Quantiles

Edexcel S1 Mathematics 2003

Introduction- what is a quantile?

Quantiles are used to divide data into intervals containing an equal number of values. For example:

• Deciles D1, …, D9 divide data into 10 parts

• Quartiles Q1, Q2, Q3 divide data into 4 parts

• Percentiles P1, …, P100 divide into 100 parts

. . . . . . . . . … . . .. .. ... ... . .. ... ...... .... .. .. . .. . … … . . . . . .D1 D2 D3 D4 D5 D6 D7 D8 D9

Ungrouped data Treat data as individual values Use textbook method of rounding to next value or next .5 th value

Grouped data Use linear interpolation to estimate quantile. Treat data as continuous within each group / class Assumes values are evenly distributed within each class.

Introduction – Grouped / Ungrouped data

Example –Ungrouped dataQuestion: The number of appointments at a doctors surgery for each of 18 days were: 6, 7, 7, 8, 8, 9, 9, 10, 10, 11, 11, 11, 11, 12, 12, 13, 15, 16

Find the median, and first and ninth deciles of the number of appointments

The median is the middle value:n/2 = 18/2 = 9 9.5th value = = 10.5 appointments

2

1110

Whole number- so round up

to .5th

6 7 7 8 8 9 9 10 10 11 11 11 11 12 12 13 15 16

Answer:

median

Find average of 9th and 10th value

Example –Ungrouped dataQuestion: The number of appointments at a doctors surgery for each of 18 days were: 6, 7, 7, 8, 8, 9, 9, 10, 10, 11, 11, 11, 11, 12, 12, 13, 15, 16

Find the median, and first and ninth deciles of the number of appointments

The median is the middle value:n/2 = 18/2 = 9 9.5th value = = 10.5 appointments

The first decile, D1, is the 1/10th value:n/10 = 18/10 = 1.8 7 appointments 2nd value =

2

1110

Not whole - so round

up

to whole

Find the 2nd value

6 7 7 8 8 9 9 10 10 11 11 11 11 12 12 13 15 16

Answer:

D1 median

Example –Ungrouped dataQuestion: The number of appointments at a doctors surgery for each of 18 days were: 6, 7, 7, 8, 8, 9, 9, 10, 10, 11, 11, 11, 11, 12, 12, 13, 15, 16

Find the median, and first and ninth deciles of the number of appointments

The median is the middle value:n/2 = 18/2 = 9 9.5th value = = 10.5 appointments

The first decile, D1, is the 1/10th value:n/10 = 18/10 = 1.8 7 appointments

The ninth decile, D9, is the 9/10th value:9n/4 = 9x18/10 = 16.2 17th value = 15 appointments

2nd value =

2

1110

Not whole - so round

up

to whole

Find the 17th value

6 7 7 8 8 9 9 10 10 11 11 11 11 12 12 13 15 16

Answer:

D1 median D9

Example –Ungrouped dataQuestion: The number of appointments at a doctors surgery for each of 18 days were: 6, 7, 7, 8, 8, 9, 9, 10, 10, 11, 11, 11, 11, 12, 12, 13, 15, 16

Find the median, and first and ninth deciles of the number of appointments

The median is the middle value:n/2 = 18/2 = 9 9.5th value = = 10.5 appointments

The first decile, D1, is the 1/10th value:n/10 = 18/10 = 1.8 7 appointments

The ninth decile, D9, is the 9/10th value:9n/4 = 9x18/10 = 16.2 17th value = 15 appointments

2nd value =

2

1110

6 7 7 8 8 9 9 10 10 11 11 11 11 12 12 13 15 16

Answer:

D1 median D9

Example – Grouped dataQuestion: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded:

Answer:

Waiting times 1 - 4 5 - 9 10 - 14 15 - 19 20 - 29 30 - 39 40 - 49 50 +

frequency 8 15 20 15 20 12 10 0

Estimate the median and interquartile range of waiting times

The median is the middle value:n/2 = 100/2 = 50th value

No rounding as interpolation is being used

lies in class 15 - 19

9 + 15 + 20 = 44 so 50th value is not in first 3 classes

Example – Grouped data

Waiting times

1 - 4 5 - 9 10 - 14 15 - 19 20 - 29 30 - 39 40 - 49 50 +

frequency 9 15 20 15 20 11 10 0

Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded:

