Question 1188777
the expected value is np or 12*0.2=2.4 of them require repairs
the variance is 2.4*0.8= (np(1-p)=1.92
sd is sqrt (V)=1.386
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exactly 4 is 12C4*0.2^4*0.8^8=0.1329
binompdf(12,0.2,4)=0.1329
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fewer than 3 is 0,1,2
for 0 it is 0.8^12=0.0687
for 1 it is 12*0.2*0.8^11=0.2062
for 2 it is 12C2*0.2^2*0.8^10=0.2835
that sum is 0.5584
may confirm by binomcdf(12,0.2,2)ENTER=0.5583 (so slight rounding error)