Lecture 12
ON
MATH-208
(Probability and Statistics)
BY
Kiran Kumar Shrestha
Department of Mathematics
School of Science
Kathmandu University
TO
CE - II – I Group Students
Topics Covered

Normal Distribution
Date: Thursday, June 3, 2021
Normal Distribution
(Notes Continued...)
#.9
Probability properties
If
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For a normal distribution 68.26% of data lie between the points of inflection, i.e.,
In the same way, 99.73% of data lie between
0.0027 lie outside
.
and only 100%-99.73% = 0.27% =
Example, If X ~ N(24, 32) , then
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#.10
The kurtosis of a normal distribution is 3, i.e.,
or it is mesokurtic.
#.11
The linear combination of normal random variables also follow normal distribution.
For example- If
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#.(i) Let Y1 = 2X, then ( )
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#.12
Let
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#.(ii) Let Y2 = 2X+3, then ( )
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#.(b)
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A normal distribution having mean 0 and variance 1 is called standard normal distribution.
Proofs:
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Thus,
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Standard Normal Distribution
A standard normal distribution is that type of normal distribution which has mean 0 and variance 1.
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A normal distribution
following transformation:
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Standard normal distribution is usually denoted as
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The probability density function of standard normal distribution is
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For a standard normal distribution
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The cumulative distribution function (cdf) of standard normal distribution is given by
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∫ ( )
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The value of above integral for different levels of 'z' are given in a table called normal probability table as
shown below-
For example, from table we see that P(Z< = 1.22) = P(Z < 1.22) = 0.8888
Using Z-table
#.1
Find following probabilities#.(a)
P(Z<=1.25) = 0.8944
#.(b)
P(Z < = 2.61) = 0.9955
#.(c) P(Z >1.12) = 1 – P(Z < 1.12)
= 1 – 0.8686 = 0.1314
#.(d)
P(Z < - 2.11) = P(Z > 2.11) (due to symmetry) = 1- P(Z < 2.11) = 1 – 0.9826=0.0174
Note: (
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#.(d) P(Z > -1.72)
P(Z>-1.72) = 1 – P(Z< -1.72) = 1 – (1 – P(Z < 1.72)) = P(Z < 1.72) = 0.9573
#.(e) P(0.75 < Z< 2.41)
P(0.75 < Z < 2.41) = P(Z < 2.41) – P(Z < 0.75) = 0.9920 – 0.7734 = 0.2186
#.(f) P(-1.96 < Z < 1.96)
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#.2
Find 'c' if
#.(a) P(Z < c) = 0.9793
From Z-table,
P(Z < c)= P(Z < 2.04)
So, c = 2.04
#.(b) P(Z < c) = 0.3594
Or, P(Z < c) = 1 – 0.6406
Or, 1 – P(Z < c) = 0.6406
Or, P(Z < - c) = 0.6406
Or, P(Z < -c) = P(Z < 0.36)
So, - c = 0.36 Or, c = - 0.36
#.(c) P(Z > c) = 0.1093
Or, 1 – P(Z < c) = 0.1093
Or, P(Z < c) = 1 – 0.1093
Or, P(Z < c) = 0.8907
Or, P(Z < c) = P(Z < 1.23)
So, c = 1.23
#.(d) P(Z > c) = 0.9920
Or, 1 – P(Z < c) = 0.9920
Or, P(Z < -c) = 0.9920
Or, P(Z < -c) = P(Z < 2.41)
c = -2.41
#.(e) P(-c < Z < c) = 0.95
Or, P(Z < c) – P(Z < -c) = 0.95
Or, P(Z < c) – {1 – P(Z < c)}=0.95
Or, 2. P(Z < c) -1 = 0.95
Or, 2.P(Z < c) = 1.95
Or, P(Z < c) = 1.95/2 = 0.975 = P(Z < 1.96)
So, c = 1.96
#.3
Given X ~ N(12 , 9), find P(X > 16).
SolutionGiven X ~ N(12, 9).
So
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