Answer:

Estimate the median and interquartile range of waiting times

The median is the middle value:n/2 = 100/2 = 50th value

Lower class boundary

(lcb)

class frequenc

y

median = 14.5 + 15

4450 (19.5 – 14.5)

Median position

Cumulative frequency to lcb

ucb

lcb

15

14.5 19.5

median

44

15 – 19 class

lies in class 15 - 19

Example – Grouped data

Waiting times

1 - 4 5 - 9 10 - 14 15 - 19 20 - 29 30 - 39 40 - 49 50 +

frequency 9 15 20 15 20 11 10 0

Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded:

Answer:

Estimate the median and interquartile range of waiting times

The median is the middle value:n/2 = 100/2 = 50th value

Lower class boundary

(lcb)

class frequenc

y

median = 14.5 + 15

4450 (19.5 – 14.5)

Median position

Cumulative frequency to lcb

ucb

lcb

9

14.5 19.5

6

median

44

15 – 19 class

Linear interpolation: Assume 15 values

are evenly distributed in classlies in class 15 - 19

Example – Grouped data

Waiting times

1 - 4 5 - 9 10 - 14 15 - 19 20 - 29 30 - 39 40 - 49 50 +

frequency 9 15 20 15 20 11 10 0

Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded:

Answer:

Estimate the median and interquartile range of waiting times

The median is the middle value:n/2 = 100/2 = 50th value

Lower class boundary

(lcb)

class frequenc

y

median = 14.5 + 15

6

Frequency in class up to median

96

median

44

15 – 19 class

14.5 19.5

Linear interpolation: Assume 15 values

are evenly distributed in classlies in class 15 - 19

(19.5 – 14.5)

ucb

lcb

Example – Grouped data

Waiting times

1 - 4 5 - 9 10 - 14 15 - 19 20 - 29 30 - 39 40 - 49 50 +

frequency 9 15 20 15 20 11 10 0

Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded:

Answer:

Estimate the median and interquartile range of waiting times

The median is the middle value:n/2 = 100/2 = 50th value

Lower class boundary

(lcb)

class frequenc

y

median = 14.5 + 15

6. (5)

Frequency in class up to median

class width

9

5

6

median

44

15 – 19 class

14.5 19.5

Linear interpolation: Assume 15 values

are evenly distributed in classlies in class 15 - 19

Example – Grouped data

Waiting times

1 - 4 5 - 9 10 - 14 15 - 19 20 - 29 30 - 39 40 - 49 50 +

frequency 9 15 20 15 20 11 10 0

Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded:

Answer:

Estimate the median and interquartile range of waiting times

The median is the middle value:n/2 = 100/2 = 50th value

median = 14.5 + 9

2

6

median

44

15 – 19 class

14.5 19.5

Linear interpolation: Assume 15 values

are evenly distributed in class

2

3

lies in class 15 - 19

Example – Grouped data

Waiting times

1 - 4 5 - 9 10 - 14 15 - 19 20 - 29 30 - 39 40 - 49 50 +

frequency 9 15 20 15 20 11 10 0

Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded:

Answer:

Estimate the median and interquartile range of waiting times

The median is the middle value:n/2 = 100/2 = 50th value

median = 14.5 + 96

median

= 16.5

44

15 – 19 class

14.5 19.5

Linear interpolation: Assume 15 values

are evenly distributed in class

2 = 16.5 minutes

lies in class 15 - 19

Example – Grouped data

Waiting times

1 - 4 5 - 9 10 - 14 15 - 19 20 - 29 30 - 39 40 - 49 50 +

frequency 9 15 20 15 20 11 10 0

Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded:

Answer:

Estimate the median and interquartile range of waiting times

The Q1 value is the 1/4th value:n/4 = 100/4 = 25th value Q1 lies in class 10 - 14

Example – Grouped data

Waiting times

1 - 4 5 - 9 10 - 14 15 - 19 20 - 29 30 - 39 40 - 49 50 +

frequency 9 15 20 15 20 11 10 0

Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded:

Answer:

