Respuesta :
Answer:
a) The mean is 8 and the standard error is 1.5.
b) 9.18% probability that the sample mean will be greater than 10.
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 [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
I call the population mean [tex]\mu[/tex] and the population standard deviation [tex]\sigma[/tex]
So
[tex]\mu = 8, \sigma = 6[/tex]
a) What is the expected mean and the standard error for a sample size of n=16
By the Central Limit Theorem
[tex]\mu = 8[/tex]
[tex]s = \frac{6}{\sqrt{16}} = 1.5[/tex]
b) What is the probability that the sample mean will be greater than 10?
This probability is 1 subtracted by the pvalue of Z when X = 10.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
Due to the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{10 - 8}{1.5}[/tex]
[tex]Z = 1.33[/tex]
[tex]Z = 1.33[/tex] has a pvalue of 0.9082.
So there is a 1-0.9082 = 0.0918 = 9.18% probability that the sample mean will be greater than 10.
