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# S1 help please ! Normal distribution! watch

1. Please can someone help me with this. I've asked some friends and they worked it out and all got different answers. We only learnt s1 for a week and I missed a few of the lessons.
I would really appreciate some guidance with this question and would appreciate any help. Thanks in advance.

The lifetime of an electrical component may be modelled by the normal distribution X with mean 2500 hours and variance 900 hours ^2

Find the value of P(X>a) = 0.8

All I've got so far is

X~N(2500,900^2)

P(X>a) = 0.8

P(Z<(a-2500)/900) = 0.8
2. (Original post by Aty100)
Please can someone help me with this. I've asked some friends and they worked it out and all got different answers. We only learnt s1 for a week and I missed a few of the lessons.
I would really appreciate some guidance with this question and would appreciate any help. Thanks in advance.

The lifetime of an electrical component may be modelled by the normal distribution X with mean 2500 hours and variance 900 hours ^2

Find the value of P(X>a) = 0.8

All I've got so far is

X~N(2500,900^2)

P(X>a) = 0.8

P(Z<(a-2500)/900) = 0.8
I suspect that the variance is 900 rather than 900^2.

Your last statement isn't correct. You should go from P(X>a) = 0.8 to P(Z> ....) = 0.8, and then apply what you know about normal distributions to solve that.
3. (Original post by SeanFM)
I suspect that the variance is 900 rather than 900^2.

Your last statement isn't correct. You should go from P(X>a) = 0.8 to P(Z> ....) = 0.8, and then apply what you know about normal distributions to solve that.
Oh sorry about that it's just because the statement said variance is 900^2 so I got a little confused
4. (Original post by SeanFM)
I suspect that the variance is 900 rather than 900^2.

Your last statement isn't correct. You should go from P(X>a) = 0.8 to P(Z> ....) = 0.8, and then apply what you know about normal distributions to solve that.
Would I do

1-P(Z>a-2500/900) = 0.8

So 1-0.8 =0.2

So

a-2500/900 = 0.2 ?

Or do I just use 0.8 ??
5. (Original post by Aty100)
Oh sorry about that it's just because the statement said variance is 900^2 so I got a little confused
If it's 900^2 then I suppose the variance is 900^2.

If it's 900 hours^2 then it's just the units that are squared.

(Original post by Aty100)
Would I do

1-P(Z>a-2500/900) = 0.8

So 1-0.8 =0.2

So

a-2500/900 = 0.2 ?
Just a note about my previous post, when you go from P(X<a) to P(Z<z) or P(X>a) to P(Z>z) the same inequality is used, you don't flip it or anything.

So you've correctly done that P(X>a) can be written as P(Z>a-2500/300), so they are both equal to 0.8.

Remember that you want it of the form P(Z<z) where z is a positive number so that you can read it off of the tables.

Once you've corrected the 0.8 bit to 0.2, you should have
1-P(Z>a-2500/300) = 0.2.

Then you can express the left hand side as something else, and then use a property of the normal distribution to find the z value that corresponds to what's there (you've already posted a question that does something like this before).

Remember that P(X<a) = 1-P(X>a) and that for the normal distribution, P(Z<z) = P(Z>-z) for a fixed z (visualise with a graph).
6. (Original post by SeanFM)
If it's 900^2 then I suppose the variance is 900^2.

If it's 900 hours^2 then it's just the units that are squared.

Just a note about my previous post, when you go from P(X<a) to P(Z<z) or P(X>a) to P(Z>z) the same inequality is used, you don't flip it or anything.

So you've correctly done that P(X>a) can be written as P(Z>a-2500/300), so they are both equal to 0.8.

Remember that you want it of the form P(Z<z) where z is a positive number so that you can read it off of the tables.

Once you've corrected the 0.8 bit to 0.2, you should have
1-P(Z>a-2500/300) = 0.2.

Then you can express the left hand side as something else, and then use a property of the normal distribution to find the z value that corresponds to what's there (you've already posted a question that does something like this before).

Remember that P(X<a) = 1-P(X>a) and that for the normal distribution, P(Z<z) = P(Z>-z) for a fixed z (visualise with a graph).
Thanks ! Shouldn't the 300 be 30?
7. (Original post by Aty100)
Thanks ! Shouldn't the 300 be 30?
Yes, sorry
8. (Original post by SeanFM)
Yes, sorry
so

a-2500 / 30 = -0.2

a-2500 / 30 = -0.8416 ?

a = 2474.75 ??
9. (Original post by Aty100)
so

a-2500 / 30 = -0.2

a-2500 / 30 = -0.8416 ?

a = 2474.75 ??
That seems to be correct, though I'm not sure what you meant by that first line.

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