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Using the Tables for the standard normal distribution
Tables have been posted for the standard normal distribution.
Namely
2
21
2
uz
ZF z e du
The values of z ranging from -3.5 to 3.5
F z
z
If X has a normal distribution with mean and standard deviation then
has a standard normal distribution. Hence
XZ
a X bP a X b P
a bP Z
Z Z
b aF F
Example: Suppose X has a normal distribution with mean =160 and standard deviation =15 then find:
140 165P X
140 160 160 165 160140 165
15 15 15
XP X P
20 51.33 0.33
15 15P Z P Z
0.33 1.33Z ZF F
.6293 .0918 .5375
This also can be explained by making a change of variable
216012 15
165
140
1140 165
2 15
x
P X e dx
160
15
xz
20140, 1.33
15x u
Make the substitution1
15dz dx
5165, 0.33
15x u when and
Thus 2 21601 1
2 15 2
165 0.33
140 1.33
1 1
2 15 2
x ze dx e dz
0.33 1.33Z ZF F
.6293 .0918 .5375
The Normal Approximation to the Binomial
Let1
1= where n nn n
x x Sx S x x
n n
The Central Limit theorem
nS nx nx nz
n nn
Then the distribution of
approaches the standard normal distribution as
If x1, x2, …, xn is a sample from a distribution with mean , and standard deviations
n
the Normal distribution with
x
or the distribution of
approaches the normal distribution with
Hence the distribution ofapproaches
1n nS x x
and x x n
and n nS Sn n
Thus The Central Limit theorem states
1 nX X X
Suppose that X has a binomial distribution with parameters n and p.
Then
That sums and averages of independent R.Vs tend to have approximately a normal distribution for large n.
2: and vari iE X p X pq Note
1, , nX Xwhere are independent Bernoulli R.V.’s
Thus for large n the Central limit Theorem states that
1 nX X X
has approximately a normal distribution with
Thus for large n
and X Xn np n n pq npq
P a X b P a Y b
where X has a binomial (n,p) distribution and Y has a normal distribution with
and Y Ynp npq
The binomial distribution
0
0.05
0.1
0.15
0.2
0.25
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
The normal distribution = np, 2 = npq
0
0 5 10 15 20
-
0.0500
0.1000
0.1500
0.2000
0.2500
0 2 4 6 8 10 12 14 16 18 20
-
0.0500
0.1000
0.1500
0.2000
0.2500
0 2 4 6 8 10 12 14 16 18 20
-
-0.5
Binomial distribution
Approximating
Normal distribution
Binomial distribution n = 20, p = 0.70
049.2
14
npq
np
Normal Approximation to the Binomial distribution
• X has a Binomial distribution with parameters n and p
2121 aYaPaXP
• Y has a Normal distribution
npq
np
correction continuity21
-
0.0500
0.1000
0.1500
0.2000
0.2500
0 2 4 6 8 10 12 14 16 18 20
-
0.0500
0.1000
0.1500
0.2000
0.2500
0 2 4 6 8 10 12 14 16 18 20
Binomial distribution
-
0.0500
0.1000
0.1500
0.2000
0.2500
a
-
-0.5
Approximating
Normal distribution
P[X = a]
21a 2
1a
-
0.0500
0.1000
0.1500
0.2000
0.2500
a-
-0.5
2121 aYaP
-
0.0500
0.1000
0.1500
0.2000
0.2500
a
-
-0.5
P[X = a]
Example
• X has a Binomial distribution with parameters n = 20 and p = 0.70
13 want We XP
13 eexact valu The XP
1643.030.070.013
20 713
Using the Normal approximation to the Binomial distribution
Where Y has a Normal distribution with:
049.230.70.20
14)70.0(20
npq
np
2121 131213 YPXP
Hence 5.135.12 YP
049.2
145.13
049.2
14
049.2
145.12 YP
= 0.4052 - 0.2327 = 0.1725
24.073.0 ZP
Compare with 0.1643
Normal Approximation to the Binomial distribution
• X has a Binomial distribution with parameters n and p
2121 bYaP
• Y has a Normal distribution
npq
np
correction continuity21
)()1()( bpapapbXaP
-
0.0500
0.1000
0.1500
0.2000
0.2500
a b
-
-0.5
21a 2
1b
bXaP
-
0.0500
0.1000
0.1500
0.2000
0.2500
a b
-
-0.5
21a 2
1b
2121 bYaP
Example
• X has a Binomial distribution with parameters n = 20 and p = 0.70
1411 want We XP 1411 eexact valu The XP
614911 30.070.014
2030.070.0
11
20
)14()13()12()11( pppp
5357.01916.01643.01144.00654.0
Using the Normal approximation to the Binomial distribution
Where Y has a Normal distribution with:
049.230.70.20
14)70.0(20
npq
np
2121 14101411 YPXP
Hence
5.145.10 YP
049.2
145.14
049.2
14
049.2
145.10 YP
= 0.5948 - 0.0436 = 0.5512
24.071.1 ZP
Compare with 0.5357
Comment:
• The accuracy of the normal appoximation to the binomial increases with increasing values of n
Example• The success rate for an Eye operation is 85%
• The operation is performed n = 2000 times
Find1. The number of successful operations is
between 1650 and 1750.2. The number of successful operations is at
most 1800.
Solution
• X has a Binomial distribution with parameters n = 2000 and p = 0.85
17201680 want We XP
5.17205.1679 YP
where Y has a Normal distribution with:
969.1515.85.200
1700)85.0(2000
npq
np
17201680 Hence XP
969.15
17005.1720
969.15
1700
969.15
17005.1679 YP
= 0.9004 - 0.0436 = 0.8008
28.128.1 ZP
5.17205.1679 YP
Solution – part 2.
1800 want We XP
5.1800 YP
969.15
17005.1800
969.15
1700YP
= 1.000
29.6 ZP