|By Hello on Wednesday, December 18, 2002 - 11:19 am: Edit|
For all you statistics people our there, I am in need of some aid. Please help me solve the following problems...
Assuming that the mean verbal score is 550 and the mean math score is 624, and the standard deviation is 100 for the random variable X (verbal) and 80 for math, the random variable Y. Assuming that the school plugged the scores into the following equation what would the new standard devieation be?
(2*math standard deviation+80)+standard deviation of verbal.
A word of caution: only variances of independant events will add. The standard deviations do not add. So we see that SAT math and verbal are not independant and therefore can't be added. If we multiply 2 times the standard deviation it is like we are multiplying 4*the variance (variance is the square of s.d and when multiplying the s.d is the square multiplied.
|By HELPME on Wednesday, December 18, 2002 - 11:27 am: Edit|
I have a question also... Lets us say that the proability of catching a flu is .3 and the probability of catching a cold is .4. Also, the probability of catching a flu or a cold is .8.
Question 1: Are the events (catching a flu, and catching a cold independant)? I think that they are not independant, at least according to the problem because the rule for getting one or the other is= p(cold)+p(flu)-(p cold and flu). By plugging the values in, I see that the p(catching flu and cold) is .1 (If they were independant the value would be .4*.3=.12).
Question 2: What is the p(not cold/not flu) (This means what is the probability of not getting he cold given the information that one does not get a flu as well). I know that this is a conditional prob. Question but i cant figure it out.
The equation for cond. prob is:
P(A/B)=(P(A) and P(B/A))/P(A)
Can someone please check my calculations and help me figure out question 2.
|By Hello on Wednesday, December 18, 2002 - 01:41 pm: Edit|
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