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Saylilbaby

  • one year ago

i need help with 3-8 anyone....i will give medals....

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  1. saylilbaby
    • one year ago
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  2. ybarrap
    • one year ago
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    In order to complete this exercise, you'll need to count the number of dots for each bin: I'll start 60 - 1 61 - 1 62 - 0 63 - 1 ... Finish the rest and we'll continue

  3. saylilbaby
    • one year ago
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    60-1 61-1 62-0 63-1 64-0 65-1 66-0 67-0 68-3 69-2 70-2 71-0 72-1 73-0 74-1 75-0 76-3 77-0 78-1 79-1 80-4 81-0 82-4 83-0 84-1 85-2 86-3 87-1 88-1 89-1 90-5 91-0 92-4 93-0 94-0 95-3 96-0 97-0 98-1 99-1 100-1 @ybarrap

  4. saylilbaby
    • one year ago
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    @Concentrationalizing

  5. saylilbaby
    • one year ago
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    @jabez177

  6. saylilbaby
    • one year ago
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    @Awolflover1

  7. saylilbaby
    • one year ago
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    @sleepyjess

  8. saylilbaby
    • one year ago
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    @ganeshie8

  9. ybarrap
    • one year ago
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    Using your values we get (see attached):

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  10. ybarrap
    • one year ago
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    You'll see how we compute mean and standard deviation in the attached above: Mean \( \bar{x}=81.72\) Standard deviation \(\sigma = 10.4\)

  11. ybarrap
    • one year ago
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    You can drill down into each cell in the spreadsheet to see the formula I used. Does this make sense?

  12. ybarrap
    • one year ago
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    We used the following definition for mean and standard deviation: $$ \operatorname{E}[X] = x_1p_1 + x_2p_2 + \dotsb + x_kp_k \;\\ \operatorname{Var}(X) = \sum_{i=1}^n p_i\cdot(x_i - \mu)^2 = \sum_{i=1}^n (p_i\cdot x_i^2) - \mu^2\\ \sigma = \sqrt{Var(X)} $$ Where \(x_i\) are the bin values, 60, 61, etc...and \(p_i\) are their probabilities.

  13. ybarrap
    • one year ago
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    And $$ \bar{x}=E[X] $$

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