Higher Engineering Mathematics

540 STATISTICS AND PROBABILITY

histogram, frequency values are represented verti-

cally and variable values horizontally. The mean

value is given by the value of the variable corre-

sponding 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

histogram to 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 proportion

to 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, sayYYshown at a time of
14 minutes, the position of the horizontal value of
the centroid can be obtained from the relationship
AM=

∑

(am), whereAis the area of the histogram,

Figure 55.1

Mis the horizontal distance of the centroid from the

axisYY,ais the area of a rectangle of the histogram

andmis the distance of the centroid of the rectangle

fromYY. The areas of the individual rectangles are

shown circled on the histogram giving a total area of

100 square units. The positions,m, of the centroids

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

YY. 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 the position of themeanwith reference to the

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

The median is the value of time corresponding to a

vertical line dividing the total area of the histogram

into two equal parts. The total area is 100 square

units, hence the vertical line must be drawn to give

50 units of area on each side. To achieve this with

reference to Fig. 55.1, rectangleABFEmust be split

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 ofABFEis split so that

24 units of area lie to the left of the line and 8 units

of area lie to the right, i.e. the vertical line must pass

through 19.5 minutes. Thus themedian valueof the

distribution is19.5 minutes.

The mode is obtained by dividing the lineAB,

which is the height of the highest rectangle, pro-

portionally to the heights of the adjacent rectangles.

With reference to Fig. 55.1, this is done by joining

ACandBDand drawing a vertical line through the

point of intersection of these two lines. This gives

themodeof the distribution and is19.3 minutes.

Now try the following exercise.

Exercise 209 Further problems on mean,

median and mode for grouped data

1. The frequency distribution given below refers
   to the heights in centimetres of 100 people.
   Determine the mean value of the distribution,
   correct to the nearest millimetre.

150–156 5, 157–163 18,

164–170 20, 171–177 27,

178–184 22, 185–191 8

[171.7 cm]