Standard deviation questions?

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    Hi, I can normally calculate the standard deviation if I were to be given grouped/raw data, but I've found these two confusing and I'm not really sure what to do, so would be grateful for some help.

    1. Three statistics students, Ali, Les and Sam spent the day fishing, and they caught three different types of fish and recorded the type and mass of each fish caught. At 4p.m, they summarised the results as follows. (I'll only enter data for Sam, as that's the data required)

    Sam. 1 perch. 1 roach 0 tench. mean mass 1kg. standard deviation.

    a) How can you deduce that each fish caught by Sam was 1kg?
    - Would this be sufficient "As the standard deviation was 0, this means that the values must've all lied on the mean, so each value must be 1kg" - Could someone help make it clearer?

    b) Before leaving the waterside, Sam catches one more fish and weighs it. He then announces that if this extra fish is included with the other two fish he caught, the standard deviation is 1.00kg. Find the mass of this extra fish.

    How would you do this?

    I've got something like 1 (s.d.) = square root (1/3 * sigmax^2 - mean^2).

    But I'm missing 2 variables, so I'm not sure if I'm doing it all wrong :/

    2. The depth of water in a lake was measured at 50 different points on the surface of the lake. The depths, x metres are summarised by sigmax = 934.5 and sigmax^2=19275.81.

    a) Find the mean and the variance of the depths.

    Mean = 934.5/50 = 18.69
    Variance = (sigmafx^2)/sigmaf - mean^2

    But I don't/can't get fx^2 data though? :/

    b) Some weeks later the water level in the lake rose by 0.23m. What would be the mean and variance of the depths taken at the same points on the lake as before?

    I'm not really sure how to approach this, but I guess doing the first bit may help.

    I'd be really grateful if someone can help me with these questions :/
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    The standard deviation represents the average distance of the masses from the mean, since sd=0 the fish must be 0 distance from the mean
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    For b) you can say

    1 = {( 12 + 12 +m2 )/3 - [( 1 + 1 + m)/3]2}

    and solve quadratically
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    (Original post by TenOfThem)
    The standard deviation represents the average distance of the masses from the mean, since sd=0 the fish must be 0 distance from the mean
    Thanks

    (Original post by the bear)
    For b) you can say

    1 = {( 12 + 12 +m2 )/3 - [( 1 + 1 + m)/3]2}

    and solve quadratically
    I tried this, but perhaps I did it wrong, or I just got the wrong answer :/

    So 1 = (1^2+1^2+x^2)/3 - (1+1+x)/3

    3=2+x^2-2-x
    0=x^2-x-3
    x=(1+/-13)/2. So x=2.3. But the answers say that the answer is 3.12kg?
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    Hi can someone help me with 1b/2a/2b please?
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    1(b)

    mean = (2+m)/3

    Difference between masses and mean

    (2+m)/3 - 1 = (m-1)/3
    (2+m)/3 - 1 = (m-1)/3
    m - (2+m)/3 = 2(m-1)/3

    squaring and adding gives

    [1/9 + 1/9 + 4/9](m-1)^2

    Dividing by 3 gives

    2/9 (m-1)^2

    and this = 1
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    (Original post by MedicalMayhem)
    Thanks



    I tried this, but perhaps I did it wrong, or I just got the wrong answer :/

    So 1 = (1^2+1^2+x^2)/3 - (1+1+x)/3

    3=2+x^2-2-x
    0=x^2-x-3
    x=(1+/-13)/2. So x=2.3. But the answers say that the answer is 3.12kg?
    You forgot to square the mean
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    (Original post by TenOfThem)
    1(b)

    mean = (2+m)/3

    Difference between masses and mean

    (2+m)/3 - 1 = (m-1)/3
    (2+m)/3 - 1 = (m-1)/3
    m - (2+m)/3 = 2(m-1)/3

    squaring and adding gives

    [1/9 + 1/9 + 4/9](m-1)^2

    Dividing by 3 gives

    2/9 (m-1)^2

    and this = 1
    So with this method, are you just making two equations and completing simultaneously? Would you say it's easier than the other method? But for the first step, why do you do Mean - 1?

    (Original post by the bear)
    You forgot to square the mean
    Erm, I tried this, but I still don't think it's right.

    So 3 = 2 + x^2 - (x+2)^2

    3 = 2 + x^2 - x^2 -4x -4.

    So 4x = -5, and x = -5/4? :/
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    (Original post by MedicalMayhem)
    Erm, I tried this, but I still don't think it's right.

    So 3 = 2 + x^2 - (x+2)^2
    You should have initially:

    1^2=\dfrac{x^2+2}{3}-\left(   \dfrac{x+2}{3}\right)^2

    giving

    1=\dfrac{x^2+2}{3}-\dfrac{x^2+4x+4}{9}

    Note the 9.
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    (Original post by MedicalMayhem)
    But for the first step, why do you do Mean - 1?
    The sd is the average distance between the mean and the values
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    (Original post by ghostwalker)
    You should have initially:

    1^2=\dfrac{x^2+2}{3}-\left(   \dfrac{x+2}{3}\right)^2

    giving

    1=\dfrac{x^2+2}{3}-\dfrac{x^2+4x+4}{9}

    Note the 9.
    (Original post by TenOfThem)
    The sd is the average distance between the mean and the values
    Thanks very much, I get it now!

    Erm...would you be able to go through 2a please?
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    (Original post by MedicalMayhem)
    Erm...would you be able to go through 2a please?
    Not sure why you've introduced "f" into your working - it's not required.

    Standard formula for n data items is all you need.
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    (Original post by ghostwalker)
    Not sure why you've introduced "f" into your working - it's not required.

    Standard formula for n data items is all you need.
    But for grouped data aren't you supposed to use, (sigma fx^2) / sigma f - mean^2? Instead of the one for raw data - sigmax^2/n - mean^2?
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    (Original post by MedicalMayhem)
    But for grouped data aren't you supposed to use, (sigma fx^2) / sigma f - mean^2? Instead of the one for raw data - sigmax^2/n - mean^2?
    .

    But your data's not grouped - check out the meaning of "grouped data".

    You're told n, \Sigma x, \Sigma x^2 and all you need to do is plug them into the formulae.
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    (Original post by ghostwalker)
    .

    But your data's not grouped - check out the meaning of "grouped data".

    You're told n, \Sigma x, \Sigma x^2 and all you need to do is plug them into the formulae.
    Thanks a lot! I think I did mix them up

    So grouped data would be a range?

    E.g. 1-10 = 5
    11-20=6

    And you would take the midpoint * frequency?
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    (Original post by MedicalMayhem)
    Thanks a lot! I think I did mix them up

    So grouped data would be a range?

    E.g. 1-10 = 5
    11-20=6

    And you would take the midpoint * frequency?
    Yes.

    You can still have a frequency distribution even if the data isn't grouped and you'd need to use the "f" formulae there.
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    (Original post by ghostwalker)
    Yes.

    You can still have a frequency distribution even if the data isn't grouped and you'd need to use the "f" formulae there.
    Hm, what do you mean by a frequency distribution?

    Like

    x=10 f=8
    x=11 f=9

    etc?
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    yes
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    (Original post by TenOfThem)
    yes
    Thanks
 
 
 
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