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Chapter 7 The Normal Distribution and Its Applications
7.1 Continuous Probability Distribution
From Frequency Distribution to Probability Distribution
height =
Probability Density
Relative Frequency Density
approximation

fi / N
Class Width
 areas = 1
area under curve = 1
pdf of continuous Distribution
2.
f(x)  0 for all value of x 
 conditions for f(x) to be a pd.


f (x)dx  1
3.

P(X = c) = 0
1.

k
4.
P(hXk)= P(h<Xk)= P(hX<k)= P(h<X<k)=
 f (x)dx
h

=
 x f ( x ) dx
and 2 = var(x) =




( x   )2 f ( x)dx =



x 2 f ( x)dx   2
7.2 Skipped (Basic Knowledge of A Continuous Probability Distribution - included in 7.1)
7.3 The Normal Distribution
Denoted by X  N(, 2)
pdf. :
1 x 2
)

 (
1
2
e
f(x) =
 2
-<x<
when  = 0, 2 = 1 then
Z  N(0, 1) it is called standard normal
Probabilities of normal distribution
X  N(, 2)
X can be transformed to standard normal by
Z
X

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P(h < X <k) = P(
h
k
<Z<
)


then table of standard normal can by used.
7.5 Normal Approximation To The Binomial (Optional)
XB  B(n, p)
generally if np > 5 and npq > 5
then
1
1
< XN < x + )
2
2
1
1
P(x1 XB  x2) = P(x - < XN < x + )
2
2
P(XB = x) = P(x -
7.6 Normal Approximation to The Poisson
Generally if  > 5, normal approximation can be used.
e.g. Calls come in to a telephone exchange at random but at a rate of 400 per hour. Find
the probability that in a given 1 hour period
a) there are 350 - 450 calls,
b) there are 360 or more calls
Solution :
The distribution is Poisson type with  = 400. Since  is large, so can
approximate it by a Normal distribution with  = 400 and  = 400 = 20.
a)
P(350  X  450)
= P(349.5 < XN < 450.5)
349.5  400
450.5  400
Z
)
= P(
20
20
= P(-2.525<Z<2.525)
= 9.884
b)
P(360  X)
359.5  400
 Z)
= P(
20
= 0.9786
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