Question

The mean monthly rent for a one-bedroom apartment without a doorman in Manhattan is $ 2,654 . Assume the standard deviation is $500. A real estate firm samples 100 apartments. What is the probability that the average rent of the sample is more than $ 2,700 ?

Answer 1

Answer: 0.1788

Step-by-step explanation:

We assume that the monthly rent for a one-bedroom apartment without a doorman in Manhattan is normally distribution.

Given : Mean monthly rent : $\mu=\ \ 2,654$

Standard deviation : $\sigma=\ \ 500$

Sample size : $n=100$

Let x be the monthly rent of randomly selected apartment.

Now. we calculate the value of z-score by the formula :

$z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

For x = $2,700

$z=\dfrac{2700-2654}{\dfrac{500}{\sqrt{100}}}=0.92$

The p-value = $P(x>2700)=P(z>0.92)$

$1-P(z<0.92)=1-0.8212136=0.1787864\approx0.1788$

Hence, the  probability that the average rent of the sample is more than $ 2,700 = 0.1788