prettylittlepixels
New member
- Joined
- Nov 15, 2013
- Messages
- 32
Suppose all of the possible samples of size n = 36 are selected from a normally distributed population with a population mean of 100 and a population standard deviation of 24. |
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Between what limits (IN TERMS OF x) would the middle 80% of the individual x values be expected to fall? What would be the lower limit? |
69.28
Between what limits (IN TERMS OF x) would the middle 80% of the individual x values be expected to fall? What would be the upper limit? |
Between what limits (IN TERMS OF x) would the middle 80% of the sample means be expected to fall? What would be the lower limit? |
94.88
Between what limits (IN TERMS OF x) would the middle 80% of the sample means be expected to fall? What would be the upper limit? |
105.12
I would appreciate a quick check to see if I configured these correctly. For the top 2 because it is individual, I calculated it by:
First I looked up the z-score for 40% because half of 80% is .4 and the z-score closest I found was .3997, which corresponded with 1.28 and -1.28. Next I used the equation:
z-score= (x-mean) / standard deviation
So I did:
-1.28= (x-100) /24
-30.72= x-100
69.28= x (LOWER FOR INDIVIDUAL)
and
1.28= (x-100) /24
30.72= x-100
130.72= x (UPPER FOR INDIVIDUAL)
Next I used the equation:
z-score= (x-mean) / (standard deviation / sqrt of n) BUT X has line on top of it.
-1.28= (x-100) / (24 / sqrt of 36)
-1.28= (x-100) / (24 / 6)
-1.28= (x-100) / 4
-5.12= x-100
94.88= x (LOWER FOR SAMPLE MEANS)
(X still has line on top of it, I just don't know how to type that on here).
and
1.28= (x-100) / (24 / sqrt of 36)
1.28= (x-100) / (24 / 6)
1.28= (x-100) / 4
5.12= x-100
105.12= x (UPPER FOR SAMPLE MEANS)
(X still has line on top of it, I just don't know how to type that on here).
I think I have the hang of it, but I'm not quite sure yet as the x with and without the line on top of it is throwing me off if i calculated these right. Please let me know. I appreciate the help.