Answer:
The standard error of the mean will be $5,656.85.
a) The expected shape of the distribution of the sample mean is bell shaped, that is, normally distributed.
b) 1.70% likelihood of selecting a sample with a mean of at least $167,000.
c) 50% likelihood of selecting a sample with a mean of more than $155,000.
d) 48.30% likelihood of selecting a sample with a mean of more than $155,000 but less than $167,000.
Step-by-step explanation:
To solve this question, it is important to know the normal probability distribution and the Central Limit Theorem.
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation , a large sample size can be approximated to a normal distribution with mean
In this problem, we have that:
If we select a random sample of 50 households, what is the standard error of the mean?
By the Central Limit Theorem, the standard error will be
The standard error of the mean will be $5,656.85.
a. What is the expected shape of the distribution of the sample mean?
By the Central Limit Theorem, the expected shape of the distribution of the sample mean is bell shaped, that is, normally distributed.
b. What is the likelihood of selecting a sample with a mean of at least $167,000?
This is 1 subtracted by the pvalue of Z when X = 167000.
So
By the central Limit Theorem
has a pvalue of 0.9830.
So there is a 1-0.9830 = 0.0170 = 1.70% likelihood of selecting a sample with a mean of at least $167,000.
c. What is the likelihood of selecting a sample with a mean of more than $155,000?
This is 1 subtracted by the pvalue of Z when X = 155000.
So
By the central Limit Theorem
has a pvalue of 0.5.
So there is a 1-0.5 = 0.5 = 50% likelihood of selecting a sample with a mean of more than $155,000.
d. Find the likelihood of selecting a sample with a mean of more than $155,000 but less than $167,000.
This is the pvalue of Z when X = 167000 subtracted by the pvalue of Z when X = 155000
X = 167000
From b., Z has a pvalue of 0.9830.
X = 155000
From c., Z has a pvalue of 0.5.
So there is a 0.9830 - 0.5 = 0.4830 = 48.30% likelihood of selecting a sample with a mean of more than $155,000 but less than $167,000.