Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Join our Q&A platform to connect with experts dedicated to providing precise answers to your questions in different areas. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
Answer:
0.8948 = 89.48% probability that the mean of a sample of 43 cars would differ from the population mean by less than 111 miles
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 normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem establishes 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.
The mean number of miles between services is 4959 miles, with a standard deviation of 448 miles
This means that [tex]\mu = 4959, \sigma = 448[/tex]
Sample of 43:
This means that [tex]n = 43, s = \frac{448}{\sqrt{43}}[/tex]
What is the probability that the mean of a sample of 43 cars would differ from the population mean by less than 111 miles?
p-value of Z when X = 4959 + 111 = 5070 subtracted by the p-value of Z when X = 4959 - 111 = 4848, that is, probability the sample mean is between these two values.
X = 5070
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{5070 - 4959}{\frac{448}{\sqrt{43}}}[/tex]
[tex]Z = 1.62[/tex]
[tex]Z = 1.62[/tex] has a p-value of 0.9474
X = 4848
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{4848 - 4959}{\frac{448}{\sqrt{43}}}[/tex]
[tex]Z = -1.62[/tex]
[tex]Z = -1.62[/tex] has a p-value of 0.0526
0.9474 - 0.0526 = 0.8948
0.8948 = 89.48% probability that the mean of a sample of 43 cars would differ from the population mean by less than 111 miles
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.