Question 1144490: Based on annual driving of 15,000 miles and fuel efficiency of 20 mpg, a car in the United States uses, on average, 750 gallons of gasoline per year. If annual automobile fuel usage is normally distributed, and if 22.96% of cars in the United States use less than 520 gallons of gasoline per year, what is the standard deviation?
Round your answer to 2 decimal places, the tolerance is +/-0.05.
please help me solve this quetion
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 15000 miles / 20 miles per gallon = 750 gallons.
this is the mean based on those numbers.
you state the mean as 710 gallons, so i'll use that, even though that doesn't look right base on the above calculations.
22.96% use less than 520 gallons.
if normally distributed, then the z-score that has that percentage to the left of it would be equal to -.7401648101.
the z-score formula is z = (x-m) / s
z is the z-score
m is the mean
x is the raw score
s is the standard deviation, in this case.
formula becomes -.7401648101 = (520-710)/s
solve for s to get:
s = (520-710)/-.7401648101 = 256.6995856
that would be your standard deviation.
here are the numbers given by the normal distribution calculator found at http://davidmlane.com/hyperstat/z_table.html
first picture gets you the z-score.
you then use the formula to get the standard deviation.
second picture shows you the raw score using the given mean and that that standard deviation.
round the standard deviation to 2 decimal places to get 256.70.
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