In this presentation different applications of normal probability curve have been discussed with the use of different practical problems. This presentation shall definitely help the viewers to solve problems based on NPC.

1. Applications of Normal
Distribution -I
Vikramjit Singh, PhD
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0.8
1
1.2
0 10 20 30 40 50 60 70
Normal Probablity Curve

2. Major Formula and Table of Use
Z=
𝑿−𝑴
σ
or Z=
𝒙
σ
In which: z = Standard Score
X = Raw Score
M = Mean of X Scores
σ = Standard Deviation of X Scores
© V. Singh
If Mean = 25 , σ = 5 and X= 30
Then Z score or standard score .
Z=
𝑿−𝑴
σ
=
𝟑𝟎−𝟐𝟓
5
= 1 σ

3. Major Formula and Table of Use
© V. Singh
Fractional parts of the total area (taken as 10,000) under the normal probability
curve, corresponding to distance on the baseline between the mean and
successive points laid off from the mean in units of standard deviation.
3413 or
34.13 %
4418 or
44.18%

4. Application : 1
Ex- In a test the mean score obtained
by the students is 60 and σ is 5.
Assuming the normality of the
distribution , Find (a) Percentage of
cases falling in between 65 and 55 , (b)
above 70 and (c) below 45.
© V. Singh
To determine percentage of cases in a
normal distribution within given
limits.

5. © V. Singh
1 σ
34.13%
-1 σ
34.13%
68.26%
13.59%
13.59% 95.44%
-2 σ 2 σ
-3 σ
2.14% 2.14%
3 σ
50%
50%
99.73%
0.14 %
0.14 %
Ex- In a test the mean
score obtained by the
students is 60 and σ is
5. Assuming the
normality of the
distribution , Find (a)
Percentage of cases
falling in between 65
and 55 , (b) above 70
and (c) below 45.

6. © V. Singh
Here
Mean (M) = 60
SD(σ) is 5
Also to convert it into Z-Score the
formula
Z=
𝑿−𝑴
σ So for X = 65 and 55
Z(X=65) =
𝑿−𝑴
σ =
=
𝟔𝟓−𝟔𝟎
5
= 1 σ
Z(X=55) =
𝑿−𝑴
σ
=
=
𝟓𝟓−𝟔𝟎
5 = -1 σ
(a) Percentage of cases falling in between 65 and 55
Mean
+1 σ
65
-1 σ
55

7. © V. Singh
65 (1σ)
55 (-1σ)
34.13%
34.13%
So 34.13% + 34.13% , i.e.
68.26% of the cases
falls in between the values
of 55 and 65.

8. © V. Singh
(b) Percentage of cases
falling above 70.
Using Z=
𝑿−𝑴
σ
Z=
𝟕𝟎−𝟔𝟎
5
= 2σ
2 σ
50%
1 σ
34.13%
13.59%
Percentage of cases below 2 σ
is = (50 + 34.13 + 13.59) %
= 97 .72%
Percentage of cases above 2 σ
Or 70 is = (100-97.72) %
= 2.28%

9. © V. Singh
(c) Find Percentage of
cases falling below 45.
Using Z=
𝑿−𝑴
σ
Z=
𝟒𝟓−𝟔𝟎
5
= -3σ
-3 σ
50%
49.86% So percentage of cases below
-3σ or 45 is
= (100-99.86) %
= 0.14%
99.86%

10. Application : 2
Ex- An intelligent test was
administered on a group of 800 cases
of class VI. The mean I.Q. of students
was found to be 100 and SD of the
scores was 15. Find how many
students of class VI having I.Q. below
90 and above 120.
© V. Singh
To determine number of cases lying
above or below a given score of
reference point.

11. © V. Singh
Here, Let the two reference
scores X1= 90, X2=120
First Case when X1= 90
Using, Z=
𝑿−𝑴
σ
=
90−100
15
=
- 0.666 σ ≈ - 0.67 σ
-0.67 σ
Area Below
Score 90 ,
Students below
score 90.

