Answer to Question #289640 in Statistics and Probability for Joys

Question #289640

The probabilities that a printer produces 0,1,2 and 3 misprints are 42% , 28%, 18% and 12% respectively. What is the mean value of the variable?

Expert's answer

"\\mu=0*0.42+1*0.28+2*0.18+3*0.12=1."

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