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Normal Approximation to the Binomial Distribution

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.

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Normal Approximation to the Binomial Distribution Examples

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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.

Solved Examples

Question 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?
Solution:
 

Here, X is N( 0.7, 0.0002)

a) P(x > 0.71) = p(z > ($\frac{(0.71-0.7)}{0.01}$))

= p(z > 0.5) = 1 – p(0 < z < 1)

= 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 > $\frac{0.025}{0.02}$)

= 2P(z > 1.25)

= 2[ -0.3944 + 0.5 ]

= 0.2112


 

Question 2: The FORD 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 FORD 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


 

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