We are given a normal distribution with mean `mu=2268"kWh"` and standard deviation `sigma=592.2`. We are asked to find the consumption rate in kWh that represents the 6th percentile, the 22nd percentile, and the third quartile (which is the 75th percentile).

Since the distribution is normal, we can convert a value for the consumption to a standard normal z-score using `z=(x-mu)/sigma` . Here, we do not know x, the amount of consumption, but we can find z. Using a little algebra, we can rewrite this equation as `x=z sigma + mu` .

To find z, we consult a standard normal chart (or use technology).

(a) For the 6th percentile, this indicates that 6% of the population has this level of consumption or lower. We look for .0600 in a z-table. Using a calculator, I got `z~~-1.555` (the table will give 2 decimal places typically).

So x=(-1.555)(592.2)+2268, which is approximately 1347kWh. (This means that 6% of the population consumes a little more than one and a half standard deviations less kWh than the mean.)

(b) Similarly, for the 22nd percentile, we get `z~~-0.772`, so

`x=-0.772(592.2)+2268 ~~1811` kWh.

(c) The third quadrant is the 75th percentile (we expect z>0). From a table (or a calculator, in this case), `z~~0.674` and `x=.674(592.2)+2268~~2667`.

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