Normal Distribution help??
Question taken from S1 jan 2008 AQA.
Q)1)b) Sawmill requires batch of logs such that there is a probability of 0.025 that any given log will have a length less than 3.1. determine the value of mu/mean. We also know that the SD is 0.16.
When i look up the Z value for 0.025 i get 1.96.
However on the mark scheme when its subbed into the equation it gives z = 3.1-μ/0.16 = -1.96 i dont understand why it becomes negative? does anyone have an answer to why??
Q)1)b) Sawmill requires batch of logs such that there is a probability of 0.025 that any given log will have a length less than 3.1. determine the value of mu/mean. We also know that the SD is 0.16.
When i look up the Z value for 0.025 i get 1.96.
However on the mark scheme when its subbed into the equation it gives z = 3.1-μ/0.16 = -1.96 i dont understand why it becomes negative? does anyone have an answer to why??
Original post by redmcq
Question taken from S1 jan 2008 AQA.
Q)1)b) Sawmill requires batch of logs such that there is a probability of 0.025 that any given log will have a length less than 3.1. determine the value of mu/mean. We also know that the SD is 0.16.
When i look up the Z value for 0.025 i get 1.96.
However on the mark scheme when its subbed into the equation it gives z = 3.1-μ/0.16 = -1.96 i dont understand why it becomes negative? does anyone have an answer to why??
Q)1)b) Sawmill requires batch of logs such that there is a probability of 0.025 that any given log will have a length less than 3.1. determine the value of mu/mean. We also know that the SD is 0.16.
When i look up the Z value for 0.025 i get 1.96.
However on the mark scheme when its subbed into the equation it gives z = 3.1-μ/0.16 = -1.96 i dont understand why it becomes negative? does anyone have an answer to why??
Presumably you're looking at the percentage points table which clearly gives .
What you want here is , in which case, you need to get this to:
which lets you state that right away.
Watch your inequality signs.
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