566 Higher Engineering Mathematics
likely that 1500 ×0.0860, i.e. 129 bottles will be
in this range.
Now try the following exercise
Exercise 215 Further problems on the
introduction to the normal distribution
1. 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]
2. The masses of 800 people are normally distributed, 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]
3. 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, calculate 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]
4. 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 component, the number likely to be rejected due to
being oversized.
[4]
5. 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]
6. The mean diameter of holes produced by a
drilling machine bit is 4.05 mm and the standard deviation of the diameters is 0.0028 mm.
For twenty holes drilled using this machine,
determine, correct to the nearest whole number, how many are likely to have diameters
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]
7. The intelligence quotients of 400 children
have a mean value of 100 and a standard deviation 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]
8. 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 continuous it automatically follows that it is normally
distributed. One way of checking that data is normally
distributed is by using normal probability paper, often
just called probability 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 determined using normal probability paper. For normally distributed data, the area beneath the standardized normal
curve and a z-value of unity (i.e. one standard 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 the z = 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 probability
likely that 1500 ×0.0860, i.e. 129 bottles will be
in this range.
Now try the following exercise
Exercise 215 Further problems on the
introduction to the normal distribution
1. 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]
2. The masses of 800 people are normally distributed, 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]
3. 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, calculate 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]
4. 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 component, the number likely to be rejected due to
being oversized.
[4]
5. 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]
6. The mean diameter of holes produced by a
drilling machine bit is 4.05 mm and the standard deviation of the diameters is 0.0028 mm.
For twenty holes drilled using this machine,
determine, correct to the nearest whole number, how many are likely to have diameters
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]
7. The intelligence quotients of 400 children
have a mean value of 100 and a standard deviation 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]
8. 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 continuous it automatically follows that it is normally
distributed. One way of checking that data is normally
distributed is by using normal probability paper, often
just called probability 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 determined using normal probability paper. For normally distributed data, the area beneath the standardized normal
curve and a z-value of unity (i.e. one standard 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 the z = 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 probability