Statistics Continuous Random Variable Answer Issue???

So i was doing this question and I'm a bit confused at the working on the back:

Q. The lengths of red pencils are normally distributed with mean 6.5 cm and standard deviation 0.23 cm.
(i) Two red pencils are chosen at random. Find the probability that their total length is greater than 12.5 cm.

The lengths of black pencils are normally distributed with mean 11.3 cm and standard deviation 0.46 cm.
(ii) Find the probability that the total length of 3 red pencils is more than 6.7 cm greater than the length of 1 black pencil.

So for part ii) I took it as 3R - B > 6.7

and according to this the mean would be 3(6.5) - 11.3 so thats 6.7 and i took variance as 3^2(0.23)^2 + (0.46)^2 which comes as 0.6877 but the answer at the back took variance for red pencils as 3 x (0.23) ^2. why is that so?????? i mean it makes sense in part i to take 2 x (0.23)^2 but i dont think that should be the case for ii or am I wrong?

Reply 1

ghostwalker

17

Original post by veeneenaa

i mean it makes sense in part i to take 2 x (0.23)^2 but i dont think that should be the case for ii or am I wrong?

The book is correct.

What's the difference in your thinking that makes you think in the first part multiplying by 2 rather than 2^2 is correct, but in the second part multplying by 3, rather than 3^2 is incorrect?

(edited 3 years ago)

Reply 2

mqb2766

21

Original post by veeneenaa

So i was doing this question and I'm a bit confused at the working on the back:

Q. The lengths of red pencils are normally distributed with mean 6.5 cm and standard deviation 0.23 cm.
(i) Two red pencils are chosen at random. Find the probability that their total length is greater than 12.5 cm.

The lengths of black pencils are normally distributed with mean 11.3 cm and standard deviation 0.46 cm.
(ii) Find the probability that the total length of 3 red pencils is more than 6.7 cm greater than the length of 1 black pencil.

So for part ii) I took it as 3R - B > 6.7

and according to this the mean would be 3(6.5) - 11.3 so thats 6.7 and i took variance as 3^2(0.23)^2 + (0.46)^2 which comes as 0.6877 but the answer at the back took variance for red pencils as 3 x (0.23) ^2. why is that so?????? i mean it makes sense in part i to take 2 x (0.23)^2 but i dont think that should be the case for ii or am I wrong?

Just look at the expression for variance. It's the sum of the squared differences from the mean so you sum variances. Just apply the two random variable case twice (or three times) if you're happy with that.
Is your mean 8.2, not 6.7?

More generally, the variance expression for a linear combination of random variables is the sum over each random variable of
c_i^2 sigma_i^2
Here c_i is the weighting applied to each random variable (+/-1 in your case, so the square is +1), not the number of occurrences.

(edited 3 years ago)

Reply 3

ghostwalker

17

Original post by veeneenaa

So for part ii) I took it as 3R - B > 6.7

That might make you think that you have one occurrence of R and are multiplying it by 3, which would lead to a multiplier of 3^2 for the red's variance.

Whereas you actually have three separate instances of R, so it would be more accurately written as R1+R2+R3 - B > 6.7.
And since the variance of R1,R2,R3 are the same, we end up with a multiplier of 3 for the red's variance.

Statistics Continuous Random Variable Answer Issue???

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