Higher Engineering Mathematics, Sixth Edition

542 Higher 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 the

median 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

6days. 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 exercise

Exercise 207 Further problems on mean,

median and mode for discrete data

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}
   [mean 7^13 , median 8, mode 8]


2. {26, 31, 21, 29, 32, 26, 25, 28}
   [mean 27.25, median 27, mode 26]


3. {4.72, 4.71, 4.74, 4.73, 4.72, 4.71, 4.73, 4.72}
   [mean 4.7225, median 4.72, mode 4.72]


4. {73.8, 126.4, 40.7, 141.7, 28.5, 237.4, 157.9}
   [mean 115.2, median 126.4, no mode]

55.3 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 dividing by the sum of the frequencies,

i.e. mean valuex=

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.

20.5–20.9 3, 21.0–21.4 10,

21.5–21.9 11, 22.0–22.4 13,

22.5–22.9 9, 23.0–23.4 2

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