Respuesta :
Answer:
66.98% probability that the mean gain for the sample was between 250 and 500.
Step-by-step explanation:
To solve this problem, 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 [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex].
In this problem, we have that:
[tex]\mu = 432, \sigma = 722, n = 40, s = \frac{722}{\sqrt{40}} = 114.16[/tex]
What is the probability that the mean gain for the sample was between 250 and 500?
This is the pvalue of Z when X = 500 subtracted by the pvalue of Z when X = 250.
So
X = 500
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{500 - 432}{114.16}[/tex]
[tex]Z = 0.6[/tex]
[tex]Z = 0.6[/tex] has a pvalue of 0.7257.
X = 250
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{250 - 432}{114.16}[/tex]
[tex]Z = -1.59[/tex]
[tex]Z = -1.59[/tex] has a pvalue of 0.0559.
So there is a 0.7257 - 0.0559 = 0.6698 = 66.98% probability that the mean gain for the sample was between 250 and 500.
