HW#2DueSep100MonLCE308-Fall 2018 [Compet bilty Model Mamoft word me Insert Page

Question

HW#2DueSep100MonLCE308-Fall 2018 [Compet bilty Model Mamoft word me Insert Page Layout References Mailings Review View w InsertFormat Tools Table Reference M ailingsWindow Help" Toolbars 2. An extrusion die is used to produce aluminum rods. Specifications are given for the length and the diameter of the rods. For eachrod, the length is classified as too short, too long, or OK, and the diameter is classified as too thin, too thick, or OK. In a population of 1,000 rods, the number of rods in each class is as follows. Given that a rod is randomly sampled from this population answer the following questions: Diameter Too thin 10 38 OK Too thick Length Too short Too long 900 25 13 a) What is the probability that this rod is too long? b) What is the probability that this rod is NOT roo long? c) What is the probability that this rod is too thick? d) What is the probability that this rod is NOT too thick?

Explanation / Answer

Total Pipes = 1000

a.) No of too long rods = 2 + 25 + 13 = 40

P(too long rods) = 40/1000 = 0.04

b.) P(not too long rods) = 1 - P(too long rods) = 1 - 0.04 = 0.96

c.) No of too thick rods = 5 + 4 + 13 = 22

P(too thick rods) = 22/1000 = 0.022

d.) P(not too long rods) = 1 - P(too long rods) = 1 - 0.022= 0.978

e.) No of rods which are both too thick and too long = 13

so, P(rods which are both too thick and too long) = 13/1000= 0.013

f.) No of rods which are either too thick or too long = No of rods which are thick + No of rods which are long - No of rods which are both thick and long

No of rods which are either too thick or too long = (2 + 25 + 13) + (5 + 4 +13) - 13

= 49

P(No of rods which are either too thick or too long) = 49/1000 = 0.049

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