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Statistics 12 Online
OpenStudy (anonymous):

10% pins are defective, if 10 pins are selected at random calculate the probability that exactly 3 are defective?

OpenStudy (johnweldon1993):

This would be a binomial process We would count a "success" as a pin being defective n = 10 p = 0.1 x = 3 So if we have a calculator *if not let me know* we could always just use the "binompdf" function \[\large binompdf(10,0.1,3)\]

OpenStudy (anonymous):

i have my final ppr on friday , i need help in studying statistics please help me out.

OpenStudy (anonymous):

this qusetion has 3 parts ? 1, exactly 3 are defective 2, more than 3 are defective 3, less than mean number are defective ?

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