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Independent random variables

  • Thread starter WY
  • Start date
WY
28
0
Hi
I'm wondering if someone can help me out on this question as to how to go about doing it:
X_1, X_2.... X_7 are independent random variables represnting a random sample of size 7 from the normal N(10, 7) distribution. Find to 3 dp probablitity that the sample total exceeds 88.

I tried to standardise this but my numbers don't seem to get me the answer of 0.005. Can someone help me out? Thanks in advance :)
 

Answers and Replies

LeonhardEuler
Gold Member
858
1
How did try to do it? Remember, the d.f. of the sum of random variables with normal distributions is another normal distribution with a mean that is the sum of the means of the individual variables and a variance that is the sum of the variances of the individual variables. Also remember when changing varibles that what appears in the the normal distribution is [tex]\frac{(x-\mu)}{\sigma}[/tex] and not [tex]\frac{(x-\mu)}{\sigma^2}[/tex], so use the standard deviation and not the variance when changing variables.
 
WY
28
0
Thanks for the help! when i originally did it i used (88-10)/7 to try and standardise it - giving me a ridiculous number. So with the normal distribution N(10,7) what should I now do with those - I'm still kind of confused....
 
LeonhardEuler
Gold Member
858
1
Remember, the mean is the sum of the means of the X_i, so that's 10+10+10+...=70. The variance is the sum of the variances. Remember to normalize with the standard deviation and not the variance. Once you do that you do get the answer you said you were supposed to.
 

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