The masses of bags of flour filled by a machine are normally distributed with a
ID: 3221859 • Letter: T
Question
The masses of bags of flour filled by a machine are normally distributed with a mean of 500 g and standard deviation of 6 g. If the weight of a bag is less than a, it is considered 'underweight' and has to be refilled. If the weight of the bag is more than b, it is considered 'overweight' and has to be refilled. a) If 3% of the bags are underweight, show that the value of a = 488.72 g. b) If 2% of the bags are overweight, show that the value of b = 512.32 g. c) We need to adjust the mean so that the machine will give us less underweight bags. (Keep the standard deviation at 6.) What should the new mean be so that the underweight bags will only be 2% of the output? d) We keep the mean at 500 g, but adjust the standard deviation. How large should the standard deviation be so that we receive only 2% underweight bags? e) If we want the underweight bags to be 2% and the overweight bags 1%, what should the mean and standard deviations be to achieve this?Explanation / Answer
mean = 500
std. dev. = 6
(a)
z-value for lower 3% i.e. 0.03 area is -1.8808
using central limit theorem,
a = mean + z*sigma = 500 - 1.8808*6 = 488.72
(b)
z-value for upper 3% i.e. 0.03 area is 1.8808
using central limit theorem,
b = mean + z*sigma = 500 + 1.8808*6 = 512.32
(c)
z-value for lower 2% i.e. 0.02 area is -2.0538
488.72 = mean - 2.0538*6
mean = 488.72 + 2.0538*6
mean = 501.04
(d)
z-value for lower 2% i.e. 0.02 area is -2.0538
488.72 = 500 - 2.0538*sigma
sigma = (488.72 - 500)/-2.0538
sigma = 5.4923 (standard deviation)
(e)
z-value for lower 2% i.e. 0.02 area is -2.0538
z-value for upper 1% i.e. 0.01 area is 2.33
488.72 = mean - 2.0538*sigma
512.32 = mean + 2.33*sigma
Solving above two equation gives,
488.72 - 512.32 = -0.2726 sigma
sigma = 5.388
mean = 488.72 + 2.0538*5.388
mean = 499.7859
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.