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yashar806

  • 2 years ago

Need help, see attachments

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  1. yashar806
    • 2 years ago
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  2. zakrocks007
    • 2 years ago
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    10

  3. zakrocks007
    • 2 years ago
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    10

  4. yashar806
    • 2 years ago
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    is it potssible?

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

  6. yashar806
    • 2 years ago
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    i got 5.1125 something

  7. yashar806
    • 2 years ago
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    I need help with last one

  8. yashar806
    • 2 years ago
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    can you help on last one , cuz I really don't get t

  9. kropot72
    • 2 years ago
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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

  10. yashar806
    • 2 years ago
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  11. yashar806
    • 2 years ago
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    i have that formula

  12. yashar806
    • 2 years ago
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    i need to use that for last one

  13. yashar806
    • 2 years ago
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    p hat is equal to = x/n

  14. kropot72
    • 2 years ago
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    The question you have posted does not involve sampling. Therefore the formula you have posted does not apply.

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