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 m and standard deviation s = 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 m is 512.00 ± 25.76.” The confidence level for this interval is:

a:over 99.9%
b: 95%
c.99%
d: 90%

Your mission is to find the z-score that corresponds with your confidence interval.

E=z\cdot\frac{\sigma}{\sqrt{n}}

You have E=25.76, \;\ {\sigma}=100, \;\ n=100

Plug them into the formula and solve for z. Then, look it up in the z-table.

Thank you soooo much!

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Galactus,
What about this one, I THINK THE ANSWER IS b, BUT I'M NOT SURE

A deck of ten cards consists of five red and five black cards. The cards are shuffled and I choose one at random, observe its color, and replace it in the set. The cards are thoroughly reshuffled, and I again choose a card at random, observe its color, and replace it in the set. This is done a total of four times. Let X be the number of red cards observed in these four trials. The random variable X has which of the following probability distributions?

a: the binomial distribution with parameters n=4 and p=1
b: the uniform distribution on 0,1,2,3,4
c: the normal distribution with mean 2 and variance 1
d: none of the above

The weights of medium oranges packaged by an orchard are normally distributed with a mean of 14 oz. and a standard deviation of 2 oz. The weights of large oranges packaged by this orchard are normally distributed with a mean of 18 oz. and a standard deviation of 3 oz. If we select a medium packaged orange at random and a large packaged orange at random, the probability that the medium orange weighs more than the large orange is


a. 0.0000
b.0.0228
c. 0.0548
d. 0.1335