sampling means question
two runners are working out on the track, with their times varying each time they run around the track. Tom's times have N(29,1.2) distribution while Jerry's times have the N(28,40) distribution in minutes. If they run the course 5 times independently on one another, find the probability that Tom's average time will be faster than Jerry's average time.
I have no idea how to do this! help?
thanks for trying to help, but Im going to see if we go over this in class tomorrow. statistics is so confusing for me...
So for ux, is it 29-28=1 and standard deviation is 1.6? (add 0.4+1.2)
see Im clueless, that part above is probably wrong...I tried reading my book, but its useless...oh well
There is nothing confusing here. I suggest you review the sum of n independent normal random variables, and then the difference of two normal random variables. The results are simple and straightforward to apply.
Originally Posted by bcahmel
Is the st dev .4 or 40
And is that the variance or st dev?
st dev is .4...not 40. typing error there..
The difference of random variables is: P(X-Y)= P(X)-P(Y)
I get that I do P(T-J)...
the mean of T-J is 1.
And correct me if I'm wrong, but variances add and not the standard deviations? So I get 1.6 for the variance, or ~ 1.26 for the standard deviation.
T-J ~ N (1,1.26/sqrt(5).
and then I just plug it into my calculator and get about 3.86! which is the right answer! :) wow you were right, not that hard...
I'm not sure what you mean by p(x-y)=p(x)-p(y)
Do you mean expected values?
And the variance of x-y is the sums of the variances IF the rvs are indep.