# Using Normal approximation to binomial distribution

Smith is a weld inspector at a ship yard. He knows from keeping track of good and substandard welds that 5% will be substandard. If he checks 300 of 7500 welds, what is the probability that he will find less than 20 substandard welds?

To solve the problems on the up-coming test we will use either the "normalcdf" or InvNorm" functions on the TI-84 Plus calculator.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

To solve this question, we use normal approximation to binomial distribution because the sample size which is 300 is too large if we use formula of binomial distribution to calculate the probability.

Now we know that n=300 and p=0.05. So mean=np=300*0.05=15 and standar deviation=sqrt[n*p*(1-p)]=sqrt[300*0.05*0.95]=3.775.

To use normal approximation to binomial distribution, z value=(x-mean)/standard deviation.

So P(less than 20 welds)=P(X<20)=(Z<(20-15)/3.775)=P(Z<1.32)=0.9066

I got 0.9066 using normalcdf function in excel.

I think that this is the standard procedure (or template) to solve this type of questions.

Hence, the probability that he will find less than 20 substandard welds is 0.9066.

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