A manufacturer has developed a new method for manufacturing heat tiles. The desi

Question

A manufacturer has developed a new method for manufacturing heat tiles. The design is that there would be no more than 3% of the tiles to be defective. A random sample of 1000 tiles is taken and inspected for defects. Let X = the number of defective heat tiles in the sample. a. If 3% are defective overall, what is P(X > 55)? Based on your answer, is 55 defective tiles an unusually large number? b. If 3% are defective overall, what is P(X > 40)? Based on your answer, is 50 defective tiles an unusually large number?

Explanation / Answer

Normal Approximation to Binomial Distribution
Mean ( np ) =1000 * 0.03 = 30
Standard Deviation ( npq )= 1000*0.03*0.97 = 5.3944
Normal Distribution = Z= X- u / sd                   
a.
P(X > 55) = (55-30)/5.3944
= 25/5.3944 = 4.6344
= P ( Z >4.634) From Standard Normal Table
= 0, it's an unusuall large number          
b.
P(X > 40) = (40-30)/5.3944
= 10/5.3944 = 1.8538
= P ( Z >1.854) From Standard Normal Table
= 0.0319                  
it's unusually small number

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