Question

The Real Estate Group NY reports that the mean monthly rent for a one-bedroom apartment without a doorman in Manhattan is $2636. Assume the standard deviation is $407. A real estate firm samples 98 apartments. What is the probability that the sample mean rent is less than $2,664.78?

Answer 1

Answer:

Probability that the sample mean rent is less than $2,664.78 is 0.75804.

Step-by-step explanation:

We are given that the Real Estate Group NY reports that the mean monthly rent for a one-bedroom apartment without a doorman in Manhattan is $2636. Assume the standard deviation is $407.

Also, a real estate firm samples 98 apartments.

Let $\bar X$ = sample mean rent

The z-score probability distribution for sample mean is given by ;

                Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean monthly rent = $2,636

            $\sigma$ = standard deviation = $407

            n = sample of apartments = 98

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, the probability that the sample mean rent is less than $2,664.78 is given by = P($\bar X$ < $2,664.78)

   P($\bar X$ < $2,664.78) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{2,664.78 - 2,636}{\frac{407}{\sqrt{98} } }$ ) = P(Z < 0.70)

                                                                        = 0.75804

The above probability is calculated using z table by looking at value of x = 0.70 in the z table which have an area of 0.75804.

Therefore, probability that the sample mean rent is less than $2,664.78 is 0.75804.