Estimate the median and interquartile range of waiting times

The Q1 value is the 1/4th value:n/4 = 100/4 = 25th value

Lower class boundary

(lcb)

class frequenc

y

Q1 = 9.5 + 20

2425 (14.5 – 9.5)

Q1 position

Cumulative frequency to lcb

ucb

lcb

19

9.5 14.5

1

Q1

24

10 – 14 class

Q1 lies in class 10 - 14

Example – Grouped data

Waiting times

1 - 4 5 - 9 10 - 14 15 - 19 20 - 29 30 - 39 40 - 49 50 +

frequency 9 15 20 15 20 11 10 0

Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded:

Answer:

Estimate the median and interquartile range of waiting times

The Q1 value is the 1/4th value:n/4 = 100/4 = 25th value

Q1 = 9.5 + 20

1. (5) 19

9.5 14.5

1

Q1

24

10 – 14 class

Q1 lies in class 10 - 14

= 9.75

Lower class boundary

(lcb)

class frequenc

y

Frequency in class up to Q1

class width

Example – Grouped data

Waiting times

1 - 4 5 - 9 10 - 14 15 - 19 20 - 29 30 - 39 40 - 49 50 +

frequency 9 15 20 15 20 11 10 0

Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded:

Answer:

Estimate the median and interquartile range of waiting times

The Q1 value is the 1/4th value:n/4 = 100/4 = 25th value

Q1 = 9.5 + 20

1. (5) 19

9.5 14.5

1

Q1

24

10 – 14 class

Q1 lies in class 10 - 14

= 9.75

Example – Grouped data

Waiting times

1 - 4 5 - 9 10 - 14 15 - 19 20 - 29 30 - 39 40 - 49 50 +

frequency 9 15 20 15 20 11 10 0

Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded:

Answer:

Estimate the median and interquartile range of waiting times

The Q3 value is the 3/4th value:n/4 = 100/4 = 75th value Q3 lies in class 20 - 29

Q1 = 9.75

Example – Grouped data

Waiting times

1 - 4 5 - 9 10 - 14 15 - 19 20 - 29 30 - 39 40 - 49 50 +

frequency 9 15 20 15 20 11 10 0

Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded:

Answer:

Estimate the median and interquartile range of waiting times

The Q3 value is the 3/4th value:n/4 = 100/4 = 75th value

Lower class boundary

(lcb)

class frequenc

y

Q3 = 19.5 + 20

5975 (19.5 – 29.5)

Q3 position

Cumulative frequency to lcb

ucb

lcb

4

19.5 29.5

16

Q3

59

20 – 29 class

Q3 lies in class 20 - 29

Q1 = 9.75

Example – Grouped data

Waiting times

1 - 4 5 - 9 10 - 14 15 - 19 20 - 29 30 - 39 40 - 49 50 +

frequency 9 15 20 15 20 11 10 0

Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded:

Answer:

Estimate the median and interquartile range of waiting times

The Q3 value is the 3/4th value:n/4 = 100/4 = 75th value

Lower class boundary

(lcb)

Q3 = 19.5 + 20

16. (10) 4

19.5 29.5

16

Q3

59

20 – 29 class

Q3 lies in class 20 - 29

Q1 = 9.75

class frequenc

y

Frequency in class up to Q3

class width

= 27.5

Example – Grouped data

Waiting times

1 - 4 5 - 9 10 - 14 15 - 19 20 - 29 30 - 39 40 - 49 50 +

frequency 9 15 20 15 20 11 10 0

Question: Waiting times, to the nearest minute, at a doctors surgery for 100 patients were recorded:

Answer:

Estimate the median and interquartile range of waiting times

Q1 = 9.75 Q3 = 27.5

IQR = Q3 – Q1 = 27.5 – 9.75 = 17.75 minutes

Grouped data - summary

frequencyclass

lcbtofreqcumulativepositionquantile )( Quantile = lcb + .(ucb – lcb)

= lcb + frequencyclass

quantiletoupclassinfreq . class width

• Use linear interpolation to estimate quantile.• Treat data as continuous within each group / class• Assumes values are evenly distributed within each class.

rest of

classfreq

lcb ucb

freq in class up to quantile

Quantile

cum. freq. to lcb

class