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# Normal Distribution help

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1. 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??
2. (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??
If the probability of it being less than 3.1 is 0.025, then you know that the area to the left is going to be very small, so you know that on the normal distribution curve it's going to be to the far left (you also know that 3.1 is going to be less than the mean, where the probability of something being less than the mean of a normal distribution is 0.5).

Ignore the writing, just look at the graph (the shaded area of 0.025 will be in the same area, just smaller and further to the left) and you see that the z value must be negative, whereas +1.96 would correspond to the opposite end where P(Z<z) would be 0.975.
3. (Original post by SeanFM)
If the probability of it being less than 3.1 is 0.025, then you know that the area to the left is going to be very small, so you know that on the normal distribution curve it's going to be to the far left (you also know that 3.1 is going to be less than the mean, where the probability of something being less than the mean of a normal distribution is 0.5).

Ignore the writing, just look at the graph (the shaded area of 0.025 will be in the same area, just smaller and further to the left) and you see that the z value must be negative, whereas +1.96 would correspond to the opposite end where P(Z<z) would be 0.975.
Thank you very much!
4. (Original post by redmcq)
Thank you very much!
No worries, do you see it now? I am happy to explain anything you might not be sure about.
5. (Original post by SeanFM)
No worries, do you see it now? I am happy to explain anything you might not be sure about.
That helped thanks but i am a little confused about another question i came across which is similar. Its from jan 2009.

Q) Assuming that the value of the SD remains unchanged, determine the mean length necessary to ensure that only 1 percent of boards have lengths less than 5 metres. (where the standard deviation is 0.05)

I understand that(x<5)=0.01
therefore P(x>5)=0.99 so P(z>5)=2.3263
But, i dont understand why when working out mean : z=5-Mu/0.05= -2.3263
I dont understand why its negative , why does it become -2.3263 when using the equation ?
6. (Original post by redmcq)
That helped thanks but i am a little confused about another question i came across which is similar. Its from jan 2009.

Q) Assuming that the value of the SD remains unchanged, determine the mean length necessary to ensure that only 1 percent of boards have lengths less than 5 metres. (where the standard deviation is 0.05)

I understand that(x<5)=0.01
therefore P(x>5)=0.99 so P(z>5)=2.3263
But, i dont understand why when working out mean : z=5-Mu/0.05= -2.3263
I dont understand why its negative , why does it become -2.3263 when using the equation ?
Again, it's because the probability of it being less than 5 is 0.01, so it's going to be to the left of the mean, and for the left of the mean you need to use negative z values. Eg P(Z<1.96) = 0.95 but P(Z<-1.96 = 0.05), so for anything where 99% of the data is to the RIGHT of it it's going to be to the left of the mean, so the z value must be negative.
7. (Original post by redmcq)
That helped thanks but i am a little confused about another question i came across which is similar. Its from jan 2009.

Q) Assuming that the value of the SD remains unchanged, determine the mean length necessary to ensure that only 1 percent of boards have lengths less than 5 metres. (where the standard deviation is 0.05)

I understand that(x<5)=0.01
therefore P(x>5)=0.99 so P(z>5)=2.3263
But, i dont understand why when working out mean : z=5-Mu/0.05= -2.3263
I dont understand why its negative , why does it become -2.3263 when using the equation ?
You should always draw a diagram similar to the one above when doing a normal distribution question. If the center represents half (as its symmetrical), any value less than a half, say P(Z<z)=0.4, z is going to be a negative value as on the standard curve the mean is 0, hence the center of the curve is at 0. If the value was P(Z<z)=0.6 it would be a positive value, as you would have passed that 0 mark on the curve, although because it is symmetrical, the value of z would be the same for both scenarios, just the sign would change based on which side of the curve the probability lies on.
It's a strange thing but if you understand the diagram you should be fine.
8. Thanks guys!
9. (Original post by SeanFM)
No worries, do you see it now? I am happy to explain anything you might not be sure about.
I have another question.... when i work out a z value, what shall in the formula table if it goes to 3 decimal places..?
E.g. i got P(z<1.125) , however in the booklet it only goes to 2 d.p. --> so i have option of either 1.12 or 1.13.
What am i supposed to do? do i round or what?
10. (Original post by redmcq)
I have another question.... when i work out a z value, what shall in the formula table if it goes to 3 decimal places..?
E.g. i got P(z<1.125) , however in the booklet it only goes to 2 d.p. --> so i have option of either 1.12 or 1.13.
What am i supposed to do? do i round or what?
Round to the nearest 2 d.p.

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