The mean monthly rent for a one-bedroom apartment without a doorman in Manhattan is $2,636. 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 $2,680?
Assuming a normal distribution:
Z = (score-mean)/SEm
SEm = SD/√n = 500/√100 = 500/10 = 50
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
To find the probability that the average rent of the sample is more than $2,680, we can use the Central Limit Theorem to approximate the sampling distribution as a normal distribution.
First, we need to standardize the sample mean using the formula for the standard error of the mean:
Standard Error (SE) = Standard Deviation (𝜎) / Square Root of Sample Size (n)
SE = 500 / √100
SE = 500 / 10
SE = 50
Next, we need to calculate the z-score, which represents the number of standard deviations away from the mean:
z = (Sample Mean - Population Mean) / Standard Error
z = (2,680 - 2,636) / 50
z = 44 / 50
z = 0.88
Now, we can find the probability using a standard normal distribution table or calculator. The z-score of 0.88 corresponds to a probability of approximately 0.8106 or 81.06%.
Therefore, the probability that the average rent of the sample is more than $2,680 is 0.8106 or 81.06%.
To solve this problem, we need to use the concept of sampling distributions and the central limit theorem. The central limit theorem states that if we have a large enough sample size, the sampling distribution of the sample means will be approximately normally distributed, regardless of the shape of the population distribution.
In this case, we are given that the population mean rent is $2,636 and the standard deviation is $500. We are also given a sample of 100 apartments.
To find the probability that the average rent of the sample is more than $2,680, we need to calculate the z-score and then use the standard normal distribution table or calculator.
The formula to calculate the z-score of a sample mean is:
z = (x - μ) / (σ / √n)
Where:
x = sample mean
μ = population mean
σ = population standard deviation
n = sample size
In this case, x = $2,680, μ = $2,636, σ = $500, and n = 100.
First, let's calculate the standard error, which is the standard deviation of the sampling distribution:
SE = σ / √n
SE = $500 / √100 = $500 / 10 = $50
Now we can calculate the z-score:
z = (x - μ) / SE
z = ($2,680 - $2,636) / $50
z = $44 / $50
z = 0.88
Now that we have the z-score, we can use the standard normal distribution table or a calculator to find the probability that a z-score is greater than 0.88.
Looking up the z-score in the standard normal distribution table, we find that the probability of z > 0.88 is approximately 0.1907.
Therefore, the probability that the average rent of the sample is more than $2,680 is approximately 0.1907, or 19.07%.