Use the confidence level and sample data to find a confidence interval for estimating the population μ. Round your answer to the same number of decimal places as the sample mean. Test scores: n = 92, = 90.6, σ = 8.9; 99% confidence

Options:
A.) 88.2 < μ < 93.0
B.) 88.4 < μ < 92.8
C.) 89.1 < μ < 92.1
D.) 88.8 < μ < 92.4

Choice A.)   88.2 < μ < 93.0

Explanation:

We have this given info:

Because n > 30 and because we know sigma, this allows us to use the Z distribution (aka standard normal distribution).

At 99% confidence, the z critical value is roughly z = 2.576; use a reference sheet, table, or calculator to determine this.

The lower bound of the confidence interval (L) is roughly

L = xbar - z*sigma/sqrt(n)

L = 90.6 - 2.576*8.9/sqrt(92)

L = 88.209757568781

L = 88.2

The upper bound (U) of this confidence interval is roughly

U = xbar + z*sigma/sqrt(n)

U = 90.6 + 2.576*8.9/sqrt(92)

U = 92.990242431219

U = 93.0

Therefore, the confidence interval in the format (L, U) is approximately (88.2, 93.0)

When converted to L < μ < U format, then we get approximately 88.2 < μ < 93.0 which shows that the final answer is choice A.

We're 99% confident that the population mean mu is somewhere between 88.2 and 93.0

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step-by-step explanation:

first you take the 6 dollars and add 10 to it then add the 50 to that then multiply by 8 with xy and your answer will be

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what is the answer?