Question 333849
Let X =number of defectives in the sample
Let n= sample size
let {{{pi}}}=population defective rate
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Since there are two outcomes to each event (each piston is either defective or not) and the we assume that the outcomes from piston to piston are independent and the probability of being defective is the same for each piston, this would define a binomial experiment. in other words x~binomial (n,{{{pi}}})
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If n =90 and {{{pi=0.03}}}  (you give this information)
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{{{mu}}}=the expected value of x = n*{{{pi}}} = 90*0.03=2.70 defective pistons
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variance of x={{{sigma^2}}=n*{{{pi*(1-pi}}}=90*0.03*0.97=2.619
standard deviation of x ={{{sigma}}}= {{{sqrt(2.619)=1.618}}}