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anonymous

  • one year ago

helllllllllllllllp please.. medal and fans A fair coin is tossed 5 times and the number of heads is recorded. (i) The random variable X is the number of heads. State the mean and the variance. (ii) The number of heads is doubled and denoted by the random variable Y. State the mean and variance of Y.

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  1. amoodarya
    • one year ago
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    |dw:1434986852277:dw| this is first step

  2. anonymous
    • one year ago
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    yes then... atleast someone to help me..

  3. amoodarya
    • one year ago
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    \[e(x)=\sum_{i=1}^{5}p_ix_i\]

  4. amoodarya
    • one year ago
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    \[e(x)=\frac{1}{32}*0+\frac{5}{32}*1+\frac{10}{32}*2+\frac{10}{32}*3+\frac{5}{32}*4+\frac{1}{32}*5\]

  5. amoodarya
    • one year ago
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    it is ok , to calculate mean ?

  6. anonymous
    • one year ago
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    yes thank you so much...

  7. amoodarya
    • one year ago
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    \[var(x)=e[x^2]-e^2[x]\\and\\e[x^2]=\sum_{i=1}^{5}p_ix_i^2\]

  8. amoodarya
    • one year ago
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    can you tell me what;s mean ? then what is the value of var (x) ?

  9. anonymous
    • one year ago
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    mean is 2.5 var(x)=1.25 right ?

  10. amoodarya
    • one year ago
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    can I see your work ?

  11. amoodarya
    • one year ago
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    \[mean=\frac{80}{32} ?\]

  12. anonymous
    • one year ago
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    yes its 80/32

  13. amoodarya
    • one year ago
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    yes it is correct

  14. amoodarya
    • one year ago
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    if you want to know more in binomial distribution mean=np=5 (1/2)=2.5 var=n p q =5 (1/2)(1/2)=5/4=1.25

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