# Thread: [SOLVED] Help!! The central limit theorem...

1. ## [SOLVED] Help!! The central limit theorem...

central limit theorem..

2. The Central Limit Theorem - $\displaystyle \frac{x - \overline{x}}{s}$ ~ N(0,1)

Where x is the observed value (in your case 23.5), x(bar) is the sample mean, and s is the sample standard deviation.

You want is P(Z<23.5), but first you need to standardize it using the CLT I mentioned above to get:

$\displaystyle P(Z < \frac{x - \overline{x}}{s})$
and then you would just use the normal table to obtain your percentage of how many student's sample mean is less than 23.5 (that's how I interpreted what the question was asking)

3. so would that be 0.4286 AKA 42.86% of the 180 students which would be 77 students?

4. Assuming you haven't made any calculation errors (as I have not caluclated it myself) then yes.

5. I tackled this question differently.. I am not sure if it is right though.

I copied this from - Central limit theorem - Wikipedia, the free encyclopedia

Let X1, X2, X3, …, Xn be a sequence of n independent and identically distributed random variables each having finite values of expectation µ and variance σ2 > 0. The central limit theorem states that as the sample size n increases the distribution of the sample average of these random variables approaches the normal distribution with a mean µ and variance σ2/n irrespective of the shape of the common distribution of the individual terms Xi.

and Zn converges in distribution to N(0,1)

Each Xi is the mean of Student i
Sn is the sum of the Student's means
X bar is the average of their averages
and n = 180

Now you need to find P(Z<Zn)

7. So I should replace n=64 to n=180? Which gives me -2.5 which is 0.0062 which gives me 1 student =S

8. or should i be doing this: 0.5 - .0062 which is 0.4938
which gives me 89 students

9. Originally Posted by xoxbabigurl
So I should replace n=64 to n=180? Which gives me -2.5 which is 0.0062 which gives me 1 student =S
Guess that's the answer, which isnt impossible (since there is a large standard error)