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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 <u>standard deviation for the population</u>, hence, the z-distribution is used. The parameters for the interval is:
- Sample mean of
- Population standard deviation of
- Sample size of
.
The margin of error is:
In which z is the critical value.
We have to find the critical value, which is z with a p-value of , in which
is the confidence level.
In this problem, , thus, z with a p-value of
, which means that it is z = 1.96.
Then:
The confidence interval is the <u>sample mean plus/minutes the margin of error</u>, hence:
The 95% confidence interval for the mean lifetime of this type of drill, in holes drilled, is (10.4, 13.94).
A similar problem is given at brainly.com/question/22596713
The expression that can be used to represent the total price is 3(p - $0.75) + 5(p - $1).
The expression that can be used to determine the total amount spent in store A is:
- number of stickers bought x ( price of the item - discount)
- number of stickers bought = 5
- price of the stickers = p
- discount = $1
5(p - $1)
The expression that can be used to determine the total amount spent in store B is:
- number of stickers bought x ( price of the item - discount)
- number of stickers bought = 3
- price of the stickers = p
- discount = $0.75
3(p - $0.75)
Total price = 3(p - $0.75) + 5(p - $1)
A similar question was solved here: brainly.com/question/25717996
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