Help on normal distribution!?

bags of cements are labelled 25kg. The bags are filled by machine and the actual weights are distributed with mean 27.5kg and standard deviation 0.50kg.


a) what is the probability that a bag selected at random will weigh less than 25.0kg?





B) in order to reduce the number of underweight bags (under 25kgs,) to 2.5% of the total, the mean is increased without changing the standard deviation.


show that the increased mean is 26.0kg.





c) it is decided to purchase a more accurate machine to fill the bags. THe requirements for this machine are that only 2.5% of bags under 25kg and that only 2.5% of bags be over 26kg.


calculate the mean and standard deviation that satisfy these requirements.





d) the cost of the new machine is 5000$. cement sells for 0.80$ per kg.


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?

Help on normal distribution!?
a) z=(25-27.5)/.5 =-5.0





Looking this up in a standard normal table (or excel since this is a very small number): 2.87E-07





B) must be a typo in either this or A because 27.5 cannot INCREASE to 26.0





ASSUMING THE TYPE IS IN A) and the problem should have said mean is 25.5:





A) z= (25-25.5)/ .5 = -1


P(%26lt;25) = .1587





B) P(%26lt;x) = .025 so from table, z = -1.96





-1.96 = (25-x)/.5


x=25.98





C) Since the amount above and below are the same percentage, the mean is obviously 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 .40 per bag





.4 x = 5000





so x=5000/.4 = 12500 bags to break even

curse of the golden flower