Higher Engineering Mathematics, Sixth Edition

568 Higher Engineering Mathematics

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 probabilityscale 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 deter-

mined from Fig. 58.6. Since a normal curve is sym-

metrical, the mean value is the value of the variable

corresponding to a 50% cumulative frequency value,

shown as pointPon the graph. This shows thatthe

mean value is 33.6kg. The standard deviation is deter-

mined using the 84% and 16% cumulative frequency

values, shown asQandRin Fig. 58.6. The variable val-

ues forQandRare 35.7 and 31.4 respectively; thus two

standard deviations correspond to 35. 7 − 31 .4, i.e. 4.3,

showing that the standard deviation of the distribution

is approximately

4. 3

2

i.e.2.15 standard deviations.

The mean value and standard deviation of the distribu-

tion can be calculated using

mean,x=

(∑

fx

)

(∑

f

)

and standard deviation,

σ=

√

√

√

√

{(∑

[f(x− ̄x)^2 ]

)

(∑

f

)

}

wherefisthefrequencyofaclassandxistheclassmid-

point value. Using these formulae gives a mean value

of the distributionof 33.6 (as obtained graphically) and

a standard deviation of 2.12, showing that the graphical

method of determining the mean and standard deviation

give quite realistic results.

Problem 6. Use normal probability paper to

determine whether the data given below is normally

distributed. Use the graph and assume a normal

distribution whether this is so or not, to find

approximate values of the mean and standard

deviation of the distribution.

Class mid-point values Frequency

5 1

15 2

25 3

35 6

Class mid-point values Frequency

45 9

55 6

65 2

75 2

85 1

95 1

To test the normality of a distribution, the upper class

boundary/percentage cumulative frequency values are

plotted on normal probability paper. The upper class

boundaryvalues are: 10, 20, 30,..., 90 and 100.The cor-

respondingcumulativefrequencyvaluesare1,1+ 2 =3,

1 + 2 + 3 =6, 12, 21, 27, 29, 31, 32 and 33. The per-

centage cumulative frequency values are

1

33

× 100 =3,

3

33

× 100 =9, 18, 36, 64, 82, 88, 94, 97 and 100.

The co-ordinates of upper class boundary values/per-

centage cumulative frequency values are plotted as

showninFig.58.7.Althoughsixofthepointslieapprox-

imately in a straight line, three points corresponding to

upper class boundary values of 50, 60 and 70 are not

close to the line and indicate thatthe distribution is

not normallydistributed. However, if a normal dis-

tribution is assumed, the mean value corresponds to

the variable value at a cumulative frequency of 50%

and, from Fig. 58.7, pointAis 48. The value of the

standard deviation of the distribution can be obtained

from the variable values corresponding to the 84% and

16% cumulative frequency values, shown asBandCin

Fig. 58.7 and give: 2σ= 69 −28, i.e. the standard devi-

ationσ= 20. 5. The calculated values of the mean and

standard deviation of the distribution are 45.9 and 19.4

respectively, showing that errors are introduced if the

graphical method of determining these values is used

for data which is not normally distributed.

Now try the following exercise

Exercise 216 Further problems on testing

for a normal distribution

1. A frequency distributionof 150measurements
   is as shown: