Question

Five percent (0.05) of the parts produced by a machine are defective. Twenty(20) parts are selected at random for study.

A. what is the probability that more than 4 parts are defective?

B. what is the probability that exactly 3 parts are defective?

C. what is the expected number of defective parts in the sample?

D. What is the variance and standard deviation of defective parts in the sample?

Answer 1

Solution Giren p=oos n=20 XBCnP) P(X=x)=Q) p (p) nok ; x=0, 1,2 p(X24) = 1-pcxct) =)-4p XEO) +PCX=1) + P(x=2) + PCX= 3) + p(x=4]} =)-{ 0.3585+ 0.3779-+0:1887+ 0.0596 + 0.033 - ) - 0.9974 = 0.0026 P(x>4) = 0.0026 / p(x=3)= ( 33) 0.05)3 (1-0.05)!7 = 0.0596 p(x-3)= 0.0596

E(X)= 4= np = 20x0.05 = ) | E(X)=1 I is the expected number of defective parts in the sample. varix) = n p q 0 - 1 9= )-P = 20X0.05x0.95 toe = 0.95 Varix) = 0.95 Standard deviation = Vnpg = 0.95 16 = 0.9747 = 0.9747