Estimation, confidence intervals and tests - S3 Edexcel

For this question, it says the mean of the individual sales of leaded petrol during this time is £8.72.

The question asks for an interval within which 90% of the sales of leaded petrol will lie.

The only problem is there is no n value, so do we just get rid of n from the formula and use it like they've done in the solution? Can we even do that?

Another question:

In S3, where we calculate confidence intervals for a population parameter (from a normal distribution).
So, we are estimating μ, we calculate a confidence interval based on z from the normal distribution.

Is it because the normal distribution and the formula, the confidence interval will always be symmetrical about the mean?

μ - lower confidence limit = upper confidence limit - μ ?

You are leaping too quickly into trying to use the formula for the distribution of a sample mean. For the leaded petrol, you are given the population mean and standard deviation. You therefore simply need to calculate mean+1.6449 x 3.25 and mean-1.6449 x 3.25to get the interval they want.

Part (d), concerning the unleaded petrol will require you to use the sample formulae.

For a unimodal symmetric distribution, it is conventional to have symmetrical confidence intervals. Certainly at S3 for anything involving the normal distribution!

Yeah, I didn't read the question properly. I was worked backwards lol. You are too good, thank you. I didn't get the sample formulae part.
The only formula I've been taught is the confidence interval (is that what you're referring to?)



P.S. You're too good. I think I get it now. Is it trying to say P(X < L < Y) = 0.90 where X and Y are the lower and upper limits respectively of the interval.

P (Z > $\frac{Y-8.72}{3.25}$) = 0.05

So, $\frac{Y - 8.72}{3.25}$ = 1.6449
Y = 8.72 + 3.25 * 1.6649

As, it is symmetrical: X = 8.72 - 3.25 * 1.6649

Can someone confirm if I've done it right here? @Gregorius @Notnek @ghostwalker

Looks fine, apart from the fact that $\Phi^{-1}(0.95)=1.6449$, not 1.6649