Respuesta :
Answer:
[tex]a=29000 -1.28*2400=25928[/tex]
Tires that wear out by 25928 miles will be replaces free of charge
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".
Solution to the problem
Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(29000,2400)[/tex]
Where [tex]\mu=29000[/tex] and [tex]\sigma=2400[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
He wants to give a guarantee for free replacement of tires that don't wear weel so then we need to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.90[/tex] (a)
[tex]P(X<a)=0.10[/tex] (b)
Because we are interested in the lower tail.
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.10 of the area on the left and 0.90 of the area on the right it's z=-1.28. On this case P(Z<-1.28)=0.10 and P(z>-1.28)=0.90
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.1[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.1[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=-1.28=<\frac{a-29000}{2400}[/tex]
And if we solve for a we got
[tex]a=29000 -1.28*2400=25928[/tex]
So the value of height that separates the bottom 10% of data from the top 90% is 25928 mi.
Tires that wear out by 25928 miles will be replaces free of charge
