Hi

*Note: 

mean of 50 and a std. deviation of 10

 z = -20/10 = -2  and z = 15/10 = 1.5

P(between 30 and 65) = P( -2 > z < 1.5) = .9332 - .0228 = .9104  

91.04% of them are between 30 and 65

Area under the graph between z = -2 and z = 1.5

 

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