Respuesta :
Answer:
For a mean oil change time of 20.51 minutes there would be a 10% chance of being at or below
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 = 21.3, \sigma = 3.9, n = 40, s = \frac{3.9}{\sqrt{40}} = 0.6166[/tex]
Treating this as a random sample, at what mean oil-change time would there be a 10% chance of being at or below?
This is the 10th percentile, which is X when Z has a pvalue of 0.1. So it is X when Z = -1.28.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]-1.28 = \frac{X - 21.3}{0.6166}[/tex]
[tex]X - 21.3 = -1.28*0.6166[/tex]
[tex]X = 20.51[/tex]
For a mean oil change time of 20.51 minutes there would be a 10% chance of being at or below
Answer:
[tex]20.51 \rm min[/tex] mean oil-change time would there be a 10% chance of being at or below.
Step-by-step explanation:
Given information:
Mean time [tex]\mu=21.3\rm min[/tex]
Standard deviation [tex]\sigma=3.9\rm min[/tex]
[tex]n=40[/tex]
standard error [tex]s=\frac{\sigma}{\sqrt n}=\frac{3.9}{40}=0.6166[/tex]
10% chance of being at or below,[tex]\widehat{p}=0.1[/tex]
By use of standard normal table
[tex]\widehat{p}=0.1[/tex],[tex]z=-1.282[/tex]
By Central limit theorem,
[tex]z=\frac{X-\mu}{s}[/tex]
On substitution,
[tex]-1.282=\frac{X-21.3}{\ 0.6166}[/tex]
[tex]X=20.51 \rm min[/tex]
Hence,[tex]20.51 \rm min[/tex] mean oil-change time would there be a 10% chance of being at or below.
For more details please refer link:
https://brainly.com/question/5686852?referrer=searchResults
