Edexcel S1 June 2005 Q6 Watch

CalculusMan
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#1
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A scientist found that the time taken, M minutes, to carry out an experiment can be modelled by a normal random variable with mean 155 minutes and standard deviation 3.5 minutes.

a) Find P(M > 160), done and got 0.0764
b) Find P(150 <= M <= 157), done and got 0.6369
c) the value of m, to 1 dp, such that P(M <= m) = 0.30

What I've done so far:

Converted to Z, P(Z <= (m - 155)/3.5) = 0.3
I can't find 0.3 in the table, have I forgotten symmetry? Can anyone help me out?

In the MS, it has:

(m-155)/3.5 = -0.5244

Where has -0.5244 come from?!?

Thanks,
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Jean
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from pg21 of the formulae booklet..
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__WiZaRD__
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which paper is this from?
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Geerththanan
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Its...if i remember right...

(m - 155) / 3.5 = z value of (1 - 0.30)
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Geerththanan
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(Original post by __WiZaRD__)
which paper is this from?
Thread title...
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CalculusMan
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The z value of 0.7 is between 0.52 and 0.53, so that would make sense, but why is it negative?
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Geerththanan
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Coz u cant get "z" values for under 0.5 probability...and when it is P(< X) and the value is less than 0.5, in this case 0.3...its changed to negative..
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Ramin Gorji
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for part C
you use the tables which cannot be for 0.3 so you get 0.5244. Since you know that that probabilty was 0.3 you have to remember this is up to and including this therefore z must be negative.

M-155/3.5=-0.5244 M=153 trick is to know z is negative then your sorted
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Thrug
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I know this thread is old, but still:

We get told to just reflect it to the positive side then deal with it as normal.

So the M<m = 0.3 to the left hand side of the normal distribution graph now gets changed to m>M = 0.3 on the right hand side.

Then proceed as normal, with the formula now being (155 - m) / 3.5 = 0.5244
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