Respuesta :
Answer:
87.27% probability that the sample mean would differ from the population mean by less than 344 miles in a sample of 121 tires if the manager is correct
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 = 34631, \sigma = 2480, n = 121, s = \frac{2480}{\sqrt{121}} = 225.45[/tex]
What is the probability that the sample mean would differ from the population mean by less than 344 miles in a sample of 121 tires if the manager is correct
This is the pvalue of Z when X = 34641 + 344 = 34985 subtracted by the pvalue of Z when X = 34641 - 344 = 34297. So
X = 34985
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{34985 - 34631}{225.45}[/tex]
[tex]Z = 1.525[/tex]
[tex]Z = 1.525[/tex] has a pvalue of 0.93635
X = 34297
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{34297 - 34631}{225.45}[/tex]
[tex]Z = -1.525[/tex]
[tex]Z = -1.525[/tex] has a pvalue of 0.06365
0.93635 - 0.06365 = 0.8727
87.27% probability that the sample mean would differ from the population mean by less than 344 miles in a sample of 121 tires if the manager is correct
Answer:
[tex] P(34641- 344 <\bar X < 34641 +344) = P(34297 <\bar X <34985)[/tex]
And for this case we can use the z score formula given by:
[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we use the formula for the limits that we have, we got:
[tex] z_1 = \frac{34297-34641}{\frac{2480}{\sqrt{121}}}= -1.526[/tex]
[tex] z_2 = \frac{34985-34641}{\frac{2480}{\sqrt{121}}}= 1.526[/tex]
So we want to find this probability and we can use the normal standard table and this difference:
[tex] P(-1.526 <Z<1.526) = P(z<1.526) -P(Z<-1.526) = 0.9365- 0.0635= 0.8730[/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".
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
Let X the random variable that represent the heights of a population, and for this case we know the following properties:
Where [tex]\mu =34641 , \sigma =2480[/tex]
Since the sample size used n =121 >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 we want to find this probability:
[tex] P(34641- 344 <\bar X < 34641 +344) = P(34297 <\bar X <34985)[/tex]
And for this case we can use the z score formula given by:
[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we use the formula for the limits that we have, we got:
[tex] z_1 = \frac{34297-34641}{\frac{2480}{\sqrt{121}}}= -1.526[/tex]
[tex] z_2 = \frac{34985-34641}{\frac{2480}{\sqrt{121}}}= 1.526[/tex]
So we want to find this probability and we can use the normal standard table and this difference:
[tex] P(-1.526 <Z<1.526) = P(z<1.526) -P(Z<-1.526) = 0.9365- 0.0635= 0.8730[/tex]
