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1. Please can someone help me with the following questions. We only started learning S1 the week before half term holidays and I was absent for a few lessons. (was ill on mon and tues and that when we have 2 lessons of stats per day) so I don't really understand the tasks ive been set as ive never done stats before . Any help will be much appreciated. Thanks in advance.

Consider the distribution X~N(16,9) to calculate the following probability:
1) P(10<X<12)

Consider the distribution X~N(85,25) to calculate the following probability:
1) P(80<X<100)
2) The value of a if P(X>a) = 0.67

Consider the distribution X~N(15,16) to calculate the following probability:
1) P(16<X<19)
2) The value of a if P(X<a) = 0.975
2. (Original post by Aty100)
x
I would suggest getting the textbook, reading through them and attempting the questions and post here with your working if you're stuck

This page seems to have a few videos on the normal distribution.
3. (Original post by SeanFM)
I would suggest getting the textbook, reading through them and attempting the questions and post here with your working if you're stuck
I don't have a textbook
4. (Original post by Aty100)
I don't have a textbook
Get a hold of one if you can, they are really useful.

See the link in the above post for some videos, or find some if that's not a useful one.
5. (Original post by SeanFM)
Get a hold of one if you can, they are really useful.

See the link in the above post for some videos, or find some if that's not a useful one.
thanks, I think we will be getting textbooks the week when we get back because they have ordered them.
6. (Original post by SeanFM)
Get a hold of one if you can, they are really useful.

See the link in the above post for some videos, or find some if that's not a useful one.
so for the first one I did

= P(10-16/3 < Z < 12-16/3)
= P(-2 < Z < -1.33)

now do I do

= 1- P(Z > 1.33) - 1 - P(Z > 2)
??
I'm stuck
7. (Original post by Aty100)
so for the first one I did

= P(10-16/3 < Z < 12-16/3)
= P(-2 < Z < -1.33)

now do I do

= 1- P(Z > 1.33) - 1 - P(Z > 2)
??
I'm stuck
Good start

So you have P(-2 < Z < -1.33), which can be written as P(Z<-1.33) - P(Z<-2). Why? You may wish to use a number line to help you.

You can split the range Z < -1.33 into two different regions: -2 < Z < -1.33 (which is what we want) and the rest is Z<-2, and combining those two regions together again gives Z < - 1.33. So if we express P(-2 < Z < -1.33) in terms of P(Z < - 1.33) then we would be counting the area 'Z < -2' but we're not interested in that, only the region -2 < Z < -1.33, so that's why it can be written as P(Z<-1.33) - P(Z<-2). That's just for your information/understanding.

Now the problem is that the tables have values P(Z<z) where z is a positive number, and you have to use properties of the normal distribution when you're dealing with things like P(Z>z) or P(Z<-z) (where z is a positive number) or even P(Z>-z).

So we've got P(Z<-1.33) - P(Z<-2) to work out. We need everything to be of the form P(Z<z), where z is a positive number.

So you use two things (one of which you've done already). P(Z<-z) = P(Z>z) by symmetry. This is because if you draw the graph it's symmetric around 0.

The other rule that you can use is that P(Z>z) = 1-P(Z<z), which you can kind of mean as (if it's not less than z then it's more than z or vice versa). Just like 'if the probability that it's going to rain is 0.2, then the probability that it won't rain is 0.8 (1-0.2)'.

So you've tried to use both of those things but you've got to keep track of how you do them.
8. (Original post by SeanFM)
Good start

So you have P(-2 < Z < -1.33), which can be written as P(Z<-1.33) - P(Z<-2). Why? You may wish to use a number line to help you.

You can split the range Z < -1.33 into two different regions: -2 < Z < -1.33 (which is what we want) and the rest is Z<-2, and combining those two regions together again gives Z < - 1.33. So if we express P(-2 < Z < -1.33) in terms of P(Z < - 1.33) then we would be counting the area 'Z < -2' but we're not interested in that, only the region -2 < Z < -1.33, so that's why it can be written as P(Z<-1.33) - P(Z<-2). That's just for your information/understanding.

Now the problem is that the tables have values P(Z<z) where z is a positive number, and you have to use properties of the normal distribution when you're dealing with things like P(Z>z) or P(Z<-z) (where z is a positive number) or even P(Z>-z).

So we've got P(Z<-1.33) - P(Z<-2) to work out. We need everything to be of the form P(Z<z), where z is a positive number.

So you use two things (one of which you've done already). P(Z<-z) = P(Z>z) by symmetry. This is because if you draw the graph it's symmetric around 0.

The other rule that you can use is that P(Z>z) = 1-P(Z<z), which you can kind of mean as (if it's not less than z then it's more than z or vice versa). Just like 'if the probability that it's going to rain is 0.2, then the probability that it won't rain is 0.8 (1-0.2)'.

So you've tried to use both of those things but you've got to keep track of how you do them.
Thanks sooo much! that explanation really helped !
9. (Original post by SeanFM)
Good start

So you have P(-2 < Z < -1.33), which can be written as P(Z<-1.33) - P(Z<-2). Why? You may wish to use a number line to help you.

