Respuesta :
Answer:
20.05% probability that the sample mean is greater than 69.5 inches
Step-by-step explanation:
To solve this question, we need 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 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.
In this problem, we have that:
[tex]\mu = 68, \sigma = 4, n = 5, s = \frac{4}{\sqrt{5}} = 1.79[/tex]
What is the probability that the sample mean is greater than 69.5 inches?
This is 1 subtracted by the pvalue of Z when X = 69.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{69.5 - 68}{1.79}[/tex]
[tex]Z = 0.84[/tex]
[tex]Z = 0.84[/tex] has a pvalue of 0.7995
1 - 0.7995 = 0.2005
20.05% probability that the sample mean is greater than 69.5 inches
Answer:
[tex] P(\bar X >69.5) =P(Z \frac{69.5-68}{\frac{4}{\sqrt{5}}}) = P(Z>0.839)[/tex]
And we can use the complement rule with the normal standard table or excel and we got:
[tex] P(Z>0.839) =1-P(z<0.839) = 1-0.7993= 0.2007[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(68,4)[/tex]
Where [tex]\mu=68[/tex] and [tex]\sigma=4[/tex]
We select a sample of size n =5. Since the distribution for X is normal then we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
We want this probability:
[tex] P(\bar X >69.5)[/tex]
And we can use the z score formula given by:
[tex] z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And using this formula we got:
[tex] P(\bar X >69.5) =P(Z \frac{69.5-68}{\frac{4}{\sqrt{5}}}) = P(Z>0.839)[/tex]
And we can use the complement rule with the normal standard table or excel and we got:
[tex] P(Z>0.839) =1-P(z<0.839) = 1-0.7993= 0.2007[/tex]
