Respuesta :
Answer:
a) $5,656.85
b) Bell-shaped(normally distributed).
c) 36.32% probability of selecting a sample with a mean of at least $112,000.
d) 96.16% probability of selecting a sample with a mean of more than $100,000.
e) 59.84% probability of selecting a sample with a mean of more than $100,000 but less than $112,000.
Step-by-step explanation:
To solve this question, it is important to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are 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, of size at least 30, can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 110000, \sigma = 40000[/tex]
a. If we select a random sample of 50 households, what is the standard error of the mean?
This is the standard deviation of the sample, that is, s, when [tex]n = 50[/tex].
So
[tex]s = \frac{\sigma}{\sqrt{n}} = \frac{40000}{\sqrt{50}} = 5656.85[/tex]
b. What is the expected shape of the distribution of the sample mean?
By the Central Limit Theorem, bell-shaped(normally distributed).
c. What is the likelihood of selecting a sample with a mean of at least $112,000?
This is 1 subtracted by the pvalue of Z when X = 112000. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{112000 - 110000}{5656.85}[/tex]
[tex]Z = 0.35[/tex]
[tex]Z = 0.35[/tex] has a pvalue of 0.6368
So 1-0.6368 = 0.3632 = 36.32% probability of selecting a sample with a mean of at least $112,000.
d. What is the likelihood of selecting a sample with a mean of more than $100,000?
This is 1 subtracted by the pvalue of Z when X = 112000. So
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{100000 - 110000}{5656.85}[/tex]
[tex]Z = -1.77[/tex]
[tex]Z = -1.77[/tex] has a pvalue of 0.0384.
So 1-0.0384 = 0.9616 = 96.16% probability of selecting a sample with a mean of more than $100,000.
e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000
This is the pvalue of Z when X = 112000 subtractex by the pvalue of Z when X = 100000.
So
X = 112000
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{112000 - 110000}{5656.85}[/tex]
[tex]Z = 0.35[/tex]
[tex]Z = 0.35[/tex] has a pvalue of 0.6368
X = 100000
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{100000 - 110000}{5656.85}[/tex]
[tex]Z = -1.77[/tex]
[tex]Z = -1.77[/tex] has a pvalue of 0.0384.
So 0.6368 - 0.0384 = 0.5984 = 59.84% probability of selecting a sample with a mean of more than $100,000 but less than $112,000.
