Respuesta :
Answer:
a) By the Central Limit Theorem, it is going to be bell-shaped, that is, normally distributed. The Central Limit Theorem states that no matter the shape of the population, the shape of the sampling distribution of the sample mean is going to be normally distributed, so we do not need to make any assumptions about the shape of the population.
b) [tex]\mu = 22, s = 0.9630[/tex]
c) There is an 1.88% probability that we will obtain a sample mean greater than 24.
d) There is a 6.30% probability that we will obtain a sample mean less than 20.253.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu 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].
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.
In this problem, we have that:
There are 69 measurements, so [tex]n = 69[/tex]
The population has a mean of 22, so [tex]\mu = 22[/tex]
The population has a standard deviation of 8, so [tex]\sigma = 8[/tex].
(a) Describe the shape of the sampling distribution of the sample mean . Do we need to make any assumptions about the shape of the population? Why or why not?
By the Central Limit Theorem, it is going to be bell-shaped, that is, normally distributed. The Central Limit Theorem states that no matter the shape of the population, the shape of the sampling distribution of the sample mean is going to be normally distributed, so we do not need to make any assumptions about the shape of the population.
(b) Find the mean and the standard deviation of the sampling distribution of the sample mean.
The mean of the sampling distribution of the sample mean is the same as the mean of the population, so [tex]\mu = 22[/tex]
The standard deviation of the sampling distribution of the sample mean is the standard deviation of the population divided by the length of the sample. So
[tex]s = \frac{\sigma}{\sqrt{n}} = \frac{8}{\sqrt{69}} = 0.9630[/tex]
(c) Calculate the probability that we will obtain a sample mean greater than 24
We are working with the sample mean, so we use [tex]s[/tex] in the place of [tex]\sigma[/tex].
This probability is 1 subtracted by the pvalue of Z when [tex]X = 24[/tex].
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{24 - 22}{0.9630}[/tex]
[tex]Z = 2.08[/tex]
[tex]Z = 2.08[/tex] has a pvalue of 0.9812.
This means that there is a 1-0.9812 = 0.0188 = 1.88% probability that we will obtain a sample mean greater than 24.
(d) Calculate the probability that we will obtain a sample mean less than 20.253
This is the pvalue of Z when [tex]X = 20.53[/tex]. So:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{20.53 - 22}{0.9630}[/tex]
[tex]Z = -1.53[/tex]
[tex]Z = -1.53[/tex] has a pvalue of 0.0630.
This means that there is a 6.30% probability that we will obtain a sample mean less than 20.253.
