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hockeychick23

  • one year ago

Find the expected value of the number of questions you'd get right by guessing. What are the variance and standard deviation?

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  1. hockeychick23
    • one year ago
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  2. hockeychick23
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  3. hockeychick23
    • one year ago
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    @jim_thompson5910 can you help me please?

  4. jim_thompson5910
    • one year ago
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    X = number of correct questions |dw:1433396801483:dw|

  5. jim_thompson5910
    • one year ago
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    multiply X and P(X) to form the X*P(X) column of values

  6. hockeychick23
    • one year ago
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  7. jim_thompson5910
    • one year ago
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    good

  8. jim_thompson5910
    • one year ago
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    now add up the values in the X*P(X) column and you will get the expected value

  9. hockeychick23
    • one year ago
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  10. jim_thompson5910
    • one year ago
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    so the expected value is 0.5 which means, you expect (on average) to get 0.5 questions right if you randomly guessed

  11. hockeychick23
    • one year ago
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    ohhh ok thanks!

  12. jim_thompson5910
    • one year ago
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    now onto the second part "What are the variance and standard deviation? "

  13. jim_thompson5910
    • one year ago
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    you're going to have 2 additional columns. I'm going to erase the X*P(X) column to make room

  14. jim_thompson5910
    • one year ago
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    instead I'm going to have X - mu and (X - mu)^2 |dw:1433397803766:dw|

  15. jim_thompson5910
    • one year ago
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    mu = expected value = 0.5

  16. jim_thompson5910
    • one year ago
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    oh wait, there's one more column needed the P(X)*(X-mu)^2 column |dw:1433397806736:dw|

  17. hockeychick23
    • one year ago
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    oh ok, so do i find the mean of p(x)?

  18. jim_thompson5910
    • one year ago
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    you're asking about the value of mu? mu = 0.5 and it was found when you computed the expected value

  19. hockeychick23
    • one year ago
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    oh ok sorry it took me a while to calculate the numbers

  20. jim_thompson5910
    • one year ago
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    this is what I'm getting

  21. jim_thompson5910
    • one year ago
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    I see the typo you made. Look at the bottom row, last column

  22. jim_thompson5910
    • one year ago
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    you should have (2-0.5) and not (1-0.5)

  23. hockeychick23
    • one year ago
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    sorry i pasted the wrong table this is what i got

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  24. hockeychick23
    • one year ago
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  25. jim_thompson5910
    • one year ago
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    btw if you have excel or open office (it should work in there too), you can type in formulas like you see attached

  26. jim_thompson5910
    • one year ago
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    that way you can automate things a bit

  27. hockeychick23
    • one year ago
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    Ok thanks

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  28. jim_thompson5910
    • one year ago
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    now add up everything in the P(X)*(X-mu)^2 column, you should get 0.375

  29. jim_thompson5910
    • one year ago
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    0.375 is the variance take the square root of the variance to get the standard deviation

  30. hockeychick23
    • one year ago
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    standard deviation= .61237243569 variance= 0.375

  31. jim_thompson5910
    • one year ago
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    looks good

  32. hockeychick23
    • one year ago
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    ok thanks so much! i have a question pertaining to the same data that i need help with if you can help me with it: Let's say that for one of the two questions you can narrow the answer choices down to three. Now your chance of getting that question right by guessing is about .33. What is the new expected value for the number of questions you'd get right by guessing? What are the new variance and standard deviation of this random variable? Drawing another tree diagram and making another probability distribution table may help you answer this question. I'll give you another medal

  33. jim_thompson5910
    • one year ago
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    were you able to get started on this one? or no?

  34. hockeychick23
    • one year ago
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    sort of but I'm not sure if i did it right, i did

  35. jim_thompson5910
    • one year ago
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    I made question 1 have P(correct) = 0.33 and question 2 has P(correct) = 0.25 |dw:1433401116557:dw|

  36. jim_thompson5910
    • one year ago
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    are you seeing how I'm getting these values?

  37. hockeychick23
    • one year ago
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    ohhh yes thanks!

  38. jim_thompson5910
    • one year ago
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    so here is getting 0 questions correct |dw:1433401932751:dw| ie getting both wrong

  39. jim_thompson5910
    • one year ago
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    0.0825 is cut off, but that's getting both correct |dw:1433401967587:dw|

  40. hockeychick23
    • one year ago
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    Ok so 0.5025 is the probability of getting 0 correct and 0.0825 is getting both correct.

  41. jim_thompson5910
    • one year ago
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    and the other values are added to get the probability of getting exactly 1 correct so that gives us this new table |dw:1433402014075:dw|

  42. jim_thompson5910
    • one year ago
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    yes

  43. hockeychick23
    • one year ago
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    Ok so would i need to find the expected value again?

  44. hockeychick23
    • one year ago
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  45. jim_thompson5910
    • one year ago
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    this is what I get

  46. jim_thompson5910
    • one year ago
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    You have 0.825 instead of 0.0825

  47. jim_thompson5910
    • one year ago
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    So I'm getting an expected value of 0.58 and a variance of 0.4086 standard deviation = sqrt(variance) = sqrt(0.4086) = 0.63921827 (approximate)

  48. hockeychick23
    • one year ago
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    ok now I'm getting that too, thanks! (I'm going to post something and tag you in it to give you another medal)

  49. jim_thompson5910
    • one year ago
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    that's fine, it doesn't matter about the medal thing lol

  50. jim_thompson5910
    • one year ago
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    I'm glad it's making sense now

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