Find the random variable coefficients y1 and y2 where P(y1 < y < y2) = 0.5. Where mean is 0.7 and standard deviation is 0.03 (not sure if you need that). I have no clue where to start with this one.
Thanks for any help
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Perhaps because there are an infinite number of answers! [itex]-\infty to 0.7[/itex] would obviously work because of the symmetry of the normal distribution about the mean. So would [itex]0.7 to \infty[/itex]. For finite values of y1 and y2, try this. Convert to the "standard" z-score using [itex]z= (y- \mu)/\sigma[/itex] which here is [itex]z= (y- 0.7)/0.03[/itex]. Pick any y1 you want, less than the mean, and calculate its z-score. [For example, choosing (just because it makes the calculation easy) y1 to be 0.67, we get z= -0.03/0.03= -1]. Look that up on a table of the normal distribution (a good one is at http://people.hofstra.edu/Stefan_Waner/RealWorld/normaltable.html [Broken]) to find P(y1) [for z= -1 I get 0.46587] If that is less than 0.5, add it to 0.5 to see how much "more" you need and look up the z corresponding to that and, finally, compute the y2 that gives. [0.46587+ 0.5= 0.96587. The table says that corresponds to z= 1.82 and then 1.82= (y2- 0.7)/0.03 gives y2= 0.7546. You can choose any y1 you want, less than 0.7, and do the same to get a different y2.
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