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Question 319:



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Lets see. If I understand the question, you have a machine that has a known probability of failure of 7.3%.  You then want to know what is the probability 5 or fewer failures will be observed from a sample of 100 and you are to use the normal approximation to the binomial. The standard rule of thumb is that you're usually ok using the normal approximation if n*p > 5 and n*q > 5, where q is just 1-p. For this example you want the probability given a proportion of .073 so n*p = 100*.073 = 7.3 so it's on the low side but OK. For q we're better with 100*.927 = 92.7. 

Proceeding, we need to get a z-score out of the observed proportion. For continuous data we usually have the mean and standard deviation. We only have the mean (7.3) and no standard deviation. Fortunately, for discrete distributions, the standard deviation is equal to the square root of n*p*q.  For this data that would be SQRT ( 100*.073*.927 ) = a standard deviation of 2.6.

We now need to make a continuity correction since we have a discrete distribution and want to get as close as we can to a continuous one. We add .5 to our value of 5 that we are testing for to get 5.5.

Now we compute the z-score with the new data (5.5 - 7.3)/2.6 = a z of -.692. Using the z-score to percentile calculator gives us a one-sided probability of 24.4%. 

If we used the binomial directly we'd have the probability of 25.4%, which is pretty close to the normal approximation of 24.4%.

See another example here.

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