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URGENT Edexcel S1 - Coding?! Watch

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    Can someone explain, in detail, how I work this question out?

    Find the mean of the following set of data (x) using the coding y = x+10/100

    120 165 98 215 145 103 68 240
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    (Original post by nousernametho)
    Can someone explain, in detail, how I work this question out?

    Find the mean of the following set of data (x) using the coding y = x+10/100

    120 165 98 215 145 103 68 240
    What is the urgency - test tomorrow?

    You can do it manually if you wish BUT...

    If x was your set of values with mean 'a' and you divided each term by 100, what would the new mean be in terms of a? This will not get you the complete answer but will send you in the right direction and hopefully boost your understanding.
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    don't trust this 100% because we covered this for like 5 minutes in stats and then moved on so... sorry

    linear coding is basically a way of using 'nicer' numbers when working out things such as mean and standard deviation.

    To work out the mean, first add 10 and divide by 100 to each term as that is what the 'coding rule' is for this question. With the coded values of x, proceed to find the mean as normal, and then at the end reverse the coding for the mean.
    Eg. If you got that your answer for the mean with coding was 4, then the mean without the coding is 4*100 - 10 = 390.
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    Also, for standard deviation, you would multiply your final result by 100 but wouldn't add 10 because standard deviation is just the spread of the data, so adding 10 wouldn't change this spread.
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    (Original post by SeanFM)
    What is the urgency - test tomorrow?

    You can do it manually if you wish BUT...

    If x was your set of values with mean 'a' and you divided each term by 100, what would the new mean be in terms of a? This will not get you the complete answer but will send you in the right direction and hopefully boost your understanding.
    Haha, nah it's homework due in for tomorrow, but my teacher said: "You will need to do independent research on coding as I'm not telling you how to do it, so don't come asking me. Google it." ~ I Googled it and the majority of it didn't make sense...

    I've done it now, thank you for your help! Let's just hope I've got it right.
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    (Original post by surina16)
    don't trust this 100% because we covered this for like 5 minutes in stats and then moved on so... sorry

    linear coding is basically a way of using 'nicer' numbers when working out things such as mean and standard deviation.

    To work out the mean, first add 10 and divide by 100 to each term as that is what the 'coding rule' is for this question. With the coded values of x, proceed to find the mean as normal, and then at the end reverse the coding for the mean.
    Eg. If you got that your answer for the mean with coding was 4, then the mean without the coding is 4*100 - 10 = 390.
    Spoiler:
    Show
    Also, for standard deviation, you would multiply your final result by 100 but wouldn't add 10 because standard deviation is just the spread of the data, so adding 10 wouldn't change this spread.
    Thank you for the help!
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    (Original post by nousernametho)
    Haha, nah it's homework due in for tomorrow, but my teacher said: "You will need to do independent research on coding as I'm not telling you how to do it, so don't come asking me. Google it." ~ I Googled it and the majority of it didn't make sense...

    I've done it now, thank you for your help! Let's just hope I've got it right.
    If you have the Edexcel textbook it should be in there if you don't have the textbook.. then your school is silly.

    What did you end up with? / how did you do it? Happy to check over your work if you post it here :borat:
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    Here's what I did - thank you for helping!

    (Original post by SeanFM)
    If you have the Edexcel textbook it should be in there if you don't have the textbook.. then your school is silly.

    What did you end up with? / how did you do it? Happy to check over your work if you post it here :borat:
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    (Original post by nousernametho)
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    Here's what I did - thank you for helping!
    That may get you the right answer but it's not the intended method.

    Let's say I had the set x, which contains 5,10,15,20,25. (The mean is 15, as you can see, the same as the median but that is a coincidence for this example).

    If I had the coding y= x /5 , what is the new mean? And what do you notice about the old mean compared to the new mean, compared to the coding?
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    (Original post by SeanFM)
    That may get you the write answer but it's not the intended method.

    Let's say I had the set 5,10,15,20,25. (The mean is 15, as you can see, the same as the median but that is a coincidence for this example).

    If I had the coding y= x /5 , what is the new mean? And what do you notice about the old mean compared to the new mean, compared to the coding?
    Wait, so do I not make, in this example, 5, 10, 15, 20 and 25 the x values and then sub. that into the equation to calculate the mean?
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    (Original post by nousernametho)
    Wait, so do I not make, in this example, 5, 10, 15, 20 and 25 the x values and then sub. that into the equation to calculate the mean?
    Yes, that is what I want you to do for the purpose of this question and then see the link between the mean of x, the mean of y and the coding (y = x/5)

    (the y values are 1,2,3,4,5 to save you time)
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    (Original post by SeanFM)
    Yes, that is what I want you to do for the purpose of this question and then see the link between the mean of x, the mean of y and the coding (y = x/5)

    (the y values are 1,2,3,4,5 to save you time)
    So the new mean would be 3 (because 1+2+3+4+5=15, and then 15/5 = 3).

    Comparing that mean to the original mean: it's smaller. Or have I gone wrong somewhere?
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    (Original post by nousernametho)
    So the new mean would be 3 (because 1+2+3+4+5=15, and then 15/5 = 3).

    Comparing that mean to the original mean: it's smaller. Or have I gone wrong somewhere?
    Not quite what I mean by comparing, but I appreciate that the question is ambiguous unless you know what you're looking for.

    The old mean of 5,10,15,20,25 is 15 and the new mean when you apply y = x/5, is 3. It is no coincidence that 15/5 = 3...

    This is because if you divide each term by a constant, then the mean will also be divided by that constant.

