Higher Engineering Mathematics, Sixth Edition

The normal distribution 563

Probability

density

Standard deviations

z 1 0 z 2 z-value

Figure 58.2

Problem 1. The mean height of 500 people is

170cm and the standard deviation is 9cm.

Assuming the heights are normally distributed,

determine the number of people likely to have

heights between 150cm and 195cm.

The mean value,x, is 170cm and corresponds to a
normal standard variate value,z, of zero on the stan-
dardized normal curve. A height of 150cm has az-value

given byz=

x−x

σ

standard deviations, i.e.

150 − 170
9
or−2.22 standard deviations. Using a table of par-
tial areas beneath the standardized normal curve (see
Table 58.1), az-value of−2.22 corresponds to an area
of 0.4868 between the mean value and the ordinate
z=− 2 .22. The negativez-value shows that it lies to
the left of thez=0 ordinate.
This area is shown shaded in Fig. 58.3(a). Similarly,

195cm has az-value of

195 − 170

9

that is 2.78 standard

deviations. From Table 58.1, this value ofzcorresponds
to an area of 0.4973, the positive value ofzshowing
that it lies to the right of thez=0 ordinate. This area
is shown shaded in Fig. 58.3(b). The total area shaded
in Figs. 58.3(a) and (b) is shown in Fig. 58.3(c) and is
0. 4868 + 0 .4973, i.e. 0.9841 of the total area beneath
the curve.
However, the area is directlyproportionalto probability.
Thus, the probability that a person will have a height
of between 150 and 195cm is 0.9841. For a group of
500 people, 500× 0 .9841, i.e.492 people are likely to
haveheightsinthisrange.Thevalueof500× 0 .9841is
492.05, but since answers based on a normal probability
distributioncan onlybe approximate, results are usually
given correct to the nearest whole number.

Problem 2. For the group of people given in

Problem 1, find the number of people likely to have

heights of less than 165cm.

0 z-value

0 z-value

z-value

2 2.22

2 2.22

(a)

2.78

(b)

0 2.78

(c)

Figure 58.3

A height of 165cm corresponds to

165 − 170

9

i.e.

−0.56 standard deviations.

The area betweenz=0andz=− 0 .56 (from Table 58.1)

is 0.2123, shown shaded in Fig. 58.4(a). The total area

under the standardized normal curve is unity and since

the curve is symmetrical, it follows that the total area

to the left of thez=0 ordinate is 0.5000. Thus the area

to the left of thez=− 0 .56 ordinate (‘left’ means ‘less

than’, ‘right’ means ‘more than’) is 0. 5000 − 0 .2123,

i.e. 0.2877 of the total area, which is shown shaded in

Fig 58.4(b). The area is directly proportional to proba-

bility and since the total area beneath the standardized

normal curveisunity,theprobabilityofaperson’sheight

being less than 165cm is 0.2877. For agroupof500peo-

ple, 500× 0 .2877, i.e.144 people are likely to have

heights of less than 165 cm.

Problem 3. For the group of people given in

Problem 1 find how many people are likely to have

heights of more than 194cm.

194cm corresponds to az-value of

194 − 170

9

that is,

2.67 standard deviations. From Table 58.1, the area