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Sagot :
Using the z-distribution, it is found that the 95% confidence interval for the mean lifetime of this type of drill, in holes drilled, is (10.4, 13.94).
We are given the standard deviation for the population, hence, the z-distribution is used. The parameters for the interval is:
Sample mean of [tex]x = 12.7[/tex]
Population standard deviation of [tex]\sigma = 6.37[/tex]
Sample size of .n = 50
The margin of error is:
[tex]M = z \frac{\sigma }{\sqrt{n} }[/tex]
In which z is the critical value.
We have to find the critical value, which is z with a p-value of [tex]\frac{1+\alpha }{2}[/tex] , in which [tex]\alpha[/tex] is the confidence level.
In this problem, , thus, z with a p-value of , which means that it is z = 1.96.
Then:
[tex]M= 1.96\frac{6.37}{\sqrt{50} } = 1.77[/tex]
The confidence interval is the sample mean plus/minutes the margin of error, hence:
x- M = 12.17 - 1.77 =10.4
x+M = 12.17 + 1.77 = 13.94
The 95% confidence interval for the mean lifetime of this type of drill, in holes drilled, is (10.4, 13.94).
Learn more about confidence level at :
brainly.com/question/22596713
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