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yashar806

  • 2 years ago

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  1. yashar806
    • 2 years ago
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  2. satellite73
    • 2 years ago
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    3 are brown and so evidently 6 are not brown use the binomial \[\binom{9}{3}(.21)^3(.79)^6\]

  3. dhruv_khatkar
    • 2 years ago
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    i think you can use the formula (n x) . p^x . q ^(n-x)

  4. yashar806
    • 2 years ago
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    what about the second one?

  5. yashar806
    • 2 years ago
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    sorry third one

  6. kropot72
    • 2 years ago
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    For the third one you use the fact that the binomial distribution can be approximated by the normal distribution when n is large and p is neither small nor near 1.\[\mu=np=175\times 0.21\] \[\sigma=\sqrt{np(1-p)}=\sqrt{175\times 0.21\times 0.79}\] Then you find the required probability by means of a standard normal distribution table.

  7. yashar806
    • 2 years ago
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    oh, fourth one?

  8. kropot72
    • 2 years ago
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    Basically you use the same method for the fourth one as for the third, but change the values of n and p accordingly.

  9. kropot72
    • 2 years ago
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    You need to find one value of z-score for the third one and two values of z-score for the fourth.

  10. yashar806
    • 2 years ago
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    because there two values 17% and 24%

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