Answer: 0.0013
Step-by-step explanation:
Given : The test scores are normally distributed with
Mean : [tex]\mu=\ 60,000[/tex]
Standard deviation :[tex]\sigma= 4,000[/tex]
Sample size : [tex]n=4[/tex]
The formula to calculate the z-score :-
[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
For x = 66,000
[tex]z=\dfrac{66000-60000}{\dfrac{4000}{\sqrt{4}}}=3[/tex]
The p-value = [tex]P(z>3)\=1-P(z<3)=1- 0.9986501\approx0.0013[/tex]
Hence, the likelihood the mean tire life of these four tires is more than 66,000 miles = 0.0013
