1.10: Measures of the Location of the Data
The common measures of location are quartiles and percentiles. Quartiles are special percentiles. The first quartile, Q 1 , is the same as the 25 th percentile, and the third quartile, Q 3 , is the same as the 75 th percentile. The median, M , is called both the second quartile and the 50 th percentile.
To calculate quartiles and percentiles, the data must be ordered from smallest to largest. Quartiles divide ordered data into quarters. Percentiles divide ordered data into hundredths. To score in the 90 th percentile of an exam does not mean, necessarily, that you received 90% on a test. It means that 90% of test scores are the same or less than your score and 10% of the test scores are the same or greater than your test score.
Percentiles are useful for comparing values. For this reason, universities and colleges use percentiles extensively. One instance in which colleges and universities use percentiles is when SAT results are used to determine a minimum testing score that will be used as an acceptance factor. For example, suppose Duke accepts SAT scores at or above the 75 th percentile. That translates into a score of at least 1220.
Percentiles are mostly used with very large populations. Therefore, if you were to say that 90% of the test scores are less (and not the same or less) than your score, it would be acceptable because removing one particular data value is not significant.
The median is a number that measures the "center" of the data. You can think of the median as the "middle value," but it does not actually have to be one of the observed values. It is a number that separates ordered data into halves. Half the values are the same number or smaller than the median, and half the values are the same number or larger. For example, consider the following data.
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 is between the seventh value, 6.8, and the eighth value, 7.2. To find the median, add the two values together and divide by two.
The median is seven. Half of the values are smaller than seven and half of the values are larger than seven.
Quartiles are numbers that separate the data into quarters. Quartiles may or may not be part of the data. To find the quartiles, first find the median or second quartile. The first quartile, Q 1 , is the middle value of the lower half of the data, and the third quartile, Q 3 , is the middle value, or median, of the upper half of the data. To get the idea, consider the same data set:
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 are 1, 1, 2, 2, 4, 6, 6.8. The middle value of the lower half is two.
1; 1; 2; 2; 4; 6; 6.8
The number two, which is part of the data, is the first quartile . One-fourth of the entire sets of values are the same as or less than two and three-fourths of the values are more than two.
The upper half of the data is 7.2, 8, 8.3, 9, 10, 10, 11.5. The middle value of the upper half is nine.
The third quartile , Q 3, is nine. Three-fourths (75%) of the ordered data set are less than nine. One-fourth (25%) of the ordered data set are greater than nine. The third quartile is part of the data set in this example.
The interquartile range is a number that indicates the spread of the middle half or the middle 50% of the data. It is the difference between the third quartile ( Q 3 ) and the first quartile ( Q 1 ).
\[IQR = Q_3 – Q_1 \tag{2.4.1}\]
The IQR can help to determine potential outliers . A value is suspected to be a potential outlier if it is less than (1.5)( IQR ) below the first quartile or more than (1.5)( IQR ) above the third quartile . Potential outliers always require further investigation.
A potential outlier is a data point that is significantly different from the other data points. These special data points may be errors or some kind of abnormality or they may be a key to understanding the data.
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
Answer
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
\[M = 488,800 \nonumber\]
\[Q_{1} = \dfrac{230,500 + 387,000}{2} = 308,750\nonumber\]
\[Q_{3} = \dfrac{639,000 + 659,000}{2} = 649,000\nonumber\]
\[IQR = 649,000 - 308,750 = 340,250\nonumber\]
\[(1.5)(IQR) = (1.5)(340,250) = 510,375\nonumber\]
\[Q_{1} - (1.5)(IQR) = 308,750 - 510,375 = –201,625\nonumber\]
\[Q_{3} + (1.5)(IQR) = 649,000 + 510,375 = 1,159,375\nonumber\]
No house price is less than –201,625. However, 5,500,000 is more than 1,159,375. Therefore, 5,500,000 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
Answer
Order the data from smallest to largest.
