S1 - Discrete random variables!

Hi,

I just can't seem to get the right answer for this question, and was hoping someone could, showing me how they got it.

A fair die is rolled and the random variable N represents the number showing. A square of side N is then drawn on a piece of paper.

a) Find the variance of the perimeter of this square

I did Var(4N)=4²Var(N) and I worked out Var(N) using (n²-1)/12, and the answer I ended up with is 2 and 11/12, although in the book it is 140/3.

Can someone please help as I don't know where I've gone wrong!

Thanks.

Reply 1

Jonny W

8

E(N) = 7/2
E(N^2) = (1/6)(1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2) = 91/6
Var(N) = E(N^2) - E(N)^2 = 35/12
Var(4N) = 4^2 Var(N) = 16*(35/12) = 140/3

Reply 2

sweet_gurl

OP

4

Thanks very much!

I just came across another question I can't do, and hope someone can help:

At a fair a roll a penny stall can be played with 1p or 2p coins. If the coin lands inside a square (without touching the edges) the player receives the coin plus 2 other coins of the same value, otherwise the coin is lost. The probability of winning the prize with 1p coin is 19/40 and the probability for a 2p coin is 11/40.

a) Find the expected winnings for each coin.

I just don't know what to do, as they have given the probabilities of winning. The answers are 1p 17/40, 2p -7/20. How do you get this.

Also, in the following part it asks 'would you play this game and why?' and in the back of the book it says 'Yes, with 1p since the expected value is positive'. I was just wondering, what does a positive expected value mean, as I can't seem to understand this explanation.

Thanks!

Reply 3

Chewwy

17

sweet_gurl

Also, in the following part it asks 'would you play this game and why?' and in the back of the book it says 'Yes, with 1p since the expected value is positive'. I was just wondering, what does a positive expected value mean, as I can't seem to understand this explanation.

the return you get for the money you put in is positive, that is you get back more than you put in.

Reply 4

sweet_gurl

OP

4

chewwy

the return you get for the money you put in is positive, that is you get back more than you put in.

Oh right, thanks! :smile:

Reply 5

sweet_gurl

OP

4

Can anyone help me with this question:

At a fair a roll a penny stall can be played with 1p or 2p coins. If the coin lands inside a square (without touching the edges) the player receives the coin plus 2 other coins of the same value, otherwise the coin is lost. The probability of winning the prize with 1p coin is 19/40 and the probability for a 2p coin is 11/40.

a) Find the expected winnings for each coin.

I just don't know what to do, as they have given the probabilities of winning. The answers are 1p 17/40, 2p -7/20. How do you get this?

Thanks.

Reply 6

JohnC

Winning no the 1p 19/40 times gives 38p over 40 goes but in those 40 you lose 21 times and so lose the coin. 38 - 21 = 17
=> 17/40

Using a similar method for the 2 pence, we get
11 * 4 = 44
44 - 2(40-11) = -14/40 = -7/20

Reply 7

sweet_gurl

OP

4

JohnC

Winning no the 1p 19/40 times gives 38p over 40 goes but in those 40 you lose 21 times and so lose the coin. 38 - 21 = 17
=> 17/40

Using a similar method for the 2 pence, we get
11 * 4 = 44
44 - 2(40-11) = -14/40 = -7/20

Cheers! I can now see how to do it. I was just wondering, how would you put this data on a frequency distribution, as 'expected' values are usually calculated from frequency distributions?

S1 - Discrete random variables!

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