Higher Engineering Mathematics

THE NORMAL DISTRIBUTION 565

J

deviation is determined using the 84% and 16%
cumulative frequency values, shown asQandR
in Fig. 58.6. The variable values forQandRare
35.7 and 31.4 respectively; thus two standard devi-
ations 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
distribution can be calculated using

mean,x=

(∑

fx

)

(∑

f

)

and standard deviation,

σ=

√

√

√

√

{(∑

[f(x− ̄x)^2 ]

)

(∑

f

)

}

wherefis the frequency of a class andxis the class
mid-point value. Using these formulae gives a mean
value of the distribution of 33.6 (as obtained graphi-
cally) 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 nor-

mally 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

51

15 2

25 3

35 6

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 boundary values are: 10, 20, 30, ..., 90 and 100.
The corresponding cumulative frequency values are
1, 1+ 2 =3, 1+ 2 + 3 =6, 12, 21, 27, 29, 31, 32 and

33. The percentage 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/percentage cumulative frequency values

are plotted as shown in Fig. 58.7. Although six of

the points lie approximately 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 distribution 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% cumula-

tive frequency values, shown asBandCin Fig. 58.7

and give: 2σ= 69 −28, i.e. the standard deviation

σ= 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 217 Further problems on testing

for a normal distribution

1. A frequency distribution of 150 measure-
   ments is as shown:

Class mid-point value Frequency

26.4 5

26.6 12

26.8 24

27.0 36

27.2 36

27.4 25

27.6 12

Use normal probability paper to show that

this data approximates to a normal distribu-

tion and hence determine the approximate

values of the mean and standard deviation of

the distribution. Use the formula for mean

and standard deviation to verify the results

obtained.

⎡

⎣

Graphically, x= 27 .1,σ= 0 .3;

by calculation, x= 27 .079,

σ= 0. 3001

⎤

⎦