$28,000; $33,000; $40,500; $42,000; $54,000; $54,000; $64,500; $68,500; $69,000; $72,000; $120,000
Median = $54,000
\[Q_{1} = $40,500\nonumber\]
\[Q_{3} = $69,000\nonumber\]
\[IQR = $69,000 - $40,500 = $28,500\nonumber\]
\[(1.5)(IQR) = (1.5)($28,500) = $42,750\nonumber\]
\[Q_{1} - (1.5)(IQR) = $40,500 - $42,750 = -$2,250\nonumber\]
\[Q_{3} + (1.5)(IQR) = $69,000 + $42,750 = $111,750\nonumber\]
No salary is less than –$2,250. However, $120,000 is more than $11,750, so $120,000 is a potential outlier.
For the two data sets in the test scores example , find the following:
- The interquartile range. Compare the two interquartile ranges.
- Any outliers in either set.
Answer
The five number summary for the day and night classes is
| Minimum | Q 1 | Median | Q 3 | Maximum | |
|---|---|---|---|---|---|
| Day | 32 | 56 | 74.5 | 82.5 | 99 |
| Night | 25.5 | 78 | 81 | 89 | 98 |
-
The
IQR
for the day group is \(Q_{3} - Q_{1} = 82.5 - 56 = 26.5\)
The IQR for the night group is \(Q_{3} - Q_{1} = 89 - 78 = 11\)
The interquartile range (the spread or variability) for the day class is larger than the night class IQR . This suggests more variation will be found in the day class’s class test scores.
-
Day class outliers are found using the IQR times 1.5 rule. So,
- \(Q_{1} - IQR(1.5) = 56 – 26.5(1.5) = 16.25\)
- \(Q_{3} + IQR(1.5) = 82.5 + 26.5(1.5) = 122.25\)
Since the minimum and maximum values for the day class are greater than 16.25 and less than 122.25, there are no outliers.
Night class outliers are calculated as:
- \(Q_{1} - IQR (1.5) = 78 – 11(1.5) = 61.5\)
- \(Q_{3} + IQR(1.5) = 89 + 11(1.5) = 105.5\)
For this class, any test score less than 61.5 is an outlier. Therefore, the scores of 45 and 25.5 are outliers. Since no test score is greater than 105.5, there is no upper end outlier.
Find the interquartile range for the following 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
Answer
Class A
Order the data from smallest to largest.
65; 66; 67; 69; 69; 76; 77; 77; 79; 80; 81; 83; 85; 89; 90; 91; 94; 96; 98; 99
\(Median = \dfrac{80 + 81}{2}\) = 80.5
\(Q_{1} = \dfrac{69 + 76}{2} = 72.5\)
\(Q_{3} = \dfrac{90 + 91}{2} = 90.5\)
\(IQR = 90.5 - 72.5 = 18\)
Class B
Order the data from smallest to largest.
68; 68; 70; 71; 72; 73; 75; 78; 79; 80; 80; 90; 90; 92; 92; 95; 95; 97; 99; 100
\(Median = \dfrac{80 + 80}{2} = 80\)
\(Q_{1} = \dfrac{72 + 73}{2} = 72.5\)
\(Q_{3} = \dfrac{92 + 95}{2} = 93.5\)
\(IQR = 93.5 - 72.5 = 21\)
The data for Class B has a larger IQR , so the scores between Q 3 and Q 1 (middle 50%) for the data for Class B are more spread out and not clustered about the median.
Fifty statistics students were asked 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 28 th percentile . Notice the 0.28 in the "cumulative relative frequency" column. Twenty-eight percent of 50 data values is 14 values. There are 14 values less than the 28 th percentile. They include the two 4s, the five 5s, and the seven 6s. The 28 th percentile is between the last six and the first seven. The 28 th percentile is 6.5.
Find the median . Look again at the "cumulative relative frequency" column and find 0.52. The median is the 50 th percentile or the second quartile. 50% of 50 is 25. There are 25 values less than the median. They include the two 4s, the five 5s, the seven 6s, and eleven of the 7s. The median or 50 th percentile is between the 25 th , or seven, and 26 th , or seven, values. The median is seven.
