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anonymous

  • 4 years ago

X has a normal distribution. μ=32 Variance=σ2 Given that the probability that X < 33.14 is 0.60406, find the variance to 2 d.p.

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  1. anonymous
    • 4 years ago
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    variance = \[\sigma^2\]

  2. dumbcow
    • 4 years ago
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    look up Z-value of 0.60406 Z = (33.14 -32)/s , where s^2 is variance

  3. anonymous
    • 4 years ago
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    Sorry, the probability is 0.6406

  4. anonymous
    • 4 years ago
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    which is 0.36 for z value

  5. dumbcow
    • 4 years ago
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    i meant you need the Z-value that corresponds to the probability of .6406 ok

  6. dumbcow
    • 4 years ago
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    0.36 = 1.14/s --> s = 1.14/0.36 then square it to get variance

  7. anonymous
    • 4 years ago
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    dumb cow how did u get that dp? did they give it to you?

  8. anonymous
    • 4 years ago
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    display picture

  9. anonymous
    • 4 years ago
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    Ah Ok. Why did you have to square it?

  10. dumbcow
    • 4 years ago
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    oh that...its the default icon that i just saved to paint and messed around with it

  11. dumbcow
    • 4 years ago
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    because variance = standard deviation ^2

  12. anonymous
    • 4 years ago
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    Ah Ok. Thanks! :)

  13. anonymous
    • 4 years ago
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    root variance=s.d

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