Random probability question

Demilio

New member
Joined
May 15, 2012
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1
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

Super Moderator
Staff member
Joined
Sep 28, 2005
Messages
7,216
\(\displaystyle z=\frac{(900-880)\sqrt{100}}{150}=4/3 = 1.33\)

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.
 
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