Normal Distribution Question

[npt]

Hi I am able to do all of it up to part d. could someone please help with the last bit? I don't understand the markscheme at all! Thanks!
7. Roastie’s Coffee is sold in packets with a stated weight of 250 g. A supermarket manager claims that the mean weight of the packets is less than the stated weight. She weighs a random sample of 90 packets from their stock and finds that their weights have a mean of 248 g and a standard deviation of 5.4 g.

(a) Using a 5% level of significance, test whether or not the manager’s claim is justified. State your hypotheses clearly.

(5)​

(c) State, with a reason, the action you would recommend the manager to take over the weight of a packet of Roastie’s Coffee.

(2)​

Roastie’s Coffee company increase the mean weight of their packets to μ g and reduce the standard deviation to 3 g. The manager takes a sample of size n from these new packets. She uses the sample mean as an estimator of μ.

(d) Find the minimum value of n such that P(| − μ | <1) > or eq. to 0.98.

 

[Romsek]

You are looking for the number of necessary samples such that the probability mass about the mean of width 2, i.e. [-1,1], is 0.98.
This means that 0.02 lies in the upper and lower tails and that 0.01 lies in the lower tail.

letting \(\displaystyle \Phi(x)\) be the CDF of the standard normal we have

\(\displaystyle \Phi\left(-\dfrac 1 \sigma_s\right) = 0.01\\~\\
\sigma_s = -\dfrac{1}{\Phi^{-1}(0.01)}\)

This is the required standard deviation of our samples.

\(\displaystyle \sigma_s = \dfrac{\sigma}{\sqrt{N}} = \dfrac{3}{\sqrt{N}}\)

\(\displaystyle N = \left(\dfrac{3}{\sigma_s}\right)^2 = 9 \left(\Phi^{-1}(0.01)\right)^2 = 48.707 \Rightarrow 49 \text{ needed samples}\)