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Demilio
05-15-2012, 08:38 AM
Suppose rents for apartments in NYC are normally distributed with a mean monthly rent of $880 with a standard deviation of $150. If 100 apartments in NYC are selected at random, what is the probability that rent will be greater than $900?

I'm coming up with two answers,
.4469
or
.0912

Which is correct?

galactus
05-15-2012, 06:48 PM
z=\frac{(900-880)\sqrt{100}}{150}=4/3 = 1.33

Looking in the table we see .9082. But this would be 'less than' 880 because the z values approach from the left.

Since it is asking for 'greater than' 900, we subtract from 1 and get

1-.9082=.0918.

In other words, if we sampled 100 apartments in NYC, there would be a little more than a 9% chance that the rent would be more than $900. This seems rather counterintuitive.