Group2 It is reported that the mean monthly rent for a onebe

Group2:

It is reported that the mean monthly rent for a one-bedroom apartment without a doorman in Manhattan is $2631 and the standard deviation is $500. A real estate firm randomly samples 100 apartments to study. Assume the distribution of monthly rents for a one-bedroom apartment without a doorman is relatively normal.

1. Is the sampling distribution normal? Why? What is the mean and what is the standard deviation of this sampling distribution.

2. What is the probability that a randomly selected Manhattan apartment has a rent more than $2700?

3. What is the probability that a sample mean rent is more than $2700?

4. What is the probability that a sample mean rent is between $2500 and $2600?

5. Find the 60^thpercentile of the sampling distribution of sample mean.

6. Would it be unusual if the sample mean were greater than $2800? Explain.

7.Would it be unusual for an individual apartment to have a rent more than $2800? Explain.

Solution

Group2:

It is reported that the mean monthly rent for a one-bedroom apartment without a doorman in Manhattan is $2631 and the standard deviation is $500. A real estate firm randomly samples 100 apartments to study. Assume the distribution of monthly rents for a one-bedroom apartment without a doorman is relatively normal.

This sampling distribution is normal because the sample size =100 which is greater tha 30.

mean of this sampling distribution= 2631

the standard deviation of this sampling distribution.=500/sqrt(100) =50

Z value for 2700, z=(2700-2631)/500 =0.14

P( x >2700) = p( z>0.14) = 0.4443

Z value for mean 2700, z=(2700-2631)/50 =1.38

P(mean x >2700) = p( z>1.38) = 0.0838

Z value for mean 2500, z=(2500-2631)/50 =-2.62

Z value for mean 2600, z=(2600-2631)/50 =-0.62

P(2500<mean x <2600) = p( -2.62 < z> -0.62 ) =

P( z< -0.62)