Normal Approximation Binomial Distribution

In this page we are going to discuss about  normal approximation to the binomial distribution concept .Normal distribution is fine approximation to the binomial distribution, in a binomial distribution one can easily confirm that the mean for a single binomial trial, where Success is scored as 1 and Failure is scored as 0, is p; where p is the probability of S. Where s is a sample space hence the mean for the binomial distribution with n trials is np. Condition to the normal approximation is good for the binomial distribution. Condition of failure for normal approximation is p (1-p), standard deviation for normal approximation (np(1-p))^5 .

Understanding R Binomial Distribution is always challenging for me but thanks to all math help websites to help me out.

Normal approximation binomial distribution examples

If the number of trials in a sample space, the n is large, the binomial distribution is just about equal to the normal distribution. This is fine, since we actually do not feel like to plainly calculate binomial probabilities when n > 100.

Example 1:

The diameter of an telephone cable is in general distributed with mean 0.7 cm and variance 0.0002 cm2 .what is the probability that the diameter will exceed 0.71 cm, the cable is measured imperfect if the diameter differs from the mean by more than 0.015 cm. what is the probability of obtaining a defective cable?

X is N( 0.7, 0.0002)

a)      P(x> 0.71) = p(z> ((0.71-0.7)/0.01))

= p(z>0.5) = 1 – p(0
= 1 – 0.381 = 0.62

The result is 0.62

b)      P[(x> 0.825) U (x< 0.785)] = 2P(x> 0.825)

= 2P(z> 0.025/0.02)

= 2P(z> 1.25)

= 2[-0.3944 + 0.5 ]

= 0.2112

Example 2:

A cricket match has 45 multiple teams, in each  team with choices of a to e. One player did not play and must guess on each team, Using normal approximation, estimate the probability that the player:
a) failure in the team or gets less than 50%
b) gets a score of at least 10%
c) gets a score between 20% and 40% inclusively

a) a 50% would be getting 22.5 success in the team. Getting less than that is player getting 22 in the team.

P(X ≤ 22 ) =

22
∑ P(X = x) = 0.9999967
x = 0

≈ Probability ( Xn < 22.5 )
= P( y < ( 22.5 - 9 ) / 2.683282 )
= P( y < 5.031153 )
= 0.9999998

b) 10% of 45 is 4.5 so we are looking to get 5 or more teams.

P( X ≥ 5 ) =

45
∑ P(X = x) = 0.961764
x = 5
≈ P( Xn ≥ 4.5 )
= P( y ≥ ( 4.5 - 9 ) / 2.683282 )
= P( y ≥ -1.677051 )
= 0.9532337

c) 20% is 9 teams, 40% is 18 teams.

P( 9 ≤ X ≤ 18 ) =

18
∑ P(X = x) = 0.5587208
x = 9
≈ P( 8.5 < Xn < 18.5 )
= P( ( 8.5 - 9 ) / 2.683282 < y < ( 18.5 - 9 ) / 2.683282 )
= P( -0.186339 < y < 3.540441 )
= P( y < 3.540441 ) - P( y < -0.186339 )
= 0.9998003 - 0.4260895
= 0.5737108

Example 3 :

The FOARD Company manufactures cars. They claim that only for .08 of MNW cars are defective. What is the probability of finding 3 defective cars in a random sample of 45 FOARD cars?

Solution : Formula for cumulative binomial distribution

P(X = r) = nCr *p r *(1-p) n-r

Given r=3, n=45, p=0.08

Now we can substitute this values for this formula

= 45C3 *(.08)3* (.92)45-3

= 45C3 *(.08)3* (.92)42

= 0.219
Simulation with a binomial experiment

The authority of the Normal approximation to the binomial distribution. Simulation with a binomial experiment is one way

to produce a normal distribution. Either entire the calculations with calculate data, or convert everything including sample

space and the standard deviation to proportions.