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smithafield
11-30-2011, 01:05 AM
Consider a sample size of 100 with a standard deviation of 10.
a) What is the probability that the sample mean will be within 1 of the value of "u"?
b) Approximately 95% of the time, the mean will be within ___________ of "u"?
c) Approximately 0.3% of the time, the mean will be farther than ___________ from "u"?

Solution.
So far I only understand up to this point:
P(90 < x < 110)

I do not know how to find a z score so I am stuck at this point.

noetsi
11-30-2011, 11:24 AM
The formula for calculating a z-score (or standard score) is:

z = (x - μ) / σ

You commonly assume u is 0 but I am not sure you can do that in this assignment.

BGM
11-30-2011, 11:28 AM
You need to know the distribution of the sample mean.

Also see the 68-95-997 empirical rule.

http://en.wikipedia.org/wiki/68-95-99.7_rule

noetsi
11-30-2011, 11:31 AM
I thought Z scores required a normal distribution.... :p

Dason
11-30-2011, 11:36 AM
Calculating a z-score doesn't require a normal distribution. Making a probability statement about z-scores requires some sort of normal distribution though. But I think in this case the sample size justifies approximate normality. Also note that we don't need to know what \mu is to answer the questions in the problem.

noetsi
11-30-2011, 11:40 AM
You need it to calculate the z score with the formula I gave. :p

Dason
11-30-2011, 11:56 AM
But you don't actually need any set value of mu to do this problem. Just assuming that there is a mu is enough.

noetsi
11-30-2011, 12:15 PM
True. I responded to a question he asked, you don't actually have to have the z scores.

BGM
11-30-2011, 12:26 PM
OP somehow needs to know

1. How to make use of the result from Central Limit Theorem.
2. The relationship between an arbitrary normal distribution and the standard normal distribution (related to z-score).
3. How to calculate the probabilities related to (standard) normal distribution.

Of course if you can directly use the empirical rule as a well-known property of normal distribution, that is fine.