2. A truck is loaded with 900 boxes of books. If the total weight of the boxes e

Question

2. A truck is loaded with 900 boxes of books. If the total weight of the boxes exceeds 36,450 pounds, the driver will be fined at the weighing station. Assume that the weight (W) of a (i.e. one) randomly selected box of books from the entire population of boxes of books has a mean of ,-40 pounds and a standard deviation of@w6 pounds. (a) Find the probability that the driver will be fined. (b) Twenty percent (20%) of the time, 900 of these boxes of books will weigh a total of at least how many pounds? What is the probability that one randomly selected box of books will weigh between 34 and 46 pounds? What is the probability that the combined weight of 5 randomly selected boxes of books will exceed 210 lbs.? (c) (d)

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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