Question #179347

Chimbazeze milling company produces bags of mealie meal. The weight in Kgs per bag varies, as indicated in the below:

Weight in Kgs: 44 45 . 46 47 . 48 49 . 50

Proportion of bags: 0.04 . 0.13 0.21. 0.29. 0.20 . 0.10. 0.03

i. Compute the mean weight per bag

ii. Compute the standard deviation of the weight per bag

iii. The company estimates the cost (in ngwee) of producing a bag of mealie meal to be 752C=+Χ, where X is the number of Kgs per bag. a). Compute the mean cost of producing a bag of mealie meal.

b). Compute the standard deviation cost of producing a bag of mealie


1
Expert's answer
2021-05-03T05:53:42-0400

i. The mean is

E[X]=x×P(X=x)=(44×0.04)+(45×0.13)+(46×0.21)+(47×0.29)+(48×0.20)+(49×0.10)+(50×0.03)=1.76+5.86+9.66+13.63+9.60+4.90+1.5=46.91E[X] = \sum x \times P(X=x) \\ = (44 \times 0.04) + (45 \times 0.13) + (46 \times 0.21) + (47 \times 0.29) + (48 \times 0.20) + (49 \times 0.10) + (50 \times 0.03) \\ = 1.76 + 5.86 + 9.66 + 13.63 + 9.60 + 4.90 + 1.5 \\ = 46.91

ii.

E[X2]=x2×P(X=x)=442×0.04+452×0.13+462×0.21+472×0.29+482×20+492×0.10+502×0.03=2201.56Var[X]=E[X2](E[X])2=2201.5646.92=1.95E[X^2] = \sum x^2 \times P(X=x) = 44^2 \times 0.04 + 45^2 \times 0.13 + 46^2 \times 0.21 + 47^2 \times 0.29 + 48^2 \times 20 + 49^2 \times 0.10 + 50^2 \times 0.03 \\ = 2201.56 \\ Var[X] = E[X^2] – (E[X])^2 \\ = 2201.56 -46.9^2 \\ = 1.95

Standard deviation is

SD[X]=Var[X]=1.3964SD[X] = \sqrt{Var[X]} = 1.3964

iii. Proper function C = 75 + 2X

a) The mean cost:

E[C]=75+2×46.91=168.82E[C] = 75 + 2 \times 46.91 = 168.82

b) Standard deviation is

SD[C]=Var[C]=Var(75+2X)=22×Var[X]=2×SD[X]=2×1.3964=2.793SD[C] = \sqrt{Var[C]} \\ = \sqrt{Var(75 + 2X)} \\ = \sqrt{2^2 \times Var[X]} \\ = 2 \times SD[X] \\ = 2 \times 1.3964 \\ = 2.793


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