Page 308 - DCOM203_DMGT204_QUANTITATIVE_TECHNIQUES_I

Unit 15: Normal Probability Distribution

Thus, we conclude that if X is a normal variate with mean and standard deviation , then Notes
X
z is a normal variate with mean zero and standard deviation unity. Since the parameters
of the distribution of z are fixed, it is a known distribution and is termed as standard normal
distribution (s.n.d.). Further, z is termed as a standard normal variate (s.n.v.).
It is obvious from the above that the distribution of any normal variate X can always be
transformed into the distribution of standard normal variate z. This fact can be utilised to
evaluate the integral given above.

X 1 X X 2
We can write P X 1 X X 2 P

X 1 X 2
z
z
P z 1 z z 2 , where 1 and 2 .
In terms of figure, this probability is equal to the area under the standard normal curve between
the ordinates at z = z and z = z . Since the distribution of z is fixed, the probabilities of z lying
1 2
in various intervals are tabulated. These tables can be used to write down the desired probability.

Example: Using the table of areas under the standard normal curve, find the following
probabilities :
1. P(0 z 1.3) 2. P( 1 z 0) 3. P( 1 z 2)
4. P( z 1.54) 5. P(|z| 2) 6. P(|z| 2)

Solution:
The required probability, in each question, is indicated by the shaded are of the corresponding
figure.

1. From the table, we can write P(0 z 1.3) = 0.4032.

Areas under the curve

(1) (2)

2. We can write P( 1 z 0) = P( z 1), because the distribution is symmetrical. From the
table, we can write P( 1 z 0) = P( z 1) = 0.3413.
3. We can write
P( 1 z 2) = P( 1 z 0) + P(0 z 2)
= P( z 1) + P(0 z 2) = 0.3413 + 0.4772 = 0.8185.

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