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Math Help - fish and prawn problem

  1. #1
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    dfsdfdf
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    Question fish and prawn problem

    Dear Sir,
    I need help some help with the below problem and would appreciate very much if can offer some help.
    thanks
    Kingman

    Fish and prawns are sold by weight. The masses, in kg, of fish and prawns are modelled as having independent normal distributions with means and standard deviations as shown in the table.

    ,
    Fish : Mean mass = 2.2 , Standard deviation= 0.5
    Prawns : Mean mass = 10.5 Standard deviation= 2.1


    Fish are sold at $3 per kg and prawns at $5 per kg.

    (i) Find the probability that a randomly chosen fish has a selling price exceeding $7. (0.395)

    (ii) Find the probability of the event that both a randomly chosen fish has a selling price exceeding $7 and a randomly chosen prawn has a selling price exceeding $55. (0.160)

    (iii) Find the probability that the total selling price of a randomly chosen fish and a randomly chosen prawn is more than $62. (0.392)

    (iv) Explain why the answer to part (iii) is greater than the answer to part (ii).
    Last edited by kingman; May 19th 2010 at 02:09 AM. Reason: misaligment
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  2. #2
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    (i)
    Let X= the price of the fish

    pr(X>7)

    Standardize the normal distribution.

    pr(Z>\frac{7-(2.2)(3)}{(0.5)(3)}

    pr(Z>0.2667)=\phi (-0.27)\approx 0.3936

    (ii)
    For this one you simply have to multiply pr(X>7) (which you found in part (i) by pr(X>55)

    pr(X>55) is found the same way that pr(X=7) was found for the fish in part (i)

    (iii) I'll get to this one if I get a chance
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  3. #3
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    dfsdfdf
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    Thanks

    Thanks very much for the answer but I would appreciate very much if you would
    explain to me how you get (2.2)*3 (why this represent the mean) and
    (.5)*3 would represent the standard deviation of ...

    what formula or reasoning skill you use?

    Also can you tell me how to write mathematical symbols on this message window

    thanks very much
    Kingman
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  4. #4
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    (2.2)(3) is the average selling price of the fish.

    \frac {\$2.2}{1kg}*3kg= \frac {(\$2.2)(3kg}{1kg}

    The kg symbols cancel each other out and you're left with:

    \frac{(\$2.2)3}{1}=\$6.6

    (0.5)(3) is the standard deviation of the selling price of the fish.

    \frac{\$0.5}{1kg}*3kg=\frac{(\$0.5)(3kg}{1kg}

    The kg symbols cancel each other out and you're left with:

    \frac{(\$0.5)3}{1}=\$1.5

    http://www.mathhelpforum.com/math-help/latex-help/

    Go to this link and read the tutorials on LaTeX to learn how to enter math equations. It takes a bit of getting used to.
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  5. #5
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    Question Thanks

    Thanks very much for the help.
    In your solution can we write the average price of fish=

    ($3/1kg)*2.2kg instead of ($2.2/1kg)*3kg

    Also can we start working R.V X=mass of chicken—if can then how we continue.
    Thanks
    Kingman
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