# Deduce the variance of a normal distribution from its $2\%$ and $95\%$ quantiles

This is a Cambridge A Level Question that I am currently trying to solve:

Metal rods produced by a machine have lengths that are normally distributed. It is known that 2% of the rods are rejected as being too short and 5% are rejected as being too long. Given that the least and greatest acceptable lengths of the rods are 6.32cm and 7.52 cm, calculate the variance of the lengths of the rods.

So far this is what I did:

I found that P(X < 7.52) = 1 - 0.05
= 0.9500
Using the Normal Distribution Table(NDT), I found that the z value in P(Z < z) = 0.9500
is 1.645

Then what I did was found that P(X < 6.32) = 0.0200
P(Z < -z) = 0.9800
z = -2.054

(7.52 - Mean)/(Standard Deviation) = 1.645
(-6.32 - Mean)/(Standard Deviation) = 3.699

Then, I simultaneously calculated the Standard Deviation which gave me 3.7416, and then
I squared it giving me a variance of 14.00


Can someone please tell me what I am doing wrong? Thanks in advance.

The solution of the system \begin{align*} \frac{7.52 - m}{s} &= 1.645 \\ \frac{6.32 - m}{s} &= -2.054 \\ \end{align*} is $s = 0.3244$ and $m = 6.9863$. Giving a variance of $0.1052$.
(Since we accept more than 90% of the parts, having about $4$ standard deviations between the acceptance cutoffs seems a bit more reasonable than having less than $1/3$, yes? Also, since small fractions of parts are rejected in each direction, having the mean of the distribution very near the mean $6.92$ of the two cutoffs is encouraging.)