Question

In: Statistics and Probability

A small manufacturing operation employs three people Ann, Bob and Cathy for assembly ofWidgets. The three...

A small manufacturing operation employs three people Ann, Bob and Cathy for assembly ofWidgets. The three work at different speeds, so out of every 1000 Widgets assembled, Ann assembles 355, Bob assembles 314 and Cathy assembles 331. Based on past data, it is known that each of the three assembles the following percentage incorrectly: Ann, 4.7%, Bob 3.1%and Cathy 2.5%.

(a) What is the probability that a Widget is assembled incorrectly? (Use a tree diagram?)

(b) Let A,B and C be the events that a randomly selected Widget is assembled is assembled by Ann, Bob or Cathy, respectively, and let I be the event that a Widget is assembled incorrectly. Label each of the given probabilities in terms of this notation.For example,P(A) =3551000.

(c) Using the notation from (b), give a formula for how you computedP(I) in terms ofthe given probabilities

Solutions

Expert Solution

(a)

Probability that a widget is assembled incorrectly

= 0.355x 0.047 +0.314x0.031+0.331x0.025=0.016685+0.0097340.008275=0.034694

Probability that a widget is assembled incorrectly = 0.034694

(b)

P(A) = Probability that a radnomly selected widget is assembled by Ann = 355/1000=0.355

P(B) = Probability that a radnomly selected widget is assembled by Bob = 314/1000=0.314

P(C) = Probability that a radnomly selected widget is assembled by Cathy = 331/1000=0.331

I be the event that a Widget is assembled incorrectly

P(I|A)

= Probability that a radnomly selected widget is assembled incorectly given that the widget is assembled by Ann

= 4.7/100 =0.047

P(I|B)

= Probability that a radnomly selected widget is assembled incorectly given that the widget is assembled by Bob

= 3.1/100 =0.031

P(I|C)

= Probability that a radnomly selected widget is assembled incorectly given that the widget is assembled by Cathy

= 2.5/100 =0.025

(c)

(c) Using the notation from (b), give a formula for how you computedP(I) in terms ofthe given probabilities

P(I) =  Probability that a radnomly selected widget is assembled incorectly

P(I) = P(A)P(I|A)+P(B)P(I|B)+P(C)P(I|C)

= 0.355x 0.047 +0.314x0.031+0.331x0.025

=0.016685+0.0097340.008275

=0.034694

P(I) = 0.034694


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