Unit 23 : Measures of Central tendency and dispersion



Unit 20 : Measures of Central Tendency and Dispersion

Learning Objectives

The students should be able to:

• determine the mean, median and mode from ungrouped data

• determine the mean, median and mode from grouped data

• determine the range, inter quartile range and standard deviation.

Activities

Teacher demonstration and student hand-on exercise.

Use MS Excel spreadsheet, internal functions and data analysis to measure central tendency and dispersion.

Reference

Suen, S.N. (1998) “Mathematics for Hong Kong 5A”; Rev. Ed.; Canotta

Measures of Central Tendency and Dispersion

1. Measure of central tendency: mean, median and mode from grouped and ungrouped data

For a set of data, we determine a quantity used to summarise the whole set of data. This quantity is termed a measure of central tendency. The most commonly used measures are mean, medium and mode.

1.1 mean

For ungrouped data,

[pic]

Example 1

Find the mean for the set data: 3, 7, 2, 1, 7

Solution

[pic]

=

For grouped data,

[pic]

Example 2

a) Find the mean of the set of data: 25, 36, 42, 38, 36

b) Find the mean from the set of grouped data

|Class mark |10.5 |30.5 |50.5 |70.5 |90.5 |110.5 |

|Frequency |19 |6 |3 |2 |1 |2 |

Solution

a) mean =

b)

|x |f |x f |

|10.5 |19 |199.5 |

|30.5 |6 | |

|50.5 |3 | |

|70.5 |2 | |

|90.5 |1 | |

|110.5 |2 | |

|sum |33 | |

mean =

Example 3

The HK Consumer Price Index B from 1997 to 2003 was as following:

1996 99.7

1997 105.5

1998 108.5

1999 103.4

2000. 99.4

2001. 97.7

2002. 94.7

2003. 92.1

Calculate the average consumer price index B:

a) For the first 4 years, (1997 – 2000).

b) For the next 3 years, (2001 – 2003)

c) For all 7 years

d) Suppose the original data was lost, and only the 4- and 3-year averages in a) and b) were available. Would it still be possible to calculate the overall 7-year average? How?

Solution

a) From 1997 − 2000, n = 4.

The average price index = (105.5 + 108.5 + 103.4 + 99.4) ( 4 =

b) From 2001− 2003, n = 3.

The average price index = (97.7 + 94.7 + 92.1) ( 3 =

c) From 1997 − 2003, n = 7.

The average price index

= (105.5 + 108.5 + 103.4 + 99.4 + 97.7 + 94.7 + 92.1) ( 7

=

d) The average price index over 7 years = ( ( 4 + ( 3 ) ( (4 + 3)

=

1.2 Median

For ungrouped data,

Median = the middle datum, when n is odd.

Median = the mean of the two middle data, when n is even.

|e.g.1 For the set of data |e.g.2 For the set of data |

|2, 4, 7, 9, 21 |3, 5, 7, 7 |

| | |

|middle datum |middle of two data |

| | |

|median = 7 |median = (5 + 7) ( 2 |

| |= 6 |

For grouped data,

Step 1: Draw the cumulative frequency polygon.

Step 2: The median is the datum corresponding to the middle value of the cumulative frequency.

Example 4

a) Find the median of 2, 3, 10, 12, 999.

b) Find the median of 2, 3, 10, 12, 22, 123.

c) The cumulative frequency polygon for maths marks of a class is given below, find the median mark.

Solution

a) Median =

b) Median =

c) Total frequency

= 40

The rank of median

= 40/2 =

From the cumulative polygon,

median =

Example 5

The provisional figures on the population by age group in Hong Kong as at 9/2001 are tabulated below. Draw a cumulative frequency polygon and determine the median age for the population.

|Age group |0 − 9 |10 − 19 |

|x < 10 |676 |0.676 |

|10 ≦ x < 20 |885 |1.561 |

|20 ≦ x < 30 |1000 | |

|30 ≦ x < 40 |1267 | |

|40 ≦ x < 50 |1208 | |

|50 ≦ x < 60 |677 | |

|60 ≦ x < 70 |503 | |

|70 < x |499 | |

The rank of median = 6.715/2 =

The median age =

1.3 mode

For ungrouped data, mode is the datum that has the highest frequency.

