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Help with Standard deviation in statistics?
A gas station earns $2.60 in revenue for each gallon of regular gas it sells, $2.75 for each gallon on midgrade gas, and $2.90 for each gallon of premium gas. Let X1, X2, and X3 denote the numbers of gallons of regular, midgrade, and premium gasoline sold in a day. Assume that X1, X2, and X3 have means μ1 = 1500, μ2 = 500, and μ3 = 300, and standard deviations σ1 = 180, σ2 =90, and σ3 = 40.
a.) Assuming X1, X2, and X3 to be independent, find the standard deviation of the daily revenue.
2 Answers
- Modus OperandiLv 69 years agoFavourite answer
standard deviation = square root of Variance, therefore
Variance = standard deviation^2, Variance of X now being written as VAR(X)
VAR(aX+BY) = a^2 VAR(X) + b^2 VAR(Y)
Revenue = 2.6X1 + 2.75X2 + 2.9X3
VAR(Revenue) = VAR(2.6X1 + 2.75X2 + 2.9X3)
= 2.6^2 VAR(X1) + 2.75^2 VAR(X2) + 2.9 VAR(X3)
= 6.76*32400 + 7.5625*8100 + 8.41*1600
= 219024 + 61256.25 + 13456
= 293736.25
Standard deviation = square root of variance, therefore
= square root (293736.25) = 541.974
- charlatanLv 79 years ago
X1+X2+X3=1500+500+300=2300
sale/day=1500*2.6+500*2.75+300*2.9=6145
σ = sqrt[(deviation)^2/n]
180^2*1500=48600000
90^2*500=4050000
40^2*300=480000
total=2300=53130000
sd consolidated. σc = sqrt[(deviation^2)/total sale]=sqrt(53130000/2300)=151.98...
am not sure of the answer or the method,tried to use what i remember of what i had
learnt in school.
hoping others may correct the answer,if i am wrong.
i wouldn't be offended.