Over the past 200 working days, the number of defective parts produced by a machine is
given in the following table:
Number
of
defective
parts
0 1 2 3 4 5 6 7 8 9 10 11 12 13
Days 50 32 22 18 12 12 10 10 10 8 6 6 2 2
Determine the probability that tomorrow’s output will have
(a) No defective part
(b) At least one defective part
(c) Not more than 5 defective parts
(d)More than 13 defective parts
(i) probability that no defective part = 50/200
= 0.25
(ii)
probability that atleast one defective part
= 1- probability that no defective part
= 1- 0.25
= 0.75
(iii)
P(not more than 5 defective part) = P(no defective part)+P(1 defective part)+P(2 defective part)+P(3 defective part)
+ P(4 defective part)+ P(5 defective part)
(iv)
P(more than 13 defective parts)
= 2/100
= 0.02
hope this helps you
= 0.25
(ii)
probability that atleast one defective part
= 1- probability that no defective part
= 1- 0.25
= 0.75
(iii)
P(not more than 5 defective part) = P(no defective part)+P(1 defective part)+P(2 defective part)+P(3 defective part)
+ P(4 defective part)+ P(5 defective part)
(iv)
P(more than 13 defective parts)
= 2/100
= 0.02
hope this helps you