Basic Engineering Mathematics, Fifth Edition

300 Basic Engineering Mathematics

The mean value is obtained by adding together the

values of the members of the set and dividing by the

number of members in the set. Thus,

mean value,x

=

2 + 3 + 7 + 5 + 5 + 13 + 1 + 7 + 4 + 8 + 3 + 4 + 3

13

=

65

13

= 5

To obtain the median value the set is ranked, that is,

placed in ascending order of magnitude, and since the

set contains an odd number of members the value of the

middle member is the median value. Ranking the set

gives {1, 2, 3, 3, 3, 4, 4, 5, 5, 7, 7, 8, 13}. The middle

term is the seventh member; i.e., 4. Thus, themedian

value is 4.

Themodal valueis the value of the most commonly

occurring member and is 3 , which occurs three times,

all other members only occurring once or twice.

Problem 2. The following set of data refers to the

amount of money in £s taken by a news vendor for

6 days. Determine the mean, median and modal

values of the set

{27.90, 34.70, 54.40, 18.92, 47.60, 39.68}

Mean value

=

27. 90 + 34. 70 + 54. 40 + 18. 92 + 47. 60 + 39. 68

6

=£37.20

The ranked set is

{18.92, 27.90, 34.70, 39.68, 47.60, 54.40}.

Since the set has an even number of members, the mean

of the middle two members is taken to give the median

value; i.e.,

median value=

34. 70 + 39. 68

2

=£37.19

Since no two members have the same value, this set has

no mode.

Now try the following Practice Exercise

PracticeExercise 125 Mean, median and

mode for discrete data (answers on

page 353)

In problems 1 to 4, determine the mean, median

and modal values for the sets given.

1. {3, 8, 10, 7, 5, 14, 2, 9, 8}


2. {26, 31, 21, 29, 32, 26, 25, 28}


3. {4.72, 4.71, 4.74, 4.73, 4.72, 4.71, 4.73, 4.72}


4. {73.8, 126.4, 40.7, 141.7, 28.5, 237.4, 157.9}

32.3 Mean, median and mode for grouped data

The mean value for a set of grouped data is found by

determiningthesumofthe(frequency×classmid-point

values) and dividingby the sum of the frequencies; i.e.,

mean value,x=

f 1 x 1 +f 2 x 2 +···fnxn

f 1 +f 2 +···+fn

=

∑

(fx)

∑

f

wherefis the frequency of the class having a mid-point

value ofx, and so on.

Problem 3. The frequency distribution for the

value of resistance in ohms of 48 resistors is as

shown. Determine the mean value of resistance

20 .5–20. 93

21 .0–21. 410

21 .5–21. 911

22 .0–22. 413

22 .5–22. 99

23 .0–23. 42

The class mid-point/frequency values are 20.7 3, 21.2

10, 21.7 11, 22.2 13, 22.7 9 and 23.2 2.

For grouped data, the mean value is given by

x=

∑

(fx)

∑

f

where fis the class frequency andxis the class mid-

point value. Hence mean value,

( 3 × 20. 7 )+( 10 × 21. 2 )+( 11 × 21. 7 )

x=

+( 13 × 22. 2 )+( 9 × 22. 7 )+( 2 × 23. 2 )

48

=

1052. 1

48

= 21. 919 ...

i.e.the mean value is 21.9 ohms, correct to 3 significant

figures.

32.3.1 Histograms

The mean, median and modal values for grouped data

may be determined from ahistogram. In a histogram,