SOLUTION: when packaging a product, a manufacturer finds that one packet in twenty is unerweight. Determine the probabilities that in a box of 72 packets (a)two (b)less than four will be u

Question 1187690: when packaging a product, a manufacturer finds that one packet in twenty is unerweight. Determine the probabilities that in a box of 72 packets
(a)two
(b)less than four will be underweight
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
p=0.05
for exactly 2 it is 72C2*0.05^2*0.95^70=0.1763
for fewer than 4, calculate 3, 2,1,0
for 3 it is 0.2165
for 1 it is 72C1*0.05*0.95^71=0.0943
for 0 it is 0.95^72=0.0249
Those 4 add to 0.5120
on the calculator, use binomcdf(72,0.05,3) and get 0.5119. The difference is in rounding, since the exact answer to more decimal places is 0.51194967, which barely fails to meet the rounding criteria to 0.5120.
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