Have you put down the full question and answer? I’m pretty sure it’s talking about the band that 90% of the values fall into. For a 2-tailed test this is 1.65 standard deviations each side of the mean. Your standard deviation is 5, so the band is 8.25 (1.65*5) units away from the mean on each side.
@mattmania: yes it is possible some part of the question got lost when I copied it. But how did you get to the 8.25 in your solution? Their solution goes like this: 100 +/- 1.65 (5) = 91.75 to 108.25
@panos.kollias: ok, so in every population problem I do not need n in the formula? It just always goes: μ +/- score x σ, like in their solution above? I am confused because I thought the difference of population or sample parameters determine the test Z or T. I do not remember coming across a formula for a confidence interval that didnt have the n in it. thanks!
Yes! Keep in mind that population parameters are the trueparameters of the population so there is no need for approximation. We don’t need to know how big the sample was, we know what the true standard deviation and mean is.
The reason why we use n in the standard error calculation is that the greater the sample size the more accurate (close to the true population) our parameters are going to be.
Eskimo: your population distribution is normal. And you have the population standard dev. No standard error necessary! In real life you wouldn’t have the population data and would work off of samples. That’s why 99% of all examples and questions would require the standard error.