars
The first quartile is the 25th percentile, so \(i = \frac{25}{100}(65 + 1) = 16.5\). Average the 16th and 17th values; both fall in the 4-car block.
$$ Q_1 = \frac{4 + 4}{2} = 4 $$Answer: first quartile = 4 cars.
Problem 13. Second quartile = median = 50th percentile =
Solution
There are \(n = 65\) values, an odd count, so the median is the single middle value — the 33rd. Positions 15–33 all hold 4 cars, so the 33rd value is 4.
Answer: second quartile = median = 50th percentile = 4 cars. Notice this equals \(Q_1\): more than a quarter of the salespersons sell exactly four cars, so the same value occupies both positions.
Problem 14. Third quartile =
Solution
The third quartile is the 75th percentile, so
$$ i = \frac{75}{100}(65 + 1) = 49.5 $$Not an integer, so average the 49th and 50th values. Positions 46–54 all hold 6 cars.
$$ Q_3 = \frac{6 + 6}{2} = 6 $$Answer: third quartile = 6 cars.
Problem 15. Interquartile range (\(IQR\)) =
Solution
Answer: the interquartile range is 2 cars. The middle half of these salespersons all sell between four and six cars a week — a very tight middle.
Problem 16. 10th percentile =
Solution
Not an integer, so average the 6th and 7th values. Positions 1–14 all hold 3 cars.
$$ P_{10} = \frac{3 + 3}{2} = 3 $$Answer: the 10th percentile is 3 cars.
Problem 17. 70th percentile =
Solution
Not an integer, so average the 46th and 47th values. Positions 46–54 all hold 6 cars.
$$ P_{70} = \frac{6 + 6}{2} = 6 $$Answer: the 70th percentile is 6 cars.
Problem 18. The median age for Black people in the U.S. is 30.9 years; for U.S. White people it is 42.3 years.
a) Based upon this information, give two reasons why the median age for Black people could be lower than the median age for White people.
b) Does the lower median age for Black people necessarily mean that Black people die younger than White people? Why or why not?
c) How might it be possible for Black people and White people to die at approximately the same age, but for the median age for White people to be higher?
Solution
The key idea here is that the median age of a living population describes the shape of that population's age distribution — it is not the age at which people die.
a) Two reasons the median age could be lower:
- A larger share of children and young adults. If Black families in the U.S. have, on average, more recent and more numerous births, then more of the population sits at the young end of the ordered list, which pulls the middle value down. Nobody has to die younger for this to happen.
- Differences in mortality at older ages. If fewer people survive into their seventies and eighties, the upper end of the ordered list is thinner, which also pulls the middle value down. This one is about age at death, and it reflects documented differences in access to health care, environmental exposure, and economic security.
b) No. Reason 1 above produces a lower median age with no difference in lifespan at all. A population with proportionally many children has a low median age even if every one of its members lives to 90. The median tells you where the middle of the current population sits, not how long anyone lives.
c) If the two groups die at approximately the same age, the medians can still differ whenever the two age distributions differ in shape. Suppose the White population has a lower birth rate and so proportionally more people in their fifties and sixties, while the Black population has a higher birth rate and so proportionally more children. Sorting each group by age puts the middle value of the first group much further up the scale than the middle value of the second — even though the two groups' members reach the same ages.
Problem 19. Six hundred adult Americans were asked by telephone poll, "What do you think constitutes a middle-class income?" The results are in Table 2.3.4. Include the left endpoint of each interval, but not the right endpoint.
| Salary (US dollars) | Relative frequency |
|---|---|
| < 20,000 | 0.02 |
| 20,000–25,000 | 0.09 |
| 25,000–30,000 | 0.19 |
| 30,000–40,000 | 0.26 |
| 40,000–50,000 | 0.18 |
| 50,000–75,000 | 0.17 |
| 75,000–99,999 | 0.02 |
| 100,000+ | 0.01 |
a) What percentage of the survey answered "not sure"?
b) What percentage think that middle-class is from $25,000 to $50,000?
c) Construct a histogram of the data.
i. Should all bars have the same width, based on the data? Why or why not?
ii. How should the < 20,000 and the 100,000+ intervals be handled? Why?
d) Find the 40th and 80th percentiles.
e) Construct a bar graph of the data.
Solution
a) The listed relative frequencies must account for everyone who gave a numeric answer; the rest answered "not sure."
$$ 1 - (0.02 + 0.09 + 0.19 + 0.26 + 0.18 + 0.17 + 0.02 + 0.01) = 1 - 0.94 = 0.06 $$6% of the survey answered "not sure."
b) Add the three intervals that span $25,000 up to $50,000:
$$ 0.19 + 0.26 + 0.18 = 0.63 $$63% think middle class runs from $25,000 to $50,000.
c) Answers may vary in appearance, but the two sub-questions have definite answers.
i. No, the bars should not all be the same width. The intervals themselves are different widths — $5,000 for 20,000–25,000, $10,000 for 30,000–40,000, $25,000 for 50,000–75,000. In a histogram the horizontal axis is a real number line, so each bar has to span its own interval. Drawing them all the same width would silently stretch the narrow intervals and squeeze the wide ones.
ii. The \(<\)20,000 and 100,000+ intervals are open-ended — they have no stated lower and upper bound, so there is no width to draw. You have to choose a reasonable boundary and say so (for example, treat them as 15,000–20,000 and 100,000–125,000), or leave them off the histogram and report them separately in a note.
d) Work down the relative frequency column, accumulating: 0.02, then 0.11, then 0.30, then 0.56, then 0.74, then 0.91.
- The 40th percentile is passed between 0.30 and 0.56, so it falls between $30,000 and $40,000.
- The 80th percentile is passed between 0.74 and 0.91, so it falls between $50,000 and $75,000.
e) Answers may vary. A bar graph treats the eight salary ranges as separate categories, so the bars are drawn with equal widths and gaps between them — which is exactly why a bar graph sidesteps the open-ended-interval problem from part (c) and a histogram does not.
Key Terms
median — the number that separates ordered data into halves; half the values are at or below it and half are at or above it. Also called the second quartile, \(Q_2\), or the 50th percentile.
quartile — one of the three numbers \(Q_1\), \(Q_2\), \(Q_3\) that separate ordered data into quarters.
first quartile \(Q_1\) — the middle value of the lower half of the ordered data; the 25th percentile.
third quartile \(Q_3\) — the middle value of the upper half of the ordered data; the 75th percentile.
percentile — a value at or below which a stated percent of the ordered data falls; percentiles divide ordered data into 100 equal parts.
index \(i\) — the position of a data value in the ordered list, calculated as \(i = \frac{k}{100}(n+1)\) when locating the \(k\)th percentile.
interquartile range (IQR) — the spread of the middle 50% of the data, \(IQR = Q_3 - Q_1\).
potential outlier — a data value more than \(1.5 \times IQR\) below \(Q_1\) or more than \(1.5 \times IQR\) above \(Q_3\); it requires further investigation before being kept or discarded.
five-number summary — the minimum, \(Q_1\), median, \(Q_3\), and maximum of a data set.