You can split the range Z < -1.33 into two different regions: -2 < Z < -1.33 (which is what we want) and the rest is Z<-2, and combining those two regions together again gives Z < - 1.33. So if we express P(-2 < Z < -1.33) in terms of P(Z < - 1.33) then we would be counting the area 'Z < -2' but we're not interested in that, only the region -2 < Z < -1.33, so that's why it can be written as P(Z<-1.33) - P(Z<-2). That's just for your information/understanding.

Now the problem is that the tables have values P(Z<z) where z is a positive number, and you have to use properties of the normal distribution when you're dealing with things like P(Z>z) or P(Z<-z) (where z is a positive number) or even P(Z>-z).

So we've got P(Z<-1.33) - P(Z<-2) to work out. We need everything to be of the form P(Z<z), where z is a positive number.

So you use two things (one of which you've done already). P(Z<-z) = P(Z>z) by symmetry. This is because if you draw the graph it's symmetric around 0.

The other rule that you can use is that P(Z>z) = 1-P(Z<z), which you can kind of mean as (if it's not less than z then it's more than z or vice versa). Just like 'if the probability that it's going to rain is 0.2, then the probability that it won't rain is 0.8 (1-0.2)'.

So you've tried to use both of those things but you've got to keep track of how you do them.
Consider the distribution X~N(15,16) to calculate the following probability:
The value of a if P(X<a) = 0.975

would I start by doing : P(Z < a - 15 / 16 ) = 0.975?

of do I divide by 4 ??
10. (Original post by Aty100)
Consider the distribution X~N(15,16) to calculate the following probability:
The value of a if P(X<a) = 0.975

would I start by doing : P(Z < a - 15 / 16 ) = 0.975?

of do I divide by 4 ??
Swap the 16 for 4 and you are correct remember that the second number is variance, and the square root of that is the standard deviation, so you could write it as N(15,(4^2)) just to remember that the standard deviation is 4.

What next?
11. (Original post by SeanFM)
Swap the 16 for 4 and you are correct remember that the second number is variance, and the square root of that is the standard deviation, so you could write it as N(15,(4^2)) just to remember that the standard deviation is 4.

What next?
the do u just do

a - 15 / 4 = 0.975
so
a - 15 = 3.9
a = 18.9 ??

sorry ive never done a question like this before
12. (Original post by Aty100)
the do u just do

a - 15 / 4 = 0.975
so
a - 15 = 3.9
a = 18.9 ??

sorry ive never done a question like this before
Remember when we're working in the opposite direction, we start with an x value, turn it into a z value (so go from P(X<x) to P(Z<z), and find the probability that matches up with that z value.

This time (in the opposite direction) we start with the probability, so what would we need to find out?
13. (Original post by SeanFM)
Remember when we're working in the opposite direction, we start with an x value, turn it into a z value (so go from P(X<x) to P(Z<z), and find the probability that matches up with that z value.

This time (in the opposite direction) we start with the probability, so what would we need to find out?
P( Z > a ) = 0.975
Z = ??

I have no idea what I'm doing
14. (Original post by Aty100)
P( Z > a ) = 0.975
Z = ??

I have no idea what I'm doing
So we've got that P(Z < (a - 15) / 4 ) = 0.975

Usually when we know what a is and we want to find out what the probability is (in this case 0.975 would be our final answer), so we take a, turn it into a z value using z = (a-mean)/standard deviation, then since we have P(Z<z) we go to the tables and find our answer, which would be 0.975 in this example.

But like I said last time we are working backwards... so we have the probability and we need to find out what a is. It is 'reversing' the steps, so what do you think we should do?
15. (Original post by SeanFM)
So we've got that P(Z < (a - 15) / 4 ) = 0.975

Usually when we know what a is and we want to find out what the probability is (in this case 0.975 would be our final answer), so we take a, turn it into a z value using z = (a-mean)/standard deviation, then since we have P(Z<z) we go to the tables and find our answer, which would be 0.975 in this example.

But like I said last time we are working backwards... so we have the probability and we need to find out what a is. It is 'reversing' the steps, so what do you think we should do?
would we find 0.975 in the table to find z which is 1.96
16. (Original post by Aty100)
would we find 0.975 in the table to find z which is 1.96
Correct so what does that z equal to?
17. (Original post by SeanFM)
Correct so what does that z equal to?
1.96.

So how to I work out a?
18. (Original post by Aty100)
1.96.

So how to I work out a?
How will z=1.96 be related to a?

Think about how you calculate a given z and work backwards.
19. (Original post by SeanFM)
How will z=1.96 be related to a?

Think about how you calculate a given z and work backwards.
so
P(1.96 < a) = 0.975 ?!
20. (Original post by Aty100)
so
P(1.96 < a) = 0.975 ?!
P(Z<1.96) = 0.975.

Remember that with an a value when we want to find P(X<a) we do (a-mean)/standard deviation to turn a into z, and find P(Z<z).

If a was 18 for example and we wanted to find P(X<18) then we would do P(Z<(18-15)/4) = P(Z<3/4).

What I was trying to hint at was that each value for a has a value for z that we calcuate, and in that example (a-15)/4 = 3/4, and in your question it's ..... = 1.96.

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