# Thread: Need help with random variable question

1. ## Need help with random variable question

I'm not sure how to work out this question. Firstly I don't know how to apply the formula and secondly I don't know what the assumptions I should make. Could someone please help.

Two statistics post-grad students went fishing at a lake during a break at a conference in England. Beforehand they agreed that the one who caught the more impressive fish would have their dinner paid for by the other. Anna caught a 40 cm long perch, whilst Joe caught a 42 cm bream.

-From long term studies of the fish populations in the lake, it is known that the length of perch is approximately normally distributed with mean of 29.6 cm and standard deviation of 9.5 cm.
The length of bream is approximately normally distributed with mean of 38.4 cm and standard deviation of 4.2 cm.

The next day Joe went fishing again and caught 1 bream and 1 perch. Use the formula for the difference of random variables to find the probability that the bream is longer than the perch. What assumption was made to do this calculation?

2. I have no idea what
Anna caught a 40 cm long perch, whilst Joe caught a 42 cm bream.
has to do with the price of rice.

Assuming independence

$P(B>P)=P(B-P>0)=P\left({(B-P)-(\mu_B-\mu_P)\over \sqrt{\sigma^2_B+\sigma^2_P}}>{0-(38.4-29.6)\over \sqrt{(4.2)^2+(9.5)^2}}\right)$

$=P\left(Z>{0-(38.4-29.6)\over \sqrt{(4.2)^2+(9.5)^2}}\right)$

3. Originally Posted by matheagle
I have no idea what
Anna caught a 40 cm long perch, whilst Joe caught a 42 cm bream.
has to do with the price of rice.
[snip]
It's a Red Herring!