



 
Hi Justin, Changing the problem to a problem using the standard normal random variable Z is the right way to go. The conversion is which in your case is . Your probability statement P(20 ≤ Y ≤ 60) = 0.50 can be represented graphically by Notice that the shaded area is symmetric about the mean 40 so P(40 ≤ Y ≤ 60) = 0.25. Now convert to the standard normal distribution and the diagram becomes (I only brought across half of the shaded region because most of the standard normal tables I have seen make it easy to deal with areas between Z = 0 and Z = some positive number.) The point s on my diagram is the converted value for Y = 60, that is . Use the normal table to determine the value of s, substitute into the equation above and solve for σ. Harley  


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