Respuesta :
Answer:
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
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 [tex]\mu = 2630, \sigma = 500[/tex]
Sample of 100:
This means that [tex]n = 100, s = \frac{500}{\sqrt{100}} = 50[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{2700 - 2630}{50}[/tex]
[tex]Z = 1.4[/tex]
[tex]Z = 1.4[/tex] 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
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{2550 - 2630}{50}[/tex]
[tex]Z = -1.6[/tex]
[tex]Z = -1.6[/tex] has a p-value 0.0548
X = 2450
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{2450 - 2630}{50}[/tex]
[tex]Z = -3.6[/tex]
[tex]Z = -3.6[/tex] 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.
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]-0.675 = \frac{X - 2630}{50}[/tex]
[tex]X - 2630 = -0.675*50[/tex]
[tex]X = 2596[/tex]
The 25th percentile of the sample mean is of $2596.
Question d and e)
We have to find the z-score when X = 2645.
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{2645 - 2630}{50}[/tex]
[tex]Z = 0.3[/tex]
|Z| = 0.3 < 2, which means it would not be unusual if the sample mean was greater than $2645.
