564 STATISTICS AND PROBABILITY30 32 34 36 38 40 420.010.050.10.20.512510203040506070809095989999.899.999.99Percentage cumulative frequencyUpper class boundaryRPQFigure 58.6Problem 5. Use normal probability paper to
determine whether the data given below, which
refers to the masses of 50 copper ingots, is
approximately normally distributed. If the data
is normally distributed, determine the mean and
standard deviation of the data from the graph
drawn.Class mid-point value (kg) Frequency29.5 2
30.5 4
31.5 6
32.5 8
33.5 9
34.5 8
35.5 6
36.5 4
37.5 2
38.5 1To test the normality of a distribution, the upper class
boundary/percentage cumulative frequency values
are plotted on normal probability paper. The upper
class boundary values are: 30, 31, 32, ..., 38, 39.
0.010.050.10.20.512510203040506070809095989999.999.9910 20 30 40 50 60 70 80 90 100 110
Upper class boundaryPercentage cumulative frequencyBACFigure 58.7The corresponding cumulative frequency values (for
‘less than’ the upper class boundary values) are:
2, (4+2)=6, (6+ 4 +2)=12, 20, 29, 37, 43, 47,
49 and 50. The corresponding percentage cumulativefrequency values are2
50× 100 =4,6
50× 100 =12,
24, 40, 58, 74, 86, 94, 98 and 100%.
The co-ordinates of upper class boundary/percen-
tage cumulative frequency values are plotted as
shown in Fig. 58.6. When plotting these values, it
will always be found that the co-ordinate for the
100% cumulative frequency value cannot be plotted,
since the maximum value on the probability scale is
99.99.Since the points plotted in Fig. 58.6 lie very
nearly in a straight line, the data is approximately
normally distributed.
The mean value and standard deviation can be
determined from Fig. 58.6. Since a normal curve
is symmetrical, the mean value is the value of the
variable corresponding to a 50% cumulative fre-
quency value, shown as pointPon the graph. This
shows thatthe mean value is 33.6 kg. The standard