Normal Distribution Properties of Normal Distribution Empirical rule of normal distribution Normality limits Standard normal distribution(z-score/ SND) Properties of SND Use of z/normal table Solved examples

1. Normal Distribution
Jagdish D. Powar
Statistician cum Tutor
Community Medicine
SMBT, IMSRC, Nashik
JDP-CM-SMBT 1

2. Competency & Learning objectives
Competency Learning Objectives
CM6.4, CM6.2
Describe and discuss the principles
and demonstrate the methods of
collection, classification, analysis,
interpretation and presentation of
statistical data
The student should be able to
 Define what is normal
distribution and normality limits
 Define and explain Properties of
normal distribution
 Find out probability or
percentage for normal variables
from normal distribution for
given examples
JDP-CM-SMBT 2

3. The Normal Distribution
A symmetrical probability distribution where most results are
located in the middle and few are spread on both sides. The normal
distribution is the most important and most widely used
distribution in statistics.
It is sometimes called the "bell curve," It is also called the
"Gaussian curve" after the mathematician Karl Friedrich Gauss.
50%50%
mean = median = mode
Normal Distribution
JDP-CM-SMBT 3

4. Normal Distribution properties
A continuous random variable X is said to follow Normal
distribution if it satisfies following property
a) It is bell shaped curve.
b) The curve is symmetrical and asymptotic (i.e. it never
touches to x-axis)
c) All measures of central tendency are equal and stable on the
highest peak axis mean = median = mode
d) Total area under the curve is equal to 1.
e) On the basis of mean, µ and standard deviation, σ the area
of normal curve is distributed as:
µ ± 1*σ=68.27% (µ - 1σ, µ -1σ)
µ ± 2*σ=95.45% (µ - 2σ, µ + 2σ)
µ ± 3*σ=99.73% (µ -3*σ, µ +3*σ)JDP-CM-SMBT 4

5. Normal Curve
Population Mean(µ) and SD(σ) Sample mean(x̅) and SD
JDP-CM-SMBT 5

6. Standard Normal Distribution
Standard Normal Variable(z-score)
Any normally distributed data can be
converted to the standardized form
using the formula:
Z =
(𝑥−µ)
σ
If we take
x = µ, then Z = 0
x = µ -1σ, then Z = -1
x = µ + 1σ, then Z = +1
x = µ ± 2*σ, then Z=-2, +2
x = µ ± 3*σ, then Z= -3, +3
JDP-CM-SMBT 6

7. Properties of Standard Normal Distribution
a) It is bell shaped curve.
b) The curve is symmetrical and asymptotic (i.e. it
never touches to x-axis)
c) All measures of central tendency are equal and
stable on the highest peak axis
mean = median = mode = 0
d) Total area under the curve is equal to 1.
e) On the basis of mean, µ= 0 and standard deviation,
σ=1 the area of normal curve is distributed as:
68% of observations lie within -1 to +1
95% of observations lie within -2 to +2
99% of the observation lie within -3 to +3
JDP-CM-SMBT 7

8. JDP-CM-SMBT 8

9. Use of Normal Curve
Normality limits
95 % confidence interval/ 5% level of significance
 x̅ ± 2*SD / x̅ ± 1.96* SD
Z- Score:-
The value of the z-score tells you how many
standard deviations you are away from the mean.
To find the probability or percentage by using
normal curve and normal table
Z =
(𝑥−µ)
𝜎
=
(𝑥−x̅ )
𝑆𝐷
JDP-CM-SMBT 9

10. 1) The average fasting blood sugar level of adult male in some area is 90
mg% with SD of 5 mg%. If fasting blood sugar levels are normally
distributed, then would you regard-Fasting blood sugar of 95 mg% as
abnormal?
a) Fasting blood sugar of 105 mg% as abnormal?
Ans:-
Let us consider variable X- fasting blood sugar level(FBSL).
Given values x̅ = mean FBSL= 90 mg%, SD=5%
Normality limits
x̅ ± 2*SD = (90-2*5, 90+2*5) = (80, 100)
a) Fasting blood sugar of 95 mg% as abnormal?
95 lie within normality limits hence it is not abnormal.
b) Fasting blood sugar of 105 mg% as abnormal?
105 lie outside of normality limits hence it is abnormal
JDP-CM-SMBT 10

11. 3) Mean height of 500 students is 160 cm and SD is 5
cm. If height follows normal distribution then calculate
a) What is the chance of height above 168 cm?
b) What percentage of students will have height above 168
cm?
c) How many of students will have height between 150 cm
to 175cm?
Ans-
Here variable X-height
Mean Height = x̅ = 160 cm
SD = 5 cm
n= number of students =500
JDP-CM-SMBT 11

12. a) What is the chance of
height above 168 cm?
Z =
(𝑥−x̅ )
𝑆𝐷
=
(168−160)
5
=1.60
A(X > 168)
=A(Z > 1.6)
= 0.5- A(Z=0 to Z=1.6)
=0.5-0.4452
=0.0548
b) What percentage of
students will have height
above 168 cm?
=A(X >168) *100
=0.0548*100
=5.48%
JDP-CM-SMBT 12

13. c) How many of students will
have height between 150 cm to
175cm?
Z1 =
(𝑥1−x̅ )
𝑆𝐷
=
(150−160)
5
=-2
Z2 =
(𝑥2−x̅ )
𝑆𝐷
=
(175−160)
5
=3
A(150< X< 175)
=A(-2< Z <3)
=A(Z=-2 to Z=0) + A(Z=0 to Z=3)
=0.4987+0.4772
=0.9759
Number of students will have
height between 150 cm to
175cm
=A(150< X< 175) *500
=0.9759*500
=487.95≈ 488 Students
JDP-CM-SMBT 13

14. 2) The mean weight of 200 students 50 kg and SD is 4 kg. If weight is
normally distributed then find the percentage of students having weight
[ A(Z=2)= 0.4772, A(Z=1.25)=0.3944 ]
a) Above 58 kg c) Above 50 kg
b) Less than 45 kg d) Below 58 kg
Ans:-
Here variable X- weight. Given values x̅ = mean weight = 50 kg, SD=4 kg
a) Percentage of students will have weight above 58 kg
Z =
(𝑥−x̅ )
𝑆𝐷
=
(58−50)
4
=2
A(Z > 2) =0.5-0.4772
=0.0228
%of students weight above 58 kg
=0.0228*100= 2.28%
JDP-CM-SMBT 14

15. b) Percentage of students
will have weight less than 45
kg
Z =
(𝑥−x̅ )
𝑆𝐷
=
(45−50)
4
=1.25
% of students having weight
less than 45 kg
A(Z < 1.25)
=A(Z < 0) +A(Z=0 to Z=1.25)
=0.5+0.3944
=0.8944
Percentage =0.8944*100
=89.44%
JDP-CM-SMBT 15

16. C) Percentage of students
will have weight above 50
kg
Z =
(𝑥−x̅ )
𝑆𝐷
=
(50−50)
4
=0
A( Z > 0) =0.5
Percentage =0.5*100
=50%
JDP-CM-SMBT 16

17. d) Percentage of students
will have weight below 58 kg
Z =
(𝑥−x̅ )
𝑆𝐷
=
(58−50)
4
=2
A(Z < 2)
=A(Z<0 )+A(Z=0 to Z=2)
=0.5+0.4772
=0.9772
%of students weight above
58 kg
=0.9772*100= 97.72%
Another Method:-
%( students below 58 kg)
=100- %( students above 58 kg)
=100- 2.28
=97.72%
JDP-CM-SMBT 17

18. JDP-CM-SMBT 18
Thank You