Hi. Regarding question 2.23 part (i) The question says that the mean and standard deviation are 20 mins and 5 mins respectfully however the solution is using a N(20,5) distribution. Is this a mistake?? Also the same problem arises in question 2.28 where the linear combination of the normal dists is adding the standard devs of both and not the variances. Am I missing something completely or is this a mistake?? Cheers James
Almost! No - it is correct (though you had me worried for a bit!) You are correct in saying that X ~ N(20, 5²) However, we are doing probabilities for X bar - which for n = 5 has a distribution of: X bar ~ N(20, 5²/5) = N(20, 5) The confusion has arisen that the sample size is the same as the standard deviation so when we divide the variance through by n it looks like we're just working with the standard deviation. Q2.28 is similar. Clear as mud?
