Westonci.ca is the premier destination for reliable answers to your questions, brought to you by a community of experts. Get quick and reliable solutions to your questions from knowledgeable professionals on our comprehensive Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
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 your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.