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# Edexcel S1 June 2015 - Question 6b watch

1. The random variable X has a normal distribution with mean 21 and standard deviation 5 (b) Find the value of w such that P(X > w | X > 28) = 0.625

So far I have done the following:

X ~ N (21 , 5^2)

P(X > w | X > 28) = P(X > w ∩ X > 28) / P (X > 28)

Z = (X - 21)/5
Z = 1.4

P(Z>1.4) = 1 - P(Z<1.4)
= 0.0808

P(X > w ∩ X > 28) / 0.080 = 0.625
P(X > w ∩ X > 28) = 0.0505

Where do I go from this point?
2. Nice work with the first part of the problem!

The way I would think about the last part is this:
Since P(X > w | X > 28) = 0.625, w must be greater than 28. (If w<28 then P(X > w | X > 28) would be 1).

Since w > 28, the event {X>w AND X>28} is included in the event {X>w>28}.

Hence P(X > w ∩ X > 28) = P(X > w).
So now you simply need to solve P(X > w) = 0.0505 using the standard normal distribution like you did before
3. (Original post by square_peg)
Nice work with the first part of the problem!

The way I would think about the last part is this:
Since P(X > w | X > 28) = 0.625, w must be greater than 28. (If w<28 then P(X > w | X > 28) would be 1).

Since w > 28, the event {X>w AND X>28} is included in the event {X>w>28}.

Hence P(X > w ∩ X > 28) = P(X > w).
So now you simply need to solve P(X > w) = 0.0505 using the standard normal distribution like you did before

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