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Given that X ~ N (15,4), find w when P(│X – 15│ < w) = 0.9 correct to 2 d.p.

Here's my working out:

https://imgur.com/a/8yZA5

The answer is 3.29, what did I do wrong?
Reply 1
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Notnek
21
Original post by Mathman69
Here's my working out:

https://imgur.com/a/8yZA5

The answer is 3.29, what did I do wrong?

Firstly I'm assuming there's a typo in your first line and it's meant to be -w instead of -2?

You need to get the probability into the form P(a < X < b) before you can normalise.

So P(-w<X-15<w) becomes P(-w + 15 < X < w + 15). Does this make sense?

Try carrying on from here and post all your working if you get stuck or get the wrong answer.
Reply 2
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Mathman69
OP
3
Original post by Notnek
Firstly I'm assuming there's a typo in your first line and it's meant to be -w instead of -2?

You need to get the probability into the form P(a < X < b) before you can normalise.

So P(-w<X-15<w) becomes P(-w + 15 < X < w + 15). Does this make sense?

Try carrying on from here and post all your working if you get stuck or get the wrong answer.


Here's my w/o: https://imgur.com/a/zOIU6

I'm not sure what to do from there.
Reply 3
Avatar for Notnek
Notnek
21
Original post by Mathman69
Here's my w/o: https://imgur.com/a/zOIU6

I'm not sure what to do from there.

P(-w/2 < Z < w/2)

This region is clearly symmetrical about Z = 0. Try drawing a bell curve and think about the regions to the left and right of the required region. Please post your thoughts if you get stuck.
Reply 4
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Mathman69
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3
Original post by Notnek
P(-w/2 < Z < w/2)

This region is clearly symmetrical about Z = 0. Try drawing a bell curve and think about the regions to the left and right of the required region. Please post your thoughts if you get stuck.


Here's what I done: https://imgur.com/a/lKQ9m

Still don't know how to find the value of w....
Reply 5
Avatar for Notnek
Notnek
21
Original post by Mathman69
Here's what I done: https://imgur.com/a/lKQ9m

Still don't know how to find the value of w....

Since the regions on either side must have equal area due to symmetry, the probability of the region on the left must be (1−0.9)÷2=0.05(1-0.9) \div 2 = 0.05.

So P(Z<−w/2)=0.05P(Z<-w/2) = 0.05. Does this make sense and can you carry on from here?
Reply 6
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Mathman69
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3
Original post by Notnek
Since the regions on either side must have equal area due to symmetry, the probability of the region on the left must be (1−0.9)÷2=0.05(1-0.9) \div 2 = 0.05.

So P(Z<−w/2)=0.05P(Z<-w/2) = 0.05. Does this make sense and can you carry on from here?

Yes that makes sense, thanks!
Reply 7
Avatar for Notnek
Notnek
21
Original post by Mathman69
Yes that makes sense, thanks!

This is a common technique that you'll need to get used to. Always draw a quick sketch and think about symmetry.

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