Respuesta :
Answer:
0.01% probability that the mean ticket price exceeds $33
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 = 29.89, \sigma = 5.28, n = 40, s = \frac{5.28}{\sqrt{40}} = 0.8348[/tex]
Find the probability that the mean ticket price exceeds $33
This is 1 subtracted by the pvalue of Z when X = 33. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{33 - 29.89}{0.8348}[/tex]
[tex]Z = 3.73[/tex]
[tex]Z = 3.73[/tex] has a pvalue of 0.9999
1 - 0.9999 = 0.0001
0.01% probability that the mean ticket price exceeds $33
Answer:
[tex]P(\bar X >33)[/tex]
And we can use the z score formula given by:
[tex] z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we find the z score for 33 we got:
[tex] z = \frac{33-29.89}{\frac{5.28}{\sqrt{40}}}= 3.725[/tex]
And we can use the complement rule and we got:
[tex] P(Z>3.725)= 1-P(Z<3.725)= 1-0.9999= 0.0001[/tex]
Step-by-step explanation:
Previous conepts
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 central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
Solution to the problem
We know the following properties for the variable of interest:
[tex]\mu = 29.89 , \sigma=5.28 [/tex]
We select a sample size of n = 40>30. From the central limit theorem 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]
And for this case we want to find this probability:
[tex]P(\bar X >33)[/tex]
And we can use the z score formula given by:
[tex] z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we find the z score for 33 we got:
[tex] z = \frac{33-29.89}{\frac{5.28}{\sqrt{40}}}= 3.725[/tex]
And we can use the complement rule and we got:
[tex] P(Z>3.725)= 1-P(Z<3.725)= 1-0.9999= 0.0001[/tex]
