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in one region, the September energy consumption levels for single-family homes are found to be normally distributed with a mean of 1050kWh and a standard deviation of 218 kWh. Find P45, which is the consumption level separating the bottom 45% from the top 55%

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Answer:


a=1050 -0.126*218=1022.53

So the value of height that separates the bottom 45% of data from the top 55% is 1022.53.

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the energy consumption of a population, and for this case we know the distribution for X is given by:


X \sim N(1050,218)

Where
\mu=1050 and
\sigma=218

For this part we want to find a value a, such that we satisfy this condition:


P(X>a)=0.55 (a)


P(X<a)=0.45 (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.45 of the area on the left and 0.55 of the area on the right it's z=-0.126. On this case P(Z<-0.126)=0.45 and P(z>-0.126)=0.55

If we use condition (b) from previous we have this:


P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.45


P(z<(a-\mu)/(\sigma))=0.45

But we know which value of z satisfy the previous equation so then we can do this:


z=-0.126<(a-1050)/(218)

And if we solve for a we got


a=1050 -0.126*218=1022.53

So the value of height that separates the bottom 45% of data from the top 55% is 1022.53.

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