Cambridge International Mathematics

288 Analysis of discrete data (Chapter 13)

Stem Leaf

2 0122

3 0014458

4 0234669

5 11458

5 j 1 represents 51

3 For the data set given, find:

a the minimum value b the maximum value

c the median d the lower quartile

e the upper quartile f the range

g the interquartile range.

When the same data appears several times we often summarise it in afrequency table. For convenience we

denote the data values byxand the frequencies of these values byf.

Data value Frequency Product

3 1 1 £3=3

4 2 2 £4=8

5 4 4 £5=20

6 14 14 £6=84

7 11 11 £7=77

8 6 6 £8=48

9 2 2 £9=18

Total 40 258

The mode

There are 14 of data value 6 which is more than any other

data value.

The mode is therefore 6.

We know the mean=

the sum of all data values

the number of data values

, so for data in a frequency table, x=

P

fx

P

f

The median

There are 40 data values, an even number, so there aretwo

middledata values.

) the median is the average of the 20 th and 21 st data values.

In the table, the blue numbers show us accumulated values.

Data Value (x) Frequency (f)

3 1 1

4 2 3

5 4 7

6 14 21

7 11 32

8 6

9 2

Total 40

one number is 3

3 numbers are 4 or less

7 numbers are 5 or less

21 numbers are 6 or less

32 numbers are 7 or less

We can see that the 20 th and 21 st

data values (in order) are both 6 s.

Notice that we have a skewed distribution for which the mean, median and mode are nearly equal. This is

why we need to be careful when we use measures of the middle to call distributions symmetric.

E DATA IN FREQUENCY TABLES [11.4]

Remember that

the median is the

middle of the

ordereddata set.

The mean

A‘Product’ column helps to add all scores.

In this case the mean=

P

fx

P

f

=

258

40

=6: 45.

As the sample size n=40,

n+1

2

=

41

2

=20: 5

) median=

6+6

2

=6.

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y:\HAESE\IGCSE01\IG01_13\288IGCSE01_13.CDR Thursday, 16 October 2008 9:40:33 AM PETER