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Mathematics 16 Online
OpenStudy (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.

OpenStudy (anonymous):

variance = \[\sigma^2\]

OpenStudy (dumbcow):

look up Z-value of 0.60406 Z = (33.14 -32)/s , where s^2 is variance

OpenStudy (anonymous):

Sorry, the probability is 0.6406

OpenStudy (anonymous):

which is 0.36 for z value

OpenStudy (dumbcow):

i meant you need the Z-value that corresponds to the probability of .6406 ok

OpenStudy (dumbcow):

0.36 = 1.14/s --> s = 1.14/0.36 then square it to get variance

OpenStudy (anonymous):

dumb cow how did u get that dp? did they give it to you?

OpenStudy (anonymous):

display picture

OpenStudy (anonymous):

Ah Ok. Why did you have to square it?

OpenStudy (dumbcow):

oh that...its the default icon that i just saved to paint and messed around with it

OpenStudy (dumbcow):

because variance = standard deviation ^2

OpenStudy (anonymous):

Ah Ok. Thanks! :)

OpenStudy (anonymous):

root variance=s.d

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