11. Bags of cement are labelled 25 kg. The bags are filled by machine and the ac

Question

11. Bags of cement are labelled 25 kg. The bags are filled by machine and the actual weights are normally distributed with mean 25.7 kg and standard deviation 0.50 kg (a) What is the probability a bag selected at random will weigh less than 25.0 kg? In order to reduce the number of underweight bags (bags weighing less than 25 kg) to 2.5% of the total, the mean is increased without changing the standard deviation. (b) Show that the increased mean is 26.0 kg. It is decided to purchase a more accurate machine for filling the bags. The requirements for this machine are that only 2.5% of bags be under 25 kg and that only 2.5% of bags be over 26 kg. (c) Calculate the mean and standard deviation that satisfy these requirements. The cost of the new machine is $5000. Cement sells for $0.80 per kg. (d) Compared to the cost of operating with a 26 kg mean, how many bags must be filled in order to recover the cost of the new equipment? Total 11 marks)

Explanation / Answer

a) given mean = 25.7 std = 0.50

z=(x-mean)/std

z=(25-25.7)/.5 =-1.4

Looking this up in a standard normal table 0.0808

B) 25.7 can INCREASE to 26.0

z= (25-26)/ .5 = -1
P(<25) = .1587

B) P(<x) = .025 so from table, z = -1.96

-1.96 = (25-x)/.5
x=25.98

C) , the mean is the midpoint of the range:

Mean = 25.5 kg

for the sd: z= .5/sd
the table value for this z should be .975 so z is 1.96

1.96= .5/sd

sd = .256

D) On avergage, each bag is .5 kg heavier which costs .80 per bag

.8 x = 5000

so x=5000/.8 = 6250 bags to break even

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