Help stats maths

the 1.96 represents the z values (0 mean, unit variance) for the 95% test, so 2.5% in each tail.
Just plug those values into your calculator so N(0,1) and the x values (inv cumulative prob) for the cumulative percentages 0.025 and 0.975.

the calculation just uses the z values +/-1.96 and applies them to the sample mean distribution which is N(123, 70^2/12144), so you get the values corresponding to the 2.5% tails.

Which part? You should have done 95% confidence interval tests for sample means? Can you say what you do understand about the question.

So you understand the mean/variance associated with the sample mean?
If so, can you use the inverse cumulative nomral on your calculator to get the x values for the 0.025 and 0.975 tails?
This is a reaonably common hypothesis test, so you must have done something similar to the above on other questions?

Theyve just broken the calculation down into two steps. Get the 0.025 and 0.975 tails for a z~N(0,1) which are +/-1.96, then rescale that back to the sample mean distribution, so mulitply by the standard deviation and add the mean.

Which part of the above are you asking about.

When you say youve done similar questions, what do you understand - the normal sample mean, how to get the 0.025 and 0.975 tails, ...? Are you talking about the
x = mu + 1.96*sigma
formula? If so, its just using the definition of z ~ N(0,1) which is defined by
z = (x - mu) / sigma
where the value of z (+/-1.96) corresponds to the 0.025 and 0.975 tails. Rearranging gives
x = mu + z*sigma
From a quick google
https://www.youtube.com/watch?v=zJ8e_wAWUzE&ab_channel=TheOrganicChemistryTutor
has a reasonable overview

As in the previous post, if you can do it using the inverse cumulative on the sample mean distribution without using the z score, that would be acceptable for this question

I can’t use the equation on the aqa formula booklet… I don’t get it 😭
why do I do 0.025 and 0.975
As the tails??