562 STATISTICS AND PROBABILITY
−0.56 (^0) z-value
−0.56 (^0) z-value
(a)
(b)
Figure 58.4
Since the standardized normal curve is symmetri-
cal, the total area to the right of thez=0 ordinate is
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 probability of a person having a height of more
than 194 cm, and for 500 people, the number of peo-
ple likely to have a height of more than 194 cm is
0. 0038 ×500, i.e.2 people.
0 2.67 z-value
(a)
0 2.67 z-value
(b)
Figure 58.5
Problem 4. A batch of 1500 lemonade bot-
tles have an average contents of 753 ml and
the standard deviation of the contents is 1.8 ml.
If the volumes of the contents are normally
distributed, find
(a) the number of bottles likely to contain less
than 750 ml,
(b) the number of bottles likely to contain
between 751 and 754 ml,
(c) the number of bottles likely to contain more
than 757 ml, and
(d) the number of bottles likely to contain
between 750 and 751 ml.
(a) Thez-value corresponding to 750 ml is given by
x−x
σ
i.e.
750 − 753
- 8
=− 1 .67 standard devia-
tions. 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 750 ml.
Thus, for a batch of 1500 bottles, it is likely that
1500 × 0 .0475, i.e.71 bottles will contain less
than 750 ml.
(b) Thez-value corresponding to 751 and 754 ml
are
751 − 753 - 8
and
754 − 753 - 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 754 ml is - 3665 + 0 .2123 (see Problem 1), i.e. 0.5788.
For 1500 bottles, it is likely that 1500× 0 .5788,
i.e.868 bottles will contain between 751 and
754 ml.
(c) The z-value corresponding to 757 ml is
757 − 753 - 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 the
z= 2 .22 ordinate is 0. 5000 − 0 .4868 (see Prob-
lem 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 757 ml.
(d) Thez-value corresponding to 750 ml is−1.67
(see part (a)), and thez-value corresponding
to 751 ml is−1.11 (see part (b)). The areas
corresponding to thesez-values are 0.4525 and
0.3665 respectively, and both these areas lie on
the left of thez=0 ordinate. The area between
z=− 1 .67 andz=− 1 .11 is 0. 4525 − 0 .3665,
i.e. 0.0860 and this is the probability of a bot-
tle having contents between 750 and 751 ml.
For 1500 bottles, it is likely that 1500× 0 .0860,
i.e.129 bottles will be in this range.