12. © V. Singh
-0.67 σ
Interpreting From NPC Table
50%
24.86% Hence % of cases below
the point - 0.67 σ (score
90).
(50- 24.86 )% = 25.14 %
Hence number of cases
or students below the
score 90 is 25.14% of 800
= 25.14% of 800 = 201.12
≈201 students
201
students
below
score 90

13. © V. Singh
Now for X2=120
when M= 100 & σ = 15
Using, Z=
𝑿−𝑴
σ =
120 −100
15
=
+1.25 σ
+1.25σ
Area Above Score
120 , Students
above score 120.

14. © V. Singh
+1.25σ
Interpreting From NPC Table
50%
39.44%
Hence % of cases above
the point + 1.25 σ (score
120).
(50- 39.44 )% = 10.56 %
Hence number of cases
or students above the
score 120 is 10.56 % of
800
= 10.56 % of 800 = 84.48
≈84 students
84 students
above
score 120

15. Application : 3
Ex- An achievement test in science
was administered to 100 students of
class VII. The value of mean and σ
was found to be 50 and 10 respectively.
Find limits of scores in which middle
75% cases lies.
© V. Singh
To determine limits of scores which
includes a given percentage of cases.

16. © V. Singh
Middle 75%
Means –
37.5% cases
each in both the
left and right of
the Mean.
75%
37.5% 37.5%
Z – Values ?

17. © V. Singh
From NPC Table Z-value at 37.5% is
1.15σ So Z1= +1.15σ , Z2= -1.15σ
M=50 and σ=10.
When Z1= +1.15σ Using the
formula Z=
𝑿−𝑴
σ , X1= Z1 σ + M
= 1.15x10+50= 61.5 ≈ 62
When Z2= -1.15σ Using the formula
Z=
𝑿−𝑴
σ , X2= Z2 σ + M
= -1.15 x 10+50= 38.5 ≈ 39
So middle 75% cases lies between
the score 39 to 62.
75%
Mean
62
39

18. Application : 4
Ex- A logical ability test was administered
on 300 science graduate boys and 200
science graduate girls. The boys mean score
is 26 with SD of 4. The girls mean and SD is
28 and 8 respectively. Find the total number
of boys who exceed the mean of girls and
total no. of girls who got score below the
mean of boys.
© V. Singh
To compare the two distribution in
terms of overlapping.

19. © V. Singh
Mean= 26 Mean= 28
No of
Girls
Below the
Mean of
Boys
No of
Boys
Above the
Mean of
Girls

20. © V. Singh
No. of Boys exceeding the Means of Girls
Boys Mean= 26 , σ=4, Point under
consideration = 28(Girls Mean)
So, Let X = 28
So using Z=
𝑿−𝑴
σ , Z Score =
𝟐𝟖−𝟐𝟔
4 = 0.5 σ
% cases from Mean of Boys and 0.5 σ from
NPC table = 19.15 %
So, % of Boys above girls mean(28 or 0.5 σ) =
(50-19.15)= 30.85%
As total Boys = 300
So No of Boys above the girls Mean =
= 300 x 30.85%
= 92.55 ≈ 93 Boys
19.15%
30.85%

21. © V. Singh
No. of Girls below the Means of Boys
Girls Mean= 28 , σ=8, Point under consideration =
26(Boys Mean)
So, Let X = 26
So using Z=
𝑿−𝑴
σ
, Z Score =
𝟐𝟔−𝟐𝟖
8
= -0.25 σ
% of cases from Mean of Girls and -0.25 σ from
NPC table = 9.87 %
So, % of Girls below boys mean(26 or -0.25 σ) = (50-
9.87)= 40.13%
As total Girls = 200
So No of Girls below the boys Mean =
= 200 x 40.13%
= 80.26 ≈ 80 Girls
9.87 %
40.13%

22. © V. Singh