(First and second solumns represent single-phase motors, thirds column represent

ID: 2948947 • Letter: #

Question

(First and second solumns represent single-phase motors, thirds column represents three-phase motors)

Assume that 10% of the motors labeled 120 V single-phase are mismarked, and that 5% of the motors marked 240 V single-phase are mismarked.

a) If a motor is selected at random, what is the probability that it is mismarked?

b) If a motor is picked at random from those marked 240 V single-phase, what is the probability that it is mismarked?

c) What is the probability that a motor selected at random is 0.5 hp and mismarked?

Horsepower 0.1 0.5 1.0 120 V ac 900 200 100 240 V ac 400 500 200 240 V 30 100 600

Given data

Now

total number of motor (n)=3000

a)No of motor that is 120v single phase=1200

no of motor that is 240v single phase =1100

total number of mismarked motor are =

10% of 120 v single phase +5% of 240 V single phase =1200*0.1+1100*0.05=120+55=175

Probablity of mismarkrd motor when randomly selected>

b)number of motor that is of 240V single phase =1100

nummber of motor mismarked within this group=55 (since we have to select within the grroup)

so the probablity of getting a mismarked motor=55/1100=0.05

c) again we have total no of sample (n)=3000

Total Number of motor of 0.5 hp(A)=800

Total number of motor missmarked(B)=175

P(A)=800/3000=0.267

P(B)=175/3000=0.0583

now probablity of getting motor that is 0.5 hp and mismarked when selected randomly i.e. P(A and B)=P(A)*P(B)=0.267*0.0583=0.0156

horsepower 120 V ac 240V ac 240V 30 0.1 900 400 0 0.5 200 500 100 1.0 100 200 600

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