Question 1172792: A machine produce 10% defective bolts. 400 bolts produced by the machine are
randomly selected. What is the probability that:
(a) 65 or more bolts are defective
(b) less than 35 bolts are defective
(c) At least 30 bolts are defective
(d) At most 60 bolts are defective
(e) Between 40 and 60 bolts are defective
Found 2 solutions by ewatrrr, ikleyn:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!
Hi
Binomial Distribution
p(def) = .10 and n = 400
Using Calculator:
(a) 65 or more bolts are defective
P(x ≥ 65) = 1 - P(x ≤ 64) = 1 - binomcdf(400,.1,64) = 1 - .9999 = .0001
(b) less than 35 bolts are defective
P(x< 35) = binomcdf(400,.1,34) = .1804
(c) At least 30 bolts are defective
P(x ≥ 30) = 1 - P(x ≤ 29) = 1 - binomcdf(400,.1,29) = 1 - .0357 = .9643
(d) At most 60 bolts are defective
P(x ≤ 60) = binomcdf(400,.1,60) = .9994
(e) Between 40 and 60 bolts are defective
P(x ≤ 40) = .542
P = .9994 - .542 =
P = .9994 - .542 = .4574
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Answer by ikleyn(52786)  (Show Source):
You can put this solution on YOUR website! .
A machine produce 10% defective bolts. 400 bolts produced by the machine are
randomly selected. What is the probability that:
(a) 65 or more bolts are defective
(b) less than 35 bolts are defective
(c) At least 30 bolts are defective
(d) At most 60 bolts are defective
(e) Between 40 and 60 bolts are defective
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To avoid an error or misunderstanding, everywhere in the post by @ewatrrr,
where you see binomcdf(100, . . . ),
you should read it as binomcdf(400, . . . ).
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