Asked by Raj
                The masses of boxes of apples are normally distributed such that 20% of the boxes are heavier than 5.08 kg and 15% of the boxes are heavier than 5.62kg. Estimate the mean and standard deviation of the masses.
            
            
        Answers
                    Answered by
            Reiny
            
    let the mean be m, and the standard deviation be s
(5.08-m)/s < .8
using the equality: m + .8s = 5.08
(5.62-m) < .85s --> m + .85s = 5.62
subtract them: .05s = .54
s = 10.8
m + .8(10.8) = 5.08
m = negative, uh uh
 
    
(5.08-m)/s < .8
using the equality: m + .8s = 5.08
(5.62-m) < .85s --> m + .85s = 5.62
subtract them: .05s = .54
s = 10.8
m + .8(10.8) = 5.08
m = negative, uh uh
                    Answered by
            Reiny
            
    rethinking this question, using ...
http://davidmlane.com/normal.html
Clicking on "value from an area", I entered .8 and got
a z-score of .841 , clicking "above"
entering .85 , my z-score was 1.036 clicking above
(5.08 - m)/s = .841 ---> m + .841s = 5.08
(5.62 - m)/s = 1.036 --- m + 1.036s = 5.62
subtract : .195s = .54
s = 2.77 , then m = 2.75
    
http://davidmlane.com/normal.html
Clicking on "value from an area", I entered .8 and got
a z-score of .841 , clicking "above"
entering .85 , my z-score was 1.036 clicking above
(5.08 - m)/s = .841 ---> m + .841s = 5.08
(5.62 - m)/s = 1.036 --- m + 1.036s = 5.62
subtract : .195s = .54
s = 2.77 , then m = 2.75
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