SOLUTION: Suppose the amount of heating oil used annually by households in Ontario is normally distributed, with a mean of 760 liters per household per year and a standard deviation of 150 l

Question 1168123: Suppose the amount of heating oil used annually by households in Ontario is normally distributed, with a mean of 760 liters per household per year and a standard deviation of 150 liters of heating oil per household per year.
a) What is the probability that a randomly selected Ontario household uses more than 570 liters of heating oil per year?
b) what is the probability that a randomly selected Ontario household uses between 680 and 1130 liters per year?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
mean is 760 liters per household.
standard deviation is 150 liters liters per household.
z-score = (x -m) / s

in this problem:
m = 760
s = 150

when x = 570, the z-score will be (570 - 760) / 150 = -1.267.
this z-score indicates that approximately 10.56% of the households will use less than 570 liters of heating oil annually.

when x1 = 680 and x2 = 1130, z1 will be (680-760)/150 = -.533 and (1130-760)/150 = 2.47.
those z-scores indicate that approximately 69.62% of the households will use between 680 and 1130 liters of heating oil manually.

on the normal distribution graph, these statistics look like the following.

the first graph shows the probability that a randomly selected household will consume less than 570 liters of fuel annually.

the second graph shows the probability that a randomly selected household will consume between 680 and 1130 liters of fuel annually.

the calculator that provided these results can be found at http://davidmlane.com/hyperstat/z_table.html

if you are using this calculator with z-scores, then the mean is 0 and the standard deviation is 1.

results from this calculator using z-scores is shown below.

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