Question

The mean monthly rent for a one-bedroom apartment without a doorman in Manhattan is 2630. Assume the standard deviation is$500 . A real estate firm samples 100 apartments. Use the TI-84 Plus calculator.a) What is the probability that the sample mean rent is greater than $27007?b) What is the probability that the sample mean rent is between $2450 and $2550? c) Find the 25th percentile of the sample mean. d) Would it be unusual if the sample mean were greater than $26457?e) Do you think it would be unusual for an individual to have a rent greater than $2645? Explain. Assume the variable is normally distributed.

Answer 1

a) 0.0808 = 8.08% probability that the sample mean rent is greater than $2700.

b) 0.0546 = 5.46% probability that the sample mean rent is between $2450 and $2550.

c) The 25th percentile of the sample mean is of $2596.

d) |Z| = 0.3 < 2, which means it would not be unusual if the sample mean was greater than $2645.

e) |Z| = 0.3 < 2, which means it would not be unusual if the sample mean was greater than $2645.

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

If |Z|>2, the measure X is considered unusual.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The mean monthly rent for a one-bedroom apartment without a doorman in Manhattan is $2630. Assume the standard deviation is $500.

This means that
`\mu = 2630, \sigma = 500`

Sample of 100:

This means that
`n = 100, s = (500)/(√(100)) = 50`

a) What is the probability that the sample mean rent is greater than $2700?

This is the 1 subtracted by the p-value of Z when X = 2700. So

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (2700 - 2630)/(50)`

`Z = 1.4`

`Z = 1.4` has a p-value 0.9192

1 - 0.9192 = 0.0808

0.0808 = 8.08% probability that the sample mean rent is greater than $2700.

b) What is the probability that the sample mean rent is between $2450 and $2550?

This is the p-value of Z when X = 2550 subtracted by the p-value of Z when X = 2450.

X = 2550

`Z = (X - \mu)/(s)`

`Z = (2550 - 2630)/(50)`

`Z = -1.6`

`Z = -1.6` has a p-value 0.0548

X = 2450

`Z = (X - \mu)/(s)`

`Z = (2450 - 2630)/(50)`

`Z = -3.6`

`Z = -3.6` has a p-value 0.0002

0.0548 - 0.0002 = 0.0546.

0.0546 = 5.46% probability that the sample mean rent is between $2450 and $2550.

c) Find the 25th percentile of the sample mean.

This is X when Z has a p-value of 0.25, so X when Z = -0.675.

`Z = (X - \mu)/(s)`

`-0.675 = (X - 2630)/(50)`

`X - 2630 = -0.675*50`

`X = 2596`

The 25th percentile of the sample mean is of $2596.

Question d and e)

We have to find the z-score when X = 2645.

`Z = (X - \mu)/(s)`

`Z = (2645 - 2630)/(50)`

`Z = 0.3`

|Z| = 0.3 < 2, which means it would not be unusual if the sample mean was greater than $2645.