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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Wish You the Best in your Studies.



 

Answer by ikleyn(52788)   (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

~~~~~~~~~~~~~~~~~







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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