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    Can someone please help me answer it: Data x1, x2 ...xn is coded by y=100x-12 given that SUM y=247 and n=40 find the mean. It is only for three marks question . I am stuck on it.
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    (Original post by Mar2000)
    Can someone please help me answer it: Data x1, x2 ...xn is coded by y=100x-12 given that SUM y=247 and n=40 find the mean. It is only for three marks question . I am stuck on it.
    Work out the mean of the coded values, i.e. the mean of y. (you have the sum of y values and the number of values)
    Then to find the mean of the original x values use the formula for y= to reverse the coding.

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    Thank you very much. I have worked out: y= 100 multiplied by 247- 12 which is 24688/ 40= 617.2 or sum of y=247 divided by n=40 which is 6.175. y=100 times 6.175 -12=605.5? Could you compare it with your answer. I am not 100 percent sure what I have to do.
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    (Original post by Mar2000)
    Thank you very much. I have worked out: y= 100 multiplied by 247- 12 which is 24688/ 40= 617.2 or sum of y=247 divided by n=40 which is 6.175. y=100 times 6.175 -12=605.5? Could you compare it with your answer. I am not 100 percent sure what I have to do.
    The mean of y = 247/40

    When you reverse the coding you are trying to find what value of x gives that value for y.

    6.175=100x-12

    Use your algebra skills to solve for x


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    The mean of y=247/40 y=6.175 6.175= 100x -12 x=0.18? Thank you. Are you doing A-level maths?
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    I have one more question- last one which I am stuck at. An Athletic Club staged a day of 'fun-runs'. In the morning 30 athlets took part. Their times are designated M. Iin the afternoon 20 athletes took part. Their times are designated A. You are given the following summary information: SUM M squared=130680 SUM M=1920 SUM A squared= 106567 SUM A= 1341. Calculate: A) The mean times of each set of runners B) the mean time of all 50 runners and C) The variance of the times of all 50 runners. This is Statistics 1- A-level. Any solutions?
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    (Original post by Mar2000)
    I have one more question- last one which I am stuck at. An Athletic Club staged a day of 'fun-runs'. In the morning 30 athlets took part. Their times are designated M. Iin the afternoon 20 athletes took part. Their times are designated A. You are given the following summary information: SUM M squared=130680 SUM M=1920 SUM A squared= 106567 SUM A= 1341. Calculate: A) The mean times of each set of runners B) the mean time of all 50 runners and C) The variance of the times of all 50 runners. This is Statistics 1- A-level. Any solutions?
    Do you understand what it means when it says "the sum of" ?
    It's the total of all of the individual values added together
    When you work out a mean you find the sum of all the values and divide by the number of values.

    So the mean of the morning runners is the sum of M divided by 30

    The mean of all of the runners would be (the sum of all the morning times plus the sum of the afternoon times) divided by (The number of athletes running in the morning plus the number of athletes in the afternoon)

    You should know how to calculate variance, again you need the total sum of the squares from morning and afternoon and the total number of athletes taking part.


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