Suppose that the population of the scores of all high school seniors that took the SAT-M (SAT math) test this year follows a Normal distribution, with mean μ and standard deviation σ = 100. You read a report that says, "On the basis of a simple random sample of 100 high school seniors that took the SAT-M test this year, a confidence interval for μ is 512.00 ± 25.76." The confidence level for this interval is

The confidence level for this interval is 99%.

Step-by-step explanation:

The margin of error M has the following equation.

In which is the standard deviation of the population and n is the size of the sample.

z is related to the confidence level.

In this problem:

So

Looking at the z table, has a pvalue of 0.995.

So the confidence level is:

The confidence level for this interval is 99%.

answer: expected value of x = 1.0

step-by-step explanation:

p(x=0) = 0.25

p(x=1) = 0.50

p(x=2) = 0.25

expected value for x = 0×0.25 + 1×0.5 + 2×0.25 = 1.0

1/4 because there is 4 grades.