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Forecasting

Forums › Ask ACCA Tutor Forums › Ask the Tutor ACCA MA – FIA FMA › Forecasting

  • This topic has 1 reply, 2 voices, and was last updated 7 years ago by John Moffat.
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  • October 21, 2015 at 12:29 pm #278166
    mujahidaslam
    Member
    • Topics: 16
    • Replies: 3
    • ☆

    Good after noon sir
    the following data represents a time series:
    x 36 y 41 34 38 42

    a series of three points moving average produced from this data has given the first two values as 38 and 39

    what are the values of x,y in the original time series?

    a)38,39 b)38,40 c)40,38 d)39,38

    sir i have seen the solution to this but still cant get through can you please explain Thank you

    October 21, 2015 at 3:59 pm #278219
    John Moffat
    Keymaster
    • Topics: 57
    • Replies: 51532
    • ☆☆☆☆☆

    X and Y are the missing figures in the list.

    Because it is a 3 point moving average, the averages are the averages of 3 items at a time.

    So……(X + 36 + Y ) / 3 = 38
    or (multiplying everything by 3)
    X + 36 + Y = 114

    For the second average,
    (36 + Y + 41) / 3 = 39
    or (multiplying everything by 3)
    36 + Y + 41 = 117
    So Y = 117 – 36 – 41 = 40.

    (In the exam you would stop here, because only one of the choices has Y = 40 🙂 )

    However, going back to the first equation, we now know that:
    X + 36 + 40 = 114
    So X = 114 – 36 – 40 = 38

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