S3 - Combinations of random variables

OP

Would that not just be N(30.8, 115.6) ?

Reply 3

Ok first, your new normal distribution is wrong. You have made the classic mistake that my teacher is always harping on about by squaring the Var's. Your distribution should b N(308,214).
So all you need to do is use the above distribution with the limits and do a standard normal calculation.
I have done the question already so let me know if you need any more help/answers!

EDIT: your limits need to be multiplied by 10 as you have 10 sweets and the numbers given are the "average" weights

Reply 4

OP

xxsarahjwxx

Ok first, your new normal distribution is wrong. You have made the classic mistake that my teacher is always harping on about by squaring the Var's. Your distribution should b N(308,214).
So all you need to do is use the above distribution with the limits and do a standard normal calculation.
I have done the question already so let me know if you need any more help/answers!

EDIT: your limits need to be multiplied by 10 as you have 10 sweets and the numbers given are the "average" weights

Thanks for your reply.

What do you mean by 'the limits need to be multiplied by 10', what are the limits?

Do you mean that I should find the probability of between 280 and 330 ?

Reply 5

exactly that! Yes by limits i meant the numbers that its got to be between. So
P(280<6X+4Y<330) and u probly no the rest...!

Reply 6

OP

Ok thanks for your help.

I'm not sure where I've gone wrong with the distrubution tho. Are you not meant to square the variance?

e.g. Var(6X + 4Y) = Var(6X) + Var(4Y) = 36Var(X) + 16Var(Y)

Reply 7

That is for when you are using 6 * something. Theres a fine line between it.
Basically wot this question asks is X + X + X + X ...and so on. You square it when the question asks 6 times something. The way i tell the difference is if it uses a plural noun in the question then it means add and if it uses a singular noun (e.g. 10 times the weight of a sweet) then it means multiply, which is where you square it.

Reply 8

OP

I'm not sure I understand that way, surely X + X + X + ... + X 6 times is 6 times X?

Reply 9

charikaar

hi

I also posted the same question on this site few days ago. The reply is this!

Let X = no. of Xtras, Y = no. of Yummies, so

(b) E(Y - X) = 2; Var(Y - X) = 41

P(Y - X > 0) = P(Z> 0-2/sqrt(41))

= P(Z>-0.31)

= 0.6217

(c) Let K = packet

K = X1 + X2 + ... + X6 + Y1 + Y2 + Y3 + Y4

K(avg) = 1/10 K

therefore K~N(308, 214)

therefore K(avg)~N(30.8, 2.14)

you can do the rest now!

To see the original post plesae visit http://www.thestudentroom.co.uk/showthread.php?p=7578570#post7578570

Reply 10

Yes technically i know it doesnt make much sense but thats the way my teacher taught us (and shes good lol).
Ok from some of my class notes (excuse any typos, i have just got a new keyboard with really stupid keys lol):
"Confusion often arises over the different random variable 2X and X1 + X2 where X1 and X2 are 2 independent observations of X. EG if X is the r.v. "the number on which a tetrahedral die lands" then 2X is the r.v. "double the number on which a tetrahedral die lands" where as X1 + X2 is the r.v "the sum of the 2 numbers when a tetrahedral die is thrown twice"."

Hope that helps!

Reply 11

OP

I understand now, its mainly due to my laziness of writing 6X rather than X1, X2...X6.

Thanks.

Reply 12