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    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
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    (Original post by Aty100)
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    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.
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    (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
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    (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.
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    (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.
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    (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
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    (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 :borat:

    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.
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    (Original post by SeanFM)
    Good start :borat:

    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 !
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    (Original post by SeanFM)
    Good start :borat:

    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 ??
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    (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?
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    (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
    add 15
    a = 18.9 ??

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

    a - 15 / 4 = 0.975
    so
    a - 15 = 3.9
    add 15
    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?
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    (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
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    (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?
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    (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
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    (Original post by Aty100)
    would we find 0.975 in the table to find z which is 1.96
    Correct :borat: so what does that z equal to?
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    (Original post by SeanFM)
    Correct :borat: so what does that z equal to?
    1.96.

    So how to I work out a?
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    (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.
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    (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 ?!
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    (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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