    See that the mean of (5/5, 10/5, 15/5, 20/5, 25/5) is 1/5 * the mean of (5, 10, 15, 20, 25).. and in general, if you are multiplying a set x by a (in this case a is a fraction, 1/5) then the new mean is a * the mean of x.You can see this now as mean(x) = 15, a = 1/5 and mean(y) = 3 (the mean of x and y you calculated...) and so you know that this formula works.

    With me so far? this is half of the challenge of working out the new mean of (x+10)/100, but it's not as simple as mean(x) / 100 because of the +10. But it's important for you to understand what happens when you have y = ax first.
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    (Original post by SeanFM)
    Not quite what I mean by comparing, but I appreciate that the question is ambiguous unless you know what you're looking for.

    The old mean of 5,10,15,20,25 is 15 and the new mean when you apply y = x/5, is 3. It is no coincidence that 15/5 = 3...

    This is because if you divide each term by a constant, then the mean will also be divided by that constant.

    See that the mean of (5/5, 10/5, 15/5, 20/5, 25/5) is 1/5 * the mean of (5, 10, 15, 20, 25).. and in general, if you are multiplying a set x by a (in this case a is a fraction, 1/5) then the new mean is a * the mean of x.You can see this now as mean(x) = 15, a = 1/5 and mean(y) = 3 (the mean of x and y you calculated...) and so you know that this formula works.

    With me so far? this is half of the challenge of working out the new mean of (x+10)/100, but it's not as simple as mean(x) / 100 because of the +10. But it's important for you to understand what happens when you have y = ax first.
    Oh right, I think I see where you're coming from now, thank you so much for helping me!
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    (Original post by nousernametho)
    Oh right, I think I see where you're coming from now, thank you so much for helping me!
    Right so if I had a set with mean 18 and coding y = x/6, what is the new mean?

    And if you had a set with mean x, and added 5 to each term (i.e y = x +5) what is the new mean then?
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    (Original post by SeanFM)
    Right so if I had a set with mean 18 and coding y = x/6, what is the new mean?

    And if you had a set with mean x, and added 5 to each term (i.e y = x +5) what is the new mean then?
    So, the old mean is 18, and the new mean when you apply y=x/6 is y=18/6 which is 3. So the new mean is 3? Right so far?

    I'm a bit confused with the bolded part, assuming that my first answer is right? Do I substitute x for numerical values (i.e. 1,2,3,4,5...?)
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    (Original post by nousernametho)
    So, the old mean is 18, and the new mean when you apply y=x/6 is y=18/6 which is 3. So the new mean is 3? Right so far?

    I'm a bit confused with the bolded part, assuming that my first answer is right? Do I substitute x for numerical values (i.e. 1,2,3,4,5...?)
    Correct, the new mean is 3 you can do that without knowing the actual points themselves, same with the +5.

    Sorry, I should have been more clear - the second part is a new set, can be anything.

    If you had a set x containing 2,4,6,8,10 and added 5 to each set (i.e y = x +5) then what is the new mean, and so what is the link between the old mean, the new mean and the coding? (Same question as before so now you know what kind of answer you need)
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    (Original post by SeanFM)
    Correct, the new mean is 3 you can do that without knowing the actual points themselves, same with the +5.

    Sorry, I should have been more clear - the second part is a new set, can be anything.

    If you had a set x containing 2,4,6,8,10 and added 5 to each set (i.e y = x +5) then what is the new mean, and so what is the link between the old mean, the new mean and the coding? (Same question as before so now you know what kind of answer you need)
    Okay, so if the set contains 2,4,6,8,10 then this makes the new set, when applying y=x+5: 7,9,11,13,15. If I calculate a mean from here I get: 55/5 = 11. Is that bit right?

    I'm really unsure about the link though.
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    (Original post by nousernametho)
    Okay, so if the set contains 2,4,6,8,10 then this makes the new set, when applying y=x+5: 7,9,11,13,15. If I calculate a mean from here I get: 55/5 = 11. Is that bit right?

    I'm really unsure about the link though.
    Correct

    Okay, no worries.

    Remember last time the coding was y = x/5, the old mean was 15 and the new mean was 3.. and the link was what x is being multiplied by (here it was multiplied by 1/5). When the set x is multiplied by a number a, the mean is multiplied by that number a as well.

    Now you have the old mean being 6 and the new mean being 11.. and the code is y = x + 5.. so what is the link?
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    (Original post by SeanFM)
    Correct

    Okay, no worries.

    Remember last time the coding was y = x/5, the old mean was 15 and the new mean was 3.. and the link was what x is being multiplied by (here it was multiplied by 1/5). When the set x is multiplied by a number a, the mean is multiplied by that number a as well.

    Now you have the old mean being 6 and the new mean being 11.. and the code is y = x + 5.. so what is the link?
    (Sorry for the late reply)

    Okay, so is the link something along the lines of, that, as x increases, the mean also increases?
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    (Original post by nousernametho)
    (Sorry for the late reply)

    Okay, so is the link something along the lines of, that, as x increases, the mean also increases?
    Yes, but how?

    You might be overcomplicating this (though admittedly I could explain it better).. if you look back to the example where the mean was 9, you added 5 to everything and the new mean was 14...
    (y=x+5).

    What I am trying to show you here is that mean(y) = mean(x) + 5, and more generally, if y = x +a where a can be any positive or negative value, then mean(y) = mean (x) + a. Can you see the link now? It is similar to the link with the mean, just a different calculation (i.e before it was multiplying the set x, now you're adding..)
 
 
 
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