Find the third quartile . The third quartile is the same as the 75 th percentile. You can "eyeball" this answer. If you look at the "cumulative relative frequency" column, you find 0.52 and 0.80. When you have all the fours, fives, sixes and sevens, you have 52% of the data. When you include all the 8s, you have 80% of the data. The 75 th percentile, then, must be an eight . Another way to look at the problem is to find 75% of 50, which is 37.5, and round up to 38. The third quartile, Q 3 , is the 38 th value, which is an eight. You can check this answer by counting the values. (There are 37 values below the third quartile and 12 values above.)
Forty bus drivers were asked how many hours they spend each day running their routes (rounded to the nearest hour). Find the 65 th 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 |
Answer
The 65 th percentile is between the last three and the first four.
The 65 th percentile is 3.5.
Using the table above in Example \(\PageIndex{3}\)
- Find the 80 th percentile.
- Find the 90 th percentile.
- Find the first quartile. What is another name for the first quartile?
Solution
Using the data from the frequency table, we have:
- The 80 th percentile is between the last eight and the first nine in the table (between the 40 th and 41 st values). Therefore, we need to take the mean of the 40 th an 41 st values. The 80 th percentile \(= \dfrac{8+9}{2} = 8.5\)
- The 90 th percentile will be the 45 th data value (location is \(0.90(50) = 45\)) and the 45 th data value is nine.
- Q 1 is also the 25 th percentile. The 25 th percentile location calculation: \(P_{25} = 0.25(50) = 12.5 \approx 13\) the 13 th data value. Thus, the 25 th percentile is six.
Refer to the table above in Exercise \(\PageIndex{3}\) . Find the third quartile. What is another name for the third quartile?
Answer
The third quartile is the 75 th percentile, which is four. The 65 th percentile is between three and four, and the 90 th percentile is between four and 5.75. The third quartile is between 65 and 90, so it must be four.
Your instructor or a member of the class will ask everyone in class how many sweaters they own. Answer the following questions:
- How many students were surveyed?
- What kind of sampling did you do?
- Construct two different histograms. For each, starting value = _____ ending value = ____.
- Find the median, first quartile, and third quartile.
-
Construct a table of the data to find the following:
- the 10 th percentile
- the 70 th percentile
- the percent of students who own less than four sweaters
A Formula for Finding the k th Percentile
If you were to do a little research, you would find several formulas for calculating the kth percentile. Here is one of them.
- \(k =\) the kth percentile. It may or may not be part of the data.
- \(i =\) the index (ranking or position of a data value)
- \(n =\) the total number of data
Order the data from smallest to largest.
Calculate \(i = \dfrac{k}{100}(n + 1)\)
If \(i\) is an integer, then the \(k^{th}\) percentile is the data value in the \(i^{th}\) position in the ordered set of data.
If \(i\) is not an integer, then round \(i\) up and round \(i\) down to the nearest integers. Average the two data values in these two positions in the ordered data set. This is easier to understand in an example.
Listed are 29 ages for Academy Award winning best actors in order from smallest to largest.
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- Find the 70 th percentile.
- Find the 83 rd percentile.
Solution
-
- \(k = 70\)
- \(i\) = the index
- \(n = 29\)
-
- \(k\) = 83 rd percentile
- \(i \)= the index
- \(n = 29\)
Listed are 29 ages for Academy Award winning best actors in order from smallest to largest.
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 20 th percentile and the 55 th percentile.
Answer
\(k = 20\). Index \(= i = \dfrac{k}{100}(n+1) = \dfrac{20}{100}(29 + 1) = 6\). The age in the sixth position is 27. The 20 th percentile is 27 years.
\(k = 55\). Index \(= i = \dfrac{k}{100}(n+1) = \dfrac{55}{100}(29 + 1) = 16.5\). Round down to 16 and up to 17. The age in the 16 th position is 52 and the age in the 17 th position is 55. The average of 52 and 55 is 53.5. The 55 th percentile is 53.5 years.
