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yashar806 Group Title

Need help, see attachments

  • one year ago
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  1. yashar806 Group Title
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    • one year ago
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  2. zakrocks007 Group Title
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    10

    • one year ago
  3. zakrocks007 Group Title
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    10

    • one year ago
  4. yashar806 Group Title
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    is it potssible?

    • one year ago
  5. kropot72 Group Title
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    (a) The mean is \[205\times 0.15=?\] The standard deviation is \[\sqrt{(205\times 0.15\times 0.85)}\]

    • one year ago
  6. yashar806 Group Title
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    i got 5.1125 something

    • one year ago
  7. yashar806 Group Title
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    I need help with last one

    • one year ago
  8. yashar806 Group Title
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    can you help on last one , cuz I really don't get t

    • one year ago
  9. kropot72 Group Title
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    (c) This part can be solved by using the Normal Approximation of the Binomial Distribution. As found in part (a) the mean = 31 and the standard deviation = 5.112. There are 27 seats available for left handed people so we need to find the probability that the number of left handed people is equal to, or fewer than 27. In this case the z-score is calculated as follows: \[z=\frac{27-31+0.5}{5.112}=-0.6847\] Note: The term 0.5 in this calculation is a correction caused by the change from a discrete to a continuous distribution. A standard normal probability table gives a cumulative probability of 0.247

    • one year ago
  10. yashar806 Group Title
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    • one year ago
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  11. yashar806 Group Title
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    i have that formula

    • one year ago
  12. yashar806 Group Title
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    i need to use that for last one

    • one year ago
  13. yashar806 Group Title
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    p hat is equal to = x/n

    • one year ago
  14. kropot72 Group Title
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    The question you have posted does not involve sampling. Therefore the formula you have posted does not apply.

    • one year ago
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