please help...i'm stuck

[ecohelp]

Question:
There are 8,000 teachers at UCLA. The average age of all the teachers is 24 years with a standard deviation of 9 years. A random sample of 36 teachers is selected.

A) Determine the standard error of the mean

B) What is the probability that the sample mean will be larger than 19.5?

C) What is the probability that the sample mean will be between 25.5 and
27 years?


I know the answers are:

A) 1.5
B) 0.9986
C) 0.1329


I would like to know how to go about solving them though, because I'm totally lost. The answers aren't any help when I don't know how to solve them... any help on walking me through on the calculations/which formulas to use would greatly be appreciated. Thanks in advance.

btw i know that for the first part A you have to use the formula:
l sample mean - population mean l
but which number goes where? i'm confused....

[JohnM]

(A)
standard error of the mean is the population standard deviation divided by the square root of the sample size - you are given both of these

--> the standard error of the mean is the standard deviation of sample means (i.e., approximately what it would be if you drew many samples, computed the mean of each sample, then computed the standard deviation of those means)

(B and C)
You need to compute a z-score (for 19.5 in B, and for both 25.5 and 27 in C) which is the sample mean minus the population mean, then this difference divided by the standard error of the mean. Don't use absolute values - it's OK to get a negative z-score. A z-score indicates how many standard deviations away from the center of the normal curve. If it's negative, it's to the left of center, if it's positive, it's to the right.....

For B, you need to find the proportion (% of the total area) of the curve above the z-score for 19.5

For C, you need to find the proportion (% of the total area) of the curve in-between the z-scores for 25.5 and 27

I'm sure there are examples in your textbook that are exactly lke this.....

[ecohelp]

okay got it that helped me out a lot, thanks...

but on this next one there is no standard deviation given so i can't solve for the standard error...i tired to solve it like the above but can't because the S.D. isn't given unless you have to solve using a different technique?


Here's how the problem reads:

10% of the items produced by a machine are defective. A random sample of 100 items is selected and checked for defects.

a) determine the standard error of the proportion

b) what is the probability that the sample will contain more than 2.5% defective units?

c) what is the probability that the sample will contain more than 13% defective units?



For part A i attempted to do square root of (N-n)/(N-1) * (o/square root of n)= standard error....but i'm missing the standard devation? how to solve? am i doing something wrong?

[ecohelp]

nevermind i was using the wrong formula should have used the (proportion) population formula. I ended up using the infinite population formula and got the answer... but b/c 10/100=.1>.05 shouldn't we have to use the finite population?



***how do you know if you have to minus the z-score from 1 or not? that is what i'm finding difficult, because sometimes you minus and other times the answer is what you look up on the z-score? is there a key word like "more" than #, or "less" than #

[quark]

Hi ecohelp,

If the sampling fraction is 5% or more then you need to use the finite population correction.

Values in the normal table represent areas under the curve to the left of Z.
So for P(Z<2) you can get directly from the table, and P(Z>2)=1-P(Z<2).

P(Z< -2)=1-P(Z<2) and P(Z> -2)= P(Z<2) by symmetry of the normal curve.

[JohnM]

ecohelp,

Unfortunately sometimes the z tables are different, depending on the textbook. Sometimes the table shows the area under the curve from negative infinity to z, and sometimes it's from 0 to z (half of a curve).

The table will have a note or diagram that says which way, however.....