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