Respuesta :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean
In this problem, we have that:
[tex]\mu = 165000, \sigma = 40000[/tex]
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 [tex]s = \frac{40000}{\sqrt{50}} = 5656.85[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{167000 - 155000}{5656.85}[/tex]
[tex]Z = 2.12[/tex]
[tex]Z = 2.12[/tex] 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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{155000 - 155000}{5656.85}[/tex]
[tex]Z = 0[/tex]
[tex]Z = 0[/tex] 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.
(1) Standard error of the mean SE is 5656.85.
(2) The likelihood of selecting a sample with a mean of at least $167,000 is 36%.
(3) The likelihood of selecting a sample with a mean of more than $155,000 is 96%
(4) The likelihood of selecting a sample with a mean of more than $155,000 but less than $167,000 is 60%
It is given that
Mean amount of life insurance per household μ = $165,000
Standard deviation σ = $40,000
Number of samples N = 50
So, standard error = σ /√N
Standard error SE = 40000/√50
Standard error SE = 5656.85
From the central limit theorem, we can say that the expected shape of the distribution of the sample-mean is bell-shaped i.e. normally distributed.
What is Z-value?
A Z-value(also called a standard value) gives you an idea of how far from the mean a data point is.
When X = 167000
Z-score = (X-μ )/SE
Z-score = (167000-165000)/5656.85
Z-score = 0.3535
Corresponding p-value = 0.3619 or ≈36%
So, the likelihood of selecting a sample with a mean of at least $167,000 is 36%.
When X = 155000
Z-score = (X-μ )/SE
Z-score = (155000-165000)/5656.85
Z-score = -1.7677
Corresponding p-value = 0.9614 or ≈96%
So, the likelihood of selecting a sample with a mean of more than $155,000 is 96%
The likelihood of selecting a sample with a mean of more than $155,000 but less than $167,000 will be 96-36 = 60%
Therefore, (1) standard error of the mean SE is 5656.85.
(2) The likelihood of selecting a sample with a mean of at least $167,000 is 36%.
(3) The likelihood of selecting a sample with a mean of more than $155,000 is 96%
(4) The likelihood of selecting a sample with a mean of more than $155,000 but less than $167,000 is 60%
To get more about the z-score visit:
brainly.com/question/25638875
