Higher Engineering Mathematics

MEASURES OF CENTRAL TENDENCY AND DISPERSION 539

J

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
hasno mode.

Now try the following exercise.

Exercise 208 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 determining the sum of the (frequency×class

mid-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

wherefis the class frequency andxis the class mid-

point value. Hence mean value,

x=

(3× 20 .7)+(10× 21 .2)+(11× 21 .7)

+(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.

Histogram

The mean, median and modal values for grouped

data may be determined from ahistogram.Ina