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.

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.

Step-by-step 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 and standard deviation , the z-score of a measure X is given by:

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 and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation .

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

Sample of 100:

This means that

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

By the Central Limit Theorem

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

has a p-value 0.0548

X = 2450

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.

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

in mathematics, a function is a relationship between two variables such that the value of one of the variables, called the dependent variable, is determined by the value of the other variable, called the independent variable. there is a specific term we use to describe the set of independent variables of a function.

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2