Higher Engineering Mathematics, Sixth Edition

The normal distribution 565

0

2 0.56 0

2 0.56

(a)

z-value

z-value

(b)

Figure 58.4

betweenz=0,z= 2 .67 and the standardized normal
curve is 0.4962, shown shaded in Fig. 58.5(a). Since
the standardized normal curve is symmetrical, the total
area to therightof thez=0 ordinateis 0.5000, hence the
shaded area shown in Fig. 58.5(b) is 0. 5000 − 0 .4962,
i.e. 0.0038. This area represents the probabilityof a per-
son having a height of more than 194cm, and for 500
people, the number of people likely to have a height of
more than 194cm is 0. 0038 ×500, i.e.2people.

0 z-value

z-value

2.67

(a)

0 2.67

(b)

Figure 58.5

Problem 4. A batch of 1500 lemonade bottles

have an average contents of 753ml and the standard

deviation of the contents is 1.8ml. If the volumes of

the contents are normally distributed, find

(a) the number of bottles likely to contain less

than 750ml,

(b) the number of bottles likely to contain

between 751 and 754ml,

(c) the number of bottles likely to contain more

than 757ml, and

(d) the number of bottles likely to contain

between 750 and 751ml.

(a) Thez-value corresponding to 750ml is given

by

x−x

σ

i.e.

750 − 753

1. 8

=− 1 .67 standard devi-

ations. From Table 58.1, the area betweenz= 0

andz=− 1 .67 is 0.4525. Thus the area to the

left of thez=− 1 .67 ordinate is 0. 5000 − 0. 4525

(see Problem 2), i.e. 0.0475. This is the prob-

ability of a bottle containing less than 750ml.

Thus, for a batch of 1500 bottles, it is likely that

1500 × 0 .0475, i.e.71 bottles will contain less

than 750ml.

(b) The z-value corresponding to 751 and 754ml

are

751 − 753

1. 8

and

754 − 753

1. 8

i.e. −1.11 and

0.56 respectively. From Table 58.1, the areas

corresponding to these values are 0.3665 and

0.2123 respectively. Thus the probability of a

bottle containing between 751 and 754ml is

0. 3665 + 0 .2123 (see Problem 1), i.e. 0.5788. For

1500 bottles, it is likely that 1500× 0 .5788, i.e.

868bottleswill containbetween751and754ml.

(c) Thez-valuecorrespondingto757ml is

757 − 753

1. 8

,

i.e. 2.22 standard deviations. From Table 58.1,

the area corresponding to az-value of 2.22 is

0.4868. The area to the right of thez= 2. 22

ordinate is 0. 5000 − 0 .4868 (see Problem 3), i.e.

0.0132. Thus, for 1500 bottles, it is likely that

1500 × 0 .0132, i.e.20 bottles will have contents

of more than 757ml.

(d) Thez-value corresponding to 750ml is−1.67

(see part (a)), and thez-value corresponding to

751ml is−1.11 (see part (b)). The areas corre-

sponding to thesez-values are 0.4525 and 0.3665

respectively, and both these areas lie on the left

of thez=0 ordinate. The area betweenz=− 1. 67

andz=− 1 .11 is 0. 4525 − 0 .3665, i.e. 0.0860 and

this is the probability of a bottle having contents

between 750 and 751ml. For 1500 bottles, it is