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?
You are not multiplying the value of the one random variable, $R$, by three, rather you are summing three independent random variables each with an identical distribution to $R$.
The variance of a sum of normal distributed random variables equals the sum of the variances of the variables.
Also $\mathsf{Var}(-B)=\mathsf {Var}(B)$ so the variance of the difference of two ND random variables equals the addition of their variances.
Thusly:
$$\mathsf {Var}(R_1+R_2+R_3-B)=3\,\mathsf{Var}(R)+\mathsf{Var}(B)$$