For grouped data, modal class is the class that has the highest frequency.

Example 6

a) Find the mode of the data:

1, 2, 2, 2, 3, 3, 9

b) Find the modal class

|Class |10 - 14 |15 - 19 |20 - 24 |25 - 29 |

|Frequency |2 |8 |7 |3 |

Solution

a) The mode is

b) The modal class is

Example 7

The temperature in degree Celsius each day cover a three week period were follow:

17, 18, 20, 21, 19, 16, 15, 18, 20, 21, 21,,22, 21, 19, 20,19, 17,16,16,17.

Compute the mean, median, and mode of these raw dates by using two-degree intervals starting with 15-16. Draw a cumulative frequency polygon.

Solution

|Temperature |Tally |Frequency |Class mark |f x |

|(℃) | |f |x | |

|15 − 16 | | | | |

|17 − 18 | | | | |

|19 − 20 | | | | |

|21 − 22 | | | | |

|23 − 24 | | | | |

| | | | | |

|Sum | | | | |

|Temperature |cumulative |

|(℃) |frquency |

|< 14.5 | |

|< 16.5 | |

|< 18.5 | |

|< 20.5 | |

|< 22.5 | |

|< 24.5 | |

The mean temperature =

The modal class of temperature is

The median temperature is

Remark

Mean seems to be the most commonly used (and often misused) quantity for measuring central tendency. If the distribution of the data set shows a strong degree of skewness, then mean is not a reliable measure as it is strongly affected by the extreme values. In this case, medium may be a better choice. Mode is used when there is reason to choose the most commonly occurring data value as the representative for the whole data set.

Measure of dispersion: Range, Inter-quartile range and Standard deviation

Apart from using a measure of central tendency to summarise a set of data, we need a quantity to measure the degree of dispersion of the set of data (so that we can determine the reliability of the set of data). Range is a measure that is very simple to use but it provides relatively little information on dispersion. Quartile deviation is used in association with the median whereas standard deviation goes with the mean.

2.1 Range

For ungrouped data, the range is the difference between the largest datum and the smallest datum.

For grouped data, the range is the difference between the highest class boundary and the lowest boundary.

Example 8

a) Find the range of the data:

1, 2, 2, 2, 3, 3, 9

b) Find the range of the grouped data

|Class |10 - 14 |15 - 19 |20 - 24 |25 - 29 |

|Frequency |2 |8 |7 |3 |

Solution

a) The range =

b) The range =

2.2 Inter quartile range

Inter quartile range = Q3 – Q1

where Q1, Q2, Q3 are called quartiles which divide the data (which have been ranked, i.e. arranged in order) into four equal parts.

Moreover,

Q2 is the median of the whole set of data,

Q1 is the median of the lower half,

Q3 is the median of the upper half.

Quartile deviation, Q.D. = ½ (Q3 − Q1)

Example 9

Find the inter quartile range of

a) 1, 2, 3, 5, 11, 12, 13.

b) 1, 2, 3, 4, 11, 12, 13, 14.

Solution

a) inter-quartile range =

b) inter-quartile range =

Example 10

The following frequency distribution gives the life hours of a sample of 50 light bulbs:

|Life hours (‘000) |Frequency |Life hours |Cumulative |

| | |Up to (‘000) |frequency |

| | |0.6 | |

|0.6 to under 0.7 |2 |0.7 | |

|0.7 to under 0.8 |4 |0.8 | |

|0.8 to under 0.9 |6 |0.9 | |

|0.9 to under 1.0 |14 |1.0 | |

|1.0 to under 1.1 |13 |1.1 | |

|1.1 to under 1.2 |7 |1.2 | |

|1.2 to under 1.3 |4 |1.3 | |

Find the median and the inter-quartile range of the data.

