THE NORMAL DISTRIBUTION 563J
Now try the following exercise.
Exercise 216 Further problems on the
introduction to the normal distribution- A component is classed as defective if it has a
diameter of less than 69 mm. In a batch of 350
components, the mean diameter is 75 mm and
the standard deviation is 2.8 mm. Assuming
the diameters are normally distributed,
determine how many are likely to be classed
as defective. [6] - The masses of 800 people are normally dis-
tributed, having a mean value of 64.7 kg and a
standard deviation of 5.4 kg. Find how many
people are likely to have masses of less than
54.4 kg. [22] - 500 tins of paint have a mean content of
1010 ml and the standard deviation of the
contents is 8.7 ml. Assuming the volumes of
the contents are normally distributed, calcu-
late the number of tins likely to have contents
whose volumes are less than (a) 1025 ml
(b) 1000 ml and (c) 995 ml.
[(a) 479 (b) 63 (c) 21] - For the 350 components in Problem 1, if those
having a diameter of more than 81.5 mm are
rejected, find, correct to the nearest compo-
nent, the number likely to be rejected due to
being oversized. [4] - For the 800 people in Problem 2, determine
how many are likely to have masses of more
than (a) 70 kg and (b) 62 kg.
[(a) 131 (b) 553] - The mean diameter of holes produced by a
drilling machine bit is 4.05 mm and the stan-
dard deviation of the diameters is 0.0028 mm.
For twenty holes drilled using this machine,
determine, correct to the nearest whole num-
ber, how many are likely to have diame-
ters of between (a) 4.048 and 4.0553 mm
and (b) 4.052 and 4.056 mm, assuming the
diameters are normally distributed.
[(a) 15 (b) 4] - The intelligence quotients of 400 children
have a mean value of 100 and a standard devi-
ation of 14. Assuming that I.Q.’s are normally
distributed, determine the number of children
likely to have I.Q.’s of between (a) 80 and 90,
(b) 90 and 110 and (c) 110 and 130.
[(a) 65 (b) 209 (c) 89]- The mean mass of active material in tablets
produced by a manufacturer is 5.00 g and the
standard deviation of the masses is 0.036 g.
In a bottle containing 100 tablets, find how
many tablets are likely to have masses of
(a) between 4.88 and 4.92 g, (b) between 4.92
and 5.04 g and (c) more than 5.04 g.
[(a) 1 (b) 85 (c) 13]
58.2 Testing for a normal distribution
It should never be assumed that because data is con-
tinuous it automatically follows that it is normally
distributed. One way of checking that data is nor-
mally distributed is by usingnormal probability
paper, often just calledprobability paper. This is
special graph paper which has linear markings on
one axis and percentage probability values from 0.01
to 99.99 on the other axis (see Figs. 58.6 and 58.7).
The divisions on the probability axis are such that
a straight line graph results for normally distributed
data when percentage cumulative frequency values
are plotted against upper class boundary values. If
the points do not lie in a reasonably straight line,
then the data is not normally distributed. The method
used to test the normality of a distribution is shown
in Problems 5 and 6. The mean value and standard
deviation of normally distributed data may be deter-
mined using normal probability paper. For normally
distributed data, the area beneath the standardized
normal curve and az-value of unity (i.e. one stan-
dard deviation) may be obtained from Table 58.1.
For one standard deviation, this area is 0.3413,
i.e. 34.13%. An area of±1 standard deviation is
symmetrically placed on either side of thez= 0
value, i.e. is symmetrically placed on either side of
the 50% cumulative frequency value. Thus an area
corresponding to±1 standard deviation extends
from percentage cumulative frequency values of
(50+ 34 .13)% to (50− 34 .13)%, i.e. from 84.13%
to 15.87%. For most purposes, these values are taken
as 84% and 16%. Thus, when using normal probabil-
ity paper, the standard deviation of the distribution
is given by:
(
variable value for 84% cumulative frequency−
variable value for 16% cumulative frequency)2