Respuesta :
Answer:
The two values of X that delineate the "82% middle pack" of this random variable are 9480 and 27520.
0.4017 = 40.17% probability that the sample average that you calculated will lie between 1,702 and 1,948.
Step-by-step explanation:
To solve the first question, we use the normal distribution, while for the second quetion, it is used with 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.
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.
First question:
Mean of 18,500:
This means that [tex]\mu = 18500[/tex]
You are told that P(X ≥ 15,000) = 0.6981.
This means that when [tex]X = 15000[/tex], Z has a o-value of 1 - 0.6981 = 0.3019, which means that when [tex]X = 15000, Z = -0.52[/tex]. We use this to find [tex]\sigma[/tex]. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.52 = \frac{15000 - 18500}{\sigma}[/tex]
[tex]0.52\sigma = 3500[/tex]
[tex]\sigma = \frac{3500}{0.52}[/tex]
[tex]\sigma = 6731[/tex]
What are the two values of X that delineate the "82% middle pack" of this random variable?
Between the 50 - (82/2) = 9th percentile and the 50 + (82/2) = 91st percentile.
9th percentile:
X when Z has a p-value of 0.09, so X when Z = -1.34.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.34 = \frac{X - 18500}{6731}[/tex]
[tex]X - 18500 = -1.34*6731[/tex]
[tex]X = 9480[/tex]
91st percentile:
X when Z has a p-value of 0.91, so X when Z = 1.34.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.34 = \frac{X - 18500}{6731}[/tex]
[tex]X - 18500 = 1.34*6731[/tex]
[tex]X = 27520[/tex]
The two values of X that delineate the "82% middle pack" of this random variable are 9480 and 27520.
Question 2:
A random variable has a population mean equal to 1,973 and population variance equal to 892,021.
This means that [tex]\mu = 1973, \sigma = \sqrt{892021} = 944.5[/tex]
Sample of 79:
This means that [tex]n = 79, s = \frac{944.5}{\sqrt{79}}[/tex]
What is the probability that the sample average that you calculated will lie between 1,702 and 1,948?
This is the p-value of Z when X = 1948 subtracted by the p-value of Z when X = 1702. So
X = 1948
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{1948 - 1973}{\frac{944.5}{\sqrt{79}}}[/tex]
[tex]Z = -0.235[/tex]
[tex]Z = -0.235[/tex] has a p-value of 0.4071
X = 1702
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{1702 - 1973}{\frac{944.5}{\sqrt{79}}}[/tex]
[tex]Z = -2.55[/tex]
[tex]Z = -2.55[/tex] has a p-value of 0.0054
0.4071 - 0.0054 = 0.4017
0.4017 = 40.17% probability that the sample average that you calculated will lie between 1,702 and 1,948.
