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# S1 Normal Distribution Question

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1. Plastic lawn edging is supplied in 9m length rolls. The actual length, Y metres of a roll may be modelled by a normal distribution with mean μ and standard deviation σ.An analysis of a batch of rolls selected at random showed that P(Y < 9.25) = 0.88

1) Use this probability to find the value of z such that 9.25 - μ = z x σ where z is a value of Z ~ N(0, 1)

2) Given also that P(Y > 8.75) = 0.975 Find values for μ and σ.

Not really sure how to approach this question as the values in the formula book for z don't go beyond 3.99, and I can't seem to figure out any other way of tackling it?
2. a) Z-transform Z = (Y-mu)/sigma
P(Y<(9.25-mu)/sigma)) = 0.88
From the tables we interpolate between 1.17 and 1.18 to 1.175.
so 9.25 - mu = 1.175 (1)

b) again P(Y>(8.75-mu)/sigma)) = 0.975
or P(Y<(1-8.75-mu)/sigma)) = 0.025
from tables
8.75-mu/sigma = -1.96 (2)

solving simultaneousy (1) and (2)
sigma = 0.159
mu = 9.06
3. (Original post by MathsSir)
a) Z-transform Z = (Y-mu)/sigma
P(Y<(9.25-mu)/sigma)) = 0.88
From the tables we interpolate between 1.17 and 1.18 to 1.175.
so 9.25 - mu = 1.175 (1)

b) again P(Y>(8.75-mu)/sigma)) = 0.975
or P(Y<(1-8.75-mu)/sigma)) = 0.025
from tables
8.75-mu/sigma = -1.96 (2)

solving simultaneousy (1) and (2)
sigma = 0.159
mu = 9.06
EDIT: I understand it now, thanks. +Repped.
4. (Original post by AmarRPM)
Not sure how you got to 9.25 - mu = 1.175
Where does the 1.175 come from?

So,

By using the table and finding where so
5. (Original post by Cryptokyo)

So,

By using the table and finding where so
I understand it now, but thanks for helping me also. +1

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