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    In a first 10 sample data, the mean is 2.4 and standard deviation is 0.8 hrs

    In second 5 sample data, the mean is 2.0 and standard deviation is 1.2 hrs

    What is the new standard deviation of the combined 15-sample data?


    how to solve it? :confused:
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    Therefore:
    √(A ∑X²/10 - mean) = 0.8 and √(B ∑X²/5 - mean) = 1.2
    So square both and you get:
    (A ∑X²/10 - mean) = 0.63 and (B ∑X²/5 - mean) = 1.44
    So add the mean and then times by n:
    (A ∑X²) = 30.3 and (B ∑X²) = 17.2
    Add those together:
    ((30.3 + 17.2)/15)-((2.4 *10) + (2*5))/15 = Var = 0.9
    Square root of Var = SD = 0.949 (3sf)
    There maybe errors as I am in a rush!
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    hi Jack, can I do it this way?

    s = √(∑(x-u)² / (n-1)

    0.8² = ∑(x-2.4)² / (10-1)
    ∑(x-2.4)² = 5.76

    1.2² = ∑(x-2)² / (5-1)
    ∑(x-2)² = 5.76


    s² = (5.76 + 5.76)/(15-1)
    s = 0.907
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    (Original post by yagmai)
    hi Jack, can I do it this way?
    If you don't get the same answers as Jack via. your way, than obviously use Jack's way.

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    up.............
 
 
 
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