Unit 20 - Central Tendency and Dispersion (Student)

7/28/2019 Unit 20 - Central Tendency and Dispersion (Student)

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CMV6120 Mathematics

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 anddispersion.

Reference

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

Unit 20 Central tendency & dispersion Page 1 of 13


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Measures of Central Tendency and Dispersion

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

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

1.1 mean

For ungrouped data,

n x x x x

x Mean n++++

= ...)( 321

Example 1

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

Solution

571273 ++++= x

=

For grouped data,

n

nn

f f f f f x f x f x f x

x Mean++++

++++=

......

)(321

332211

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 90.5 110.5Frequency 19 6 3 2 1 2

Solution

a) mean =



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b) x f x f

10.5 19 199.530.5 650.5 3

70.5 290.5 1

110.5 2sum 33

mean =

Example 3

The HK Consumer Price Index B from 1997 to 2003 was as following:1996 99.71997 105.51998 108.51999 103.42000 99.42001 97.72002 94.72003 92.1Calculate 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 yearsd) 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)=



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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 data2, 4, 7, 9, 21

middle datum

median = 7

e.g.2 For the set of data3, 5, 7, 7

middle of two data

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 =

Unit 20 Central tendency & dispersion Page 4 of 13 0

5

10

15

20

25

30

35

40

f r e q u e n

c y

9 .

5 1 9

. 5 2 9 . 5 3 9

. 5 4 9 . 5 5 9

. 5 6 9 . 5 7 9

. 5 8 9

. 5 9 9 . 5

Cumulative frequency polygon for marks in maths

Marks


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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 20 29 30 39 40 49 50 59 60 69 > 70Population(000)

676 885 1000 1267 1208 677 503 499

Solution

Age group x

Population('000)

cumulative population

(000 000) x < 10 676 0.676

10 x < 20 885 1.56120 x < 30 100030 x < 40 126740 x < 50 120850 x < 60 67760 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 - 29Frequency 2 8 7 3

Solution

a) The mode is


0

1.0

2.0

3.0

4.0

5.0

6.07.0

8.0

P o p ul a

t i on

( 0 0 0 0 0 0 )

Less than Cumulative frequency polygon for population

1 0

2 0

3 0

4 0

5 0

6 0

7 0

7 0

+

Age group


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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 hoursUp to (000)

Cumulativefrequency

0.60.6 to under 0.7 2 0.70.7 to under 0.8 4 0.80.8 to under 0.9 6 0.90.9 to under 1.0 14 1.01.0 to under 1.1 13 1.11.1 to under 1.2 7 1.21.2 to under 1.3 4 1.3

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

The rank of median= 50 =

The median of life hours is hrs.

The rank of upper quartile= 50 == 38 , to the nearest integer

The upper quartile Q 3 is hrs

The rank of lower quartile= 50 == 13 , to the nearest integer

The lower quartile Q 1 is hrs.

The inter-quartile range = Q 3 Q1 =

Quartile deviation = (Q 3 Q1)=


0

8

16

24

32

40

48

56F r e

q u e n

c y

Less than cumulative frequency polygonfor life hours of 50 sample light bulbs

0 .

6 0 .

7 0 .

8 0 .

9 1 .

0 1 .

1 1 .

2 1 .

3

Life hour (000)


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Example 11Find the range, inter-quartile range and quartile deviation for the data in example 4 and example 7respectively.

2.3 Standard deviationFor ungrouped data x1, x2,, xn, with a mean x , the standard deviation ( ) is

For grouped data with class marks x1, x2,, xn; corresponding frequencies f 1, f 2,, f n, and a mean x ,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 =i

i

n

x x = standard deviation

n

x xi

i

=

2)(

=

Calculator key-in method:Model 3600 Model 3900 Model 506R

Set Statistic mode MODE 3 MODE 2 2ndF MODE 30

Clear memory KAC KAC 2ndF CAKey-in data 8 DATA 9 DATA

10 DATA 10 DATA11 DATA

8 DATA 9 DATA10 DATA 10 DATA11 DATA

8 DATA 9 DATA10 DATA 10 DATA11 DATA

mean SHIFT 1 SHIFT 4 2ndF 4s.d. SHIFT 2 SHIFT 5 2ndF 6


n

nn

f f f f f x x f x x f x x f x x

++++

++++

...)(...)()()(

321

23

232

221

21

n

x x x x x x x x n22

3

2

2

2

1 )(...)()()( ++++


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b) mean =i

i

n

x x = standard deviation

n

x xi

i =

2)(

=

Calculator key-in method:Model 3600 Model 3900 Model 506R

Set Statistic mode MODE 3 MODE 2 2ndF MODE 30

Clear memory KAC KAC 2ndF CAKey-in data 17 2 DATA

22 4 DATA27 7 DATA32 8 DATA37 7 DATA32 4 DATA47 2 DATA

17 2 DATA22 4 DATA27 7 DATA32 8 DATA37 7 DATA32 4 DATA47 2 DATA

17 2 DATA22 4 DATA27 7 DATA32 8 DATA37 7 DATA42 4 DATA47 2 DATA

mean SHIFT 1 SHIFT 4 2ndF 4s.d. SHIFT 2 SHIFT 5 2ndF 6

Example 13

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

Life hours (000) Class mark Frequency0.6 to under 0.7 0.65 20.7 to under 0.8 0.75 40.8 to under 0.9 0.85 60.9 to under 1.0 0.95 141.0 to under 1.1 1.05 131.1 to under 1.2 1.15 71.2 to under 1.3 1.25 4

Solution

mean = ii

n

x x

= standard deviation n

x xi

i =

2)( =



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Example 14

The height of Basil team members at the 2002 FIFA World Cup is listed as following:Marcos 1.93

Cafu 1.76Lucio 1.88Roque Junior 1.86Edmilson 1.85Carlos 1.68Richardino 1.76Silva 1.85Ronaldo 1.83Rivaldo 1.86Ronaldinho 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 88525 100035 126745 120855 677

65 50375 499

Solution



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Practice

1. The Hong Kong unemployment rate in the year of 4/2003 4/2004 was as following:5/2003 8.36/2003 8.67/2003 8.78/2003 8.69/2003 8.310/2003 8.011/2003 7.512/2003 7.31/2004 7.32/2004 7.23/2004 7.24/2004 7.15/2004 7.0

Calculate the average, median, mode and the standard deviation of unemployment rate:a) For 5/2003 12/2003

b) For1/2004 5/2004c) 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 CEnglish 78 63 55Chinese 80 85 72

Mathematics 59 71 95

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

length (mm) Frequency

20 24 625 29 1030 34 1835 39 2540 44 2245 49 1550 54 4



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

Height (cm) Frequency160 164 8165 169 12

170 174 14175 179 7180 184 6185 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 is12.5. 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 6 . Find thevalues 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; 0.652) 15: 4.5; 4 3) 72; 73; 74 4) 37.4 5) 30; 7.14 6) 11 7) 158) x 1; 3.16 9) (3, 5); (5, 3)


Unit 20 - Central Tendency and Dispersion (Student)

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