You can calculate percentiles using calculators and computers. There are a variety of online calculators.
A Formula for Finding the Percentile of a Value in a Data Set
- Order the data from smallest to largest.
- \(x =\) the number of data values counting from the bottom of the data list up to but not including the data value for which you want to find the percentile.
- \(y =\) the number of data values equal to the data value for which you want to find the percentile.
- \(n =\) the total number of data.
- Calculate \(\dfrac{x + 0.5y}{n}(100)\). Then round to the nearest integer.
Listed are 29 ages for Academy Award winning best actors in order from smallest to largest.
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- Find the percentile for 58.
- Find the percentile for 25.
Solution
-
Counting from the bottom of the list, there are 18 data values less than 58. There is one value of 58.
\(x = 18\) and \(y = 1\). \(\dfrac{x + 0.5y}{n}(100) = \dfrac{18 + 0.5(1)}{29}(100) = 63.80\). 58 is the 64 th percentile.
-
Counting from the bottom of the list, there are three data values less than 25. There is one value of 25.
\(x = 3\) and \(y = 1\). \(\dfrac{x + 0.5y}{n}(100) = \dfrac{3 + 0.5(1)}{29}(100) = 12.07\). Twenty-five is the 12 th percentile.
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.
Answer
Percentile for 47: Counting from the bottom of the list, there are 15 data values less than 47. There is one value of 47.
\(x = 15\) and \(y = 1\). \(\dfrac{x + 0.5y}{n}(100) = \dfrac{15 + 0.5(1)}{30}(100) = 51.67\). 47 is the 52 nd percentile.
Percentile for 31: Counting from the bottom of the list, there are eight data values less than 31. There are two values of 31.
\(x = 8\) and \(y = 2\). \(\dfrac{x + 0.5y}{n}(100) = \dfrac{8 + 0.5(2)}{30}(100) = 30\). 31 is the 30 th percentile.
Interpreting Percentiles, Quartiles, and Median
A percentile indicates the relative standing of a data value when data are 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 15 th percentile.
- Low percentiles always correspond to lower data values.
- High percentiles always correspond to higher data values.
A percentile may or may not correspond to a value judgment about whether it is "good" or "bad." The interpretation of whether a certain percentile is "good" or "bad" depends on the context of the situation to which the data applies. In some situations, a low percentile would be considered "good;" in other contexts a high percentile might be considered "good". In many situations, there is no value judgment that applies.
Understanding how to interpret percentiles properly is important not only when describing data, but also when calculating probabilities in later chapters of this text.
GUIDELINE
When writing the interpretation of a percentile in the context of the given data, the sentence should contain the following information.
- 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.
Example 2.4.7
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.
Answer
- 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, as finishing more quickly on a timed exam is desirable. (If you take too long, you might not be able to finish.)
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.
Answer
Twenty-five percent of runners finished the race in 11.5 seconds or more. Seventy-five percent of runners finished the race in 11.5 seconds or less. A lower percentile is good because finishing a race more quickly is desirable.
Example 2.4.8
On a 20 question math test, the 70 th percentile for number of correct answers was 16. Interpret the 70 th percentile in the context of this situation.
Answer
- Seventy percent of students answered 16 or fewer questions correctly.
- Thirty percent of students answered 16 or more questions correctly.
- A higher percentile could be considered good, as answering more questions correctly is desirable.
On a 60 point written assignment, the 80 th percentile for the number of points earned was 49. Interpret the 80 th percentile in the context of this situation.
Answer
Eighty percent of students earned 49 points or fewer. Twenty percent of students earned 49 or more points. A higher percentile is good because getting more points on an assignment is desirable.
At a community college, it was found that the 30 th percentile of credit units that students are enrolled for is seven units. Interpret the 30 th percentile in the context of this situation.
Answer
- Thirty percent of students are enrolled in seven or fewer credit units.
- Seventy percent of students are enrolled in seven or more credit units.
