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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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variance = \[\sigma^2\]
look up Z-value of 0.60406 Z = (33.14 -32)/s , where s^2 is variance
Sorry, the probability is 0.6406
which is 0.36 for z value
i meant you need the Z-value that corresponds to the probability of .6406 ok
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0.36 = 1.14/s --> s = 1.14/0.36 then square it to get variance
dumb cow how did u get that dp? did they give it to you?
display picture
Ah Ok. Why did you have to square it?
oh that...its the default icon that i just saved to paint and messed around with it
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because variance = standard deviation ^2
Ah Ok. Thanks! :)
root variance=s.d
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