Edexcel A Level Statistics and Mechanics Year 1/AS Ex. 2E p.33 Q8 mean/standard deviation

Question

I agree with the answer to (a). The problem I have is that if the answer to (c) is correct, then the mean would be (4+17)/2 = 10.5kn, which is different from the value 8.1kn from part (a), so therefore the answer given to question (c) (and therefore (b)) seems wrong.

So my point is that there is not enough information to answer (b) (should we assume a skewed normal distribution? Well we are only in chapter 2 of the book and we haven't got to normal distribution yet...) And question (b) is contingent on (c). More information about the distribution of windspeed should be given in order to answer the question, like if they gave us the frequencies of different classes (groups). Maybe if we gathered grouped data and made a grouped frequency table ourselves we could answer the question, but they didn't ask for that in the question. So I think (b) and (c) are bad questions to ask the student, and on top of that the answer to (b) is probably wrong because the answer to (c) seems wrong...

Answer 1

I will offer some clues and comments because this problem is either unclear or extremely unrealistic, and alleged answers are puzzling.

It seems you already have $n = 30,$ sample mean $$\bar X = \frac 1 n\sum_i X_i = 243/30 = 8.1$$ and sample standard deviation $$S = \sqrt{\frac{1}{n}\left[\sum_i X^2 - (\sum_i X)^2/n\right]} = 3.41,$$ although the usual formula would have $\frac{1}{n-1}$ instead of $\frac 1 n$ as the first factor under the square-root sign.

If (daily average) wind speeds $W$ are normal, then the expected number of days out of 30 with wind speeds above $8.1 + 3.41 = 11.51$ is about 4.8, according to the following computation in R.

(1 - pnorm(11.51, 8.1, 3.41)) * 30
[1] 4.759658

Under the highly dubious assumption that wind speeds are uniform, I suppose you are expected to use $\mathsf{Unif}(4, 17)$ as the model, then find $P(8.1+3.41 < W < 17) = \frac{5.49}{17-4} = 0.4223$ as an estimate of the proportion of days with windspeeds in $(11.82,17).$ And finally, the estimate the number of days out of 30 with $W$ in that interval is between 12 and 13 days, which someone has rounded down to 12. [If the usual estimate of $\sigma$ is used, a slightly smaller number results; it still rounds down to 12 results.]

A major difficulty is that the distribution $\mathsf{Unif}(4,17)$ has $\mu =(4 + 17)/2 = 10.5$ and $\sigma =\sqrt{(17-4)^2/12}=3.753.$ If we are going to use a uniform distribution to model wind speed [as seems to be suggested in the answer to (c)], then a minimum of 4 in 30 days and a maximum of 17 in 30 days very nearly estimates these endpoints for the support. Using the $\mu$ and $\sigma$ for this uniform distribution leads to a quite different answer.

Technically speaking, if the model is uniform then (a) asks you to find 'method-of-moments estimators' of $\mu$ and $\sigma.$ Providing the max and min over 30 days invites you to use 'maximum likelihood estimation' to find support interval endpoints and to derive $\mu$ and $\sigma$ from them.

If the distribution of $W$ is taken to be normal, then it is unlikely (probability below 2.5%) to observe the min and max values reported, as shown by the following simulation in R.

set.seed(1234); mn = replicate(10^6, min(rnorm(30, 8.1, 3.41)))
set.seed(1234); mx = replicate(10^6, max(rnorm(30, 8.1, 3.41)))
mean(mn >= 4 & mx <= 17)
[1] 0.022274