The rank of median

= ½ × 50 =

The median of life hours is hrs.

The rank of upper quartile

= ¾ × 50 =

= 38 , to the nearest integer

The upper quartile Q3 is hrs

The rank of lower quartile

= ¼ × 50 =

= 13 , to the nearest integer

The lower quartile Q1 is hrs.

The inter-quartile range = Q3 − Q1

=

Quartile deviation = ½ (Q3 − Q1)

=

Example 11

Find the range, inter-quartile range and quartile deviation for the data in example 4 and example 7 respectively.

2.3 Standard deviation

For ungrouped data x1, x2,…,xn, with a mean [pic], the standard deviation (() is

For grouped data with class marks x1, x2,…,xn; corresponding frequencies f1,f2,…,fn, and a mean [pic], the standard deviation (() is

Example 12

Find the standard deviation for

a) the ungrouped data 8, 9, 10, 10, 11

b) the grouped data

|x |17 |22 |27 |32 |37 |42 |47 |

|f |2 |4 |7 |8 |7 |4 |2 |

Solution

a) mean [pic] = standard deviation [pic] =

Calculator key-in method:

| |Model 3600 |Model 3900 |Model 506R |

|Set Statistic mode |MODE 3 |MODE 2 |2ndF MODE 3 |

| | | |0 |

|Clear memory |KAC |KAC |2ndF CA |

|Key-in data |8 DATA 9 DATA |8 DATA 9 DATA |8 DATA 9 DATA |

| |10 DATA 10 DATA |10 DATA 10 DATA |10 DATA 10 DATA |

| |11 DATA |11 DATA |11 DATA |

|mean |SHIFT 1 |SHIFT 4 |2ndF 4 |

|s.d. |SHIFT 2 |SHIFT 5 |2ndF 6 |

b) mean [pic] = standard deviation [pic] =

Calculator key-in method:

| |Model 3600 |Model 3900 |Model 506R |

|Set Statistic mode |MODE 3 |MODE 2 |2ndF MODE 3 |

| | | |0 |

|Clear memory |KAC |KAC |2ndF CA |

|Key-in data |17 ( 2 DATA |17 ( 2 DATA |17 ， 2 DATA |

| |22 ( 4 DATA |22 ( 4 DATA |22 ， 4 DATA |

| |27 ( 7 DATA |27 ( 7 DATA |27 ， 7 DATA |

| |32 ( 8 DATA |32 ( 8 DATA |32 ， 8 DATA |

| |37 ( 7 DATA |37 ( 7 DATA |37 ， 7 DATA |

| |32 ( 4 DATA |32 ( 4 DATA |42 ， 4 DATA |

| |47 ( 2 DATA |47 ( 2 DATA |47 ， 2 DATA |

|mean |SHIFT 1 |SHIFT 4 |2ndF 4 |

|s.d. |SHIFT 2 |SHIFT 5 |2ndF 6 |

Example 13

The life hours of 50 light bulbs has the following frequency distribution. Complete the table with class marks. Calculate the mean and standard deviation.

|Life hours (‘000) |Class mark |Frequency |

|0.6 to under 0.7 |0.65 |2 |

|0.7 to under 0.8 |0.75 |4 |

|0.8 to under 0.9 |0.85 |6 |

|0.9 to under 1.0 |0.95 |14 |

|1.0 to under 1.1 |1.05 |13 |

|1.1 to under 1.2 |1.15 |7 |

|1.2 to under 1.3 |1.25 |4 |

Solution

mean [pic] = standard deviation [pic] =

Example 14

The height of Basil team members at the 2002 FIFA World Cup is listed as following:

