a truck is loaded with 900 boxes of books. if the total weight of the boxes exce

Question

a truck is loaded with 900 boxes of books. if the total weight of the boxes exceeds 36450 pounds the driver will get fined. assume that the weight (W) of a randomly selected box of books has a mean of mu w = 40 pounds and a standard diviation of sigma w = 6 pounds. (a) find the probability that the driver will be fined (b) twenty percent of the time, 900 of these boxes will weigh a total of at least how many pounds? (c) What is the probability that one randomly selected box of books will weigh between 34 and 46 pounds? (d) What is the probability that the combined weight of 5 randomly selected boxes of books will exceed 210 pounds?

Explanation / Answer

a) overload weight per box =36450/900 =40.5

So,we need to find P(x>40.5)

=40

=6

z=(x-)/

z=(40.5-40)/6

=0.5/6

=0.08

P(x>40.5 ) =P(z>0.08)

               = 1- P(z<0.08)

               = 1- 0.5319

               =0.4681

b)

P=20% =0.2

z score for P=0.2 is -0.84

so x=z* +

    =-0.84*6 +40

    =-5.04+40

    =34.96

so total weight of 900 boxes would be atleast =900*34.96 =31464

c)

n=1

For x=34

z=(x-)/(/n)

=(34-40)/(6/1)

=-6/6

=-1

for x=46

z=(x-)/(/n)

=(46-40)/(6/1)

=6/6

=1

So P(34 <x< 46) =P(-1 <z< 1)

                              =P(z<1) -P(z<-1)

                              =0.8413 - 0.1587

                              =0.6826

d)

mean for 5 boxes, =np =5*40 =200

=6

z=(x-)/

=(210-200)/6

=10/6

=1.67

P(x>210) =P(z>1.67)

               =1-P(z<1.67)

               =1- 0.9525

               =0.0475

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