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 After, I made two formulas: (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.