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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.
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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0variance = \[\sigma^2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0look up Zvalue of 0.60406 Z = (33.14 32)/s , where s^2 is variance

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry, the probability is 0.6406

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0which is 0.36 for z value

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i meant you need the Zvalue that corresponds to the probability of .6406 ok

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.00.36 = 1.14/s > s = 1.14/0.36 then square it to get variance

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dumb cow how did u get that dp? did they give it to you?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ah Ok. Why did you have to square it?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh that...its the default icon that i just saved to paint and messed around with it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0because variance = standard deviation ^2
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