- In this example, there is no "good" or "bad" value judgment associated with a higher or lower percentile. Students attend community college for varied reasons and needs, and their course load varies according to their needs.
During a season, the 40 th percentile for points scored per player in a game is eight. Interpret the 40 th percentile in the context of this situation.
Answer
Forty percent of players scored eight points or fewer. Sixty percent of players scored eight points or more. A higher percentile is good because getting more points in a basketball game is desirable.
Sharpe Middle School is applying for a grant that will be used to add fitness equipment to the gym. The principal surveyed 15 anonymous students to determine how many minutes a day the students spend exercising. The results from the 15 anonymous students are shown.
0 minutes; 40 minutes; 60 minutes; 30 minutes; 60 minutes
10 minutes; 45 minutes; 30 minutes; 300 minutes; 90 minutes;
30 minutes; 120 minutes; 60 minutes; 0 minutes; 20 minutes
Determine the following five values.
- Min = 0
- Q 1 = 20
- Med = 40
- Q 3 = 60
- Max = 300
If you were the principal, would you be justified in purchasing new fitness equipment? 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. This seems a reasonable amount of time spent exercising, so the principal would be justified in purchasing the new equipment.
However, the principal needs to be careful. The value 300 appears to be a potential outlier.
\[Q_{3} + 1.5(IQR) = 60 + (1.5)(40) = 120\].
The value 300 is greater than 120 so it is a potential outlier. If we delete it and calculate the five values, we get the following values:
- Min = 0
- Q 1 = 20
- Q 3 = 60
- Max = 120
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. However, 15 students is a small sample and the principal should survey more students to be sure of his survey results.
References
- Cauchon, Dennis, Paul Overberg. “Census data shows minorities now a majority of U.S. births.” USA Today, 2012. Available online at usatoday30.usatoday.com/news/...sus/55029100/1 (accessed April 3, 2013).
- Data from the United States Department of Commerce: United States Census Bureau. Available online at http://www.census.gov/ (accessed April 3, 2013).
- “1990 Census.” United States Department of Commerce: United States Census Bureau. Available online at http://www.census.gov/main/www/cen1990.html (accessed April 3, 2013).
- Data from San Jose Mercury News .
- Data from Time Magazine ; survey by Yankelovich Partners, Inc.
Review
The values that divide a rank-ordered set of data into 100 equal parts are called percentiles. Percentiles are used to compare and interpret data. For example, an observation at the 50 th percentile would be greater than 50 percent of the other obeservations in the set. Quartiles divide data into quarters. The first quartile ( Q 1 ) is the 25 th percentile,the second quartile ( Q 2 or median) is 50 th percentile, and the third quartile ( Q 3 ) is the the 75 th percentile. The interquartile range, or IQR , is the range of the middle 50 percent of the data values. The IQR is found by subtracting Q 1 from Q 3 , and can help determine outliers by using the following two expressions.
- \(Q_{3} + IQR(1.5)\)
- \(Q_{1} - IQR(1.5)\)
Formula Review
\[i = \dfrac{k}{100}(n+1) \nonumber\]
where \(i\) = the ranking or position of a data value,
- \(k\) = the k th percentile,
- \(n\) = total number of data.
Expression for finding the percentile of a data value: \(\left(\dfrac{x + 0.5y}{n}\right)(100)\)
where \(x =\) the number of values counting from the bottom of the data list up to but not including the data value for which you want to find the percentile,
\(y =\) the number of data values equal to the data value for which you want to find the percentile,
\(n =\) total number of data
Glossary
- Interquartile Range
- or IQR , is the range of the middle 50 percent of the data values; the IQR is found by subtracting the first quartile from the third quartile.
- Outlier
- an observation that does not fit the rest of the data
- Percentile
- a number that divides ordered data into hundredths; percentiles may or may not be part of the data. The median of the data is the second quartile and the 50 th percentile. The first and third quartiles are the 25 th and the 75 th percentiles, respectively.
- Quartiles
- the numbers that separate the data into quarters; quartiles may or may not be part of the data. The second quartile is the median of the data.