anonymous
  • anonymous
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.
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
variance = \[\sigma^2\]
dumbcow
  • dumbcow
look up Z-value of 0.60406 Z = (33.14 -32)/s , where s^2 is variance
anonymous
  • anonymous
Sorry, the probability is 0.6406

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anonymous
  • anonymous
which is 0.36 for z value
dumbcow
  • dumbcow
i meant you need the Z-value that corresponds to the probability of .6406 ok
dumbcow
  • dumbcow
0.36 = 1.14/s --> s = 1.14/0.36 then square it to get variance
anonymous
  • anonymous
dumb cow how did u get that dp? did they give it to you?
anonymous
  • anonymous
display picture
anonymous
  • anonymous
Ah Ok. Why did you have to square it?
dumbcow
  • dumbcow
oh that...its the default icon that i just saved to paint and messed around with it
dumbcow
  • dumbcow
because variance = standard deviation ^2
anonymous
  • anonymous
Ah Ok. Thanks! :)
anonymous
  • anonymous
root variance=s.d

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