Higher Engineering Mathematics, Sixth Edition

Measures of central tendency and dispersion 543

where fis 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 valueis 21.9ohms,correct to 3 significant
figures.

Histogram

The mean, median and modal values for grouped data
may be determined from ahistogram. In a histogram,
frequency values are represented vertically and vari-
able values horizontally. The mean value is given by
the value of the variable corresponding to a vertical
line drawn through the centroid of the histogram. The
median value is obtained by selecting a variable value
such that the area of the histogramto the left of a vertical
line drawn through the selected variable value is equal
to the area of the histogram on the right of the line. The
modal value is the variable value obtained by dividing
the width of the highest rectangle in the histogram in
proportionto the heights of the adjacent rectangles. The
method of determining the mean, median and modal
values from a histogram is shown in Problem 4.

Problem 4. The time taken in minutes to

assemble a device is measured 50 times and the

results are as shown. Draw a histogram depicting

this data and hence determine the mean, median

and modal values of the distribution.

14.5–15.5 5, 16.5–17.5 8,

18.5–19.5 16, 20.5–21.5 12,

22.5–23.5 6, 24.5–25.5 3

The histogram is shown in Fig. 55.1. The mean value
lies at the centroid of the histogram. With reference to
any arbitrary axis, sayYYshownat a timeof 14minutes,
theposition ofthehorizontalvalueofthecentroidcan be
obtained from the relationshipAM=

∑
(am),whereA
is the area of the histogram,Mis the horizontal distance
of the centroid from the axisYY,ais the area of a rectan-
gle of the histogram andmis thedistanceof the centroid
of the rectangle fromYY. The areas of the individual
rectangles are shown circled on the histogram giving a

14 15 16 17 18 19 20 21 22 23 24 25

6

26 27

4

2

Frequency 6

10

8

12

16

14

Time in minutes

12

24

32

16

10

E

D

Y A

Y

5.6

Mode

Median Mean

F

C

B

Figure 55.1

total area of 100square units. The positions,m,ofthe

centroids of the individual rectangles are 1, 3 , 5 ,...units

fromYY. Thus

100 M=( 10 × 1 )+( 16 × 3 )+( 32 × 5 )

+( 24 × 7 )+( 12 × 9 )+( 6 × 11 )

i.e. M=

560

100

= 5 .6 units fromYY

Thus thepositionof themeanwithreference to the time

scale is 14+ 5 .6, i.e.19.6minutes.

The median is the value of time correspondingto a verti-

cal line dividing the total area of the histogram into two

equal parts. The total area is 100square units, hence the

vertical line must be drawn to give 50units of area on

each side. To achieve this with reference to Fig. 55.1,

rectangleABFEmust besplit so that 50−( 10 + 16 )units

of area lie on one side and 50−( 24 + 12 + 6 )units of

area lie on the other. This shows that the area ofABFE

is split so that 24units of area lie to the left of the line

and 8units of area lie to the right, i.e. the vertical line

must pass through19.5minutes. Thus themedianvalue

of the distribution is19.5minutes.

The mode is obtained by dividing the lineAB,which

is the height of the highest rectangle, proportionally to

the heights of the adjacent rectangles. With reference to

Fig. 55.1, this is done byjoiningACandBDand drawing

a vertical line through the point of intersection of these

two lines. This gives themodeof the distributionand is

19.3minutes.