Marcos 1.93

Cafu 1.76

Lucio 1.88

Roque Junior 1.86

Edmilson 1.85

Carlos 1.68

Richardino 1.76

Silva 1.85

Ronaldo 1.83

Rivaldo 1.86

Ronaldinho 1.80

Calculate the average height, and the standard deviation:

Solution

Example 15

Find the mean and standard deviation for the data given below:

|Age group x |Population ('000) |

|5 |676 |

|15 |885 |

|25 |1000 |

|35 |1267 |

|45 |1208 |

|55 |677 |

|65 |503 |

|75 |499 |

Solution

Practice

1. The Hong Kong unemployment rate in the year of 4/2003 – 4/2004 was as following:

5/2003 8.3

6/2003 8.6

7/2003 8.7

8/2003 8.6

9/2003 8.3

10/2003 8.0

11/2003 7.5

12/2003 7.3

1/2004 7.3

2/2004 7.2

3/2004 7.2

4/2004 7.1

5/2004 7.0

Calculate the average, median, mode and the standard deviation of unemployment rate:

a) For 5/2003 – 12/2003

b) For1/2004 – 5/2004

c) For all 13 months.

2. Find the mean, median, mode of the following:

10, 13, 14, 14, 14, 15, 15,16, 17, 22

3. Which student has the highest average mark?

|Student |A |B |C |

|English |78 |63 |55 |

|Chinese |80 |85 |72 |

|Mathematics |59 |71 |95 |

4. The frequency distribution of the lengths of 100 leaves from a certain species of plant is given below:

|length (mm) |Frequency |

|20 – 24 |6 |

|25 – 29 |10 |

|30 – 34 |18 |

|35 – 39 |25 |

|40 – 44 |22 |

|45 – 49 |15 |

|50 – 54 |4 |

5. The following table shows the distribution of heights of 50 students:

|Height (cm) |Frequency |

|160 – 164 |8 |

|165 – 169 |12 |

|170 – 174 |14 |

|175 – 179 |7 |

|180 – 184 |6 |

|185 – 189 |3 |

Find the range and standard deviation of heights.

6. The mean of one set of six numbers is 9 and the mean of a second set of eight numbers is 12.5. Calculate the mean of the combined set of fourteen numbers.

7. The mean of the numbers a, b, c, d is 8 and the mean of the numbers a, b, c, d, e, f, g is 11. What is the mean of the numbers e, f, g?

8. Find the mean and standard deviation of the 5 numbers in term of x:

x−5, x-3, x−2, x+1, x+4.

9. The mean of the five numbers 6, 9, 2, x, y is 5 and the standard deviation is[pic]. Find the values of x and y.

Answer:

1.a) 8.16: 8.3; undefined; 0.52 b) 7.16: 7.2; 7.2; 0.11 c) 7.78: 7.5; undefined; 0.65

2) 15: 4.5; 4 3) 72; 73; 74 4) 37.4 5) 30; 7.14 6) 11 7) 15

8) x−1; 3.16 9) (3, 5); (5, 3)

-----------------------

Temperature (℃)

Age group

[pic]

[pic]

Life hour (‘000)

24.5

22.5

20.5

18.5

16.5

14.5

Cumulative frequency polygon

for temperature

Frequency

1.3

1.2

24

20

16

12

8

10

Marks

Cumulative frequency polygon for marks in maths

99.5

89.5

79.5

69.5

59.5

49.5

39.5

29.5

19.5

9.5

frequency

40

35

30

25

20

15

10

5

0

20

Less than Cumulative frequency polygon for population

30

40

50

60

70

70+

Population (‘000 000)

8.0

7.0

6.0

5.0

4.0

3.0

2.0

1.0

0

4

0

1.1

1.0

0.9

0.8

0.7

0.6

Less than cumulative frequency polygon

for life hours of 50 sample light bulbs

Frequency

56

48

40

32

24

16

8

0

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