I got the same answer for Q3 but for the conditional probability i did what shrz09 saidQ3
E(X) was 150, Var (X) was 30,000 (I accidentally wrote the s.d. tho)
In the last q I'm afraid you had to divide P(X<4) by P(X>=1) :S
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I got the same answer for Q3 but for the conditional probability i did what shrz09 saidQ3
E(X) was 150, Var (X) was 30,000 (I accidentally wrote the s.d. tho)
In the last q I'm afraid you had to divide P(X<4) by P(X>=1) :S
Yet the diference is very small in our workings,not more than 1 mark will be lost by whoever among us did it wrong,lets seeI just realised you're probably right.
My guess is 37, as it had 2 tricky qs and that is the gt papers with tricky qs getWhat do you guys say the GT will be? I say It'll be lower this time..a 40.
Thanks for the encouragement, broYet the diference is very small in our workings,not more than 1 mark will be lost by whoever among us did it wrong,lets see
yep, exactly what I did for both questionsI also got E(x) 150 and Var(x) 30000, as far as q5 is considered it was regarding confidence intervals,part 3 was (0.96)^4 as per binomial process.While as per my concepts the conditional probability was not p(x<4)/p(x is greater or equal to 1).Rather it was p(x=1,2,3)/p(x is greater or equal to 1) as zero is not an intersection between p(x is greater or equal to 1) and p(x<4)
Of course it canHow can you get 30,000 as variance? That makes the mean 150, and standard deviation 173.2 (if variance = 30,000). Do you really think the standard deviation can be greater than the mean? Think again.![]()
You canYes it is possible, you can get a variance greater than mean, but you can't get a sd greater than the mean
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