Normalize 515 and 585, namely
z(515)=(515-550)/35=-1
z(585)=(585-550)/35=1
The percent is the difference of probabilities of one-tail z-values between -1 and +1. It should be a little less than 70%.
The mean of a set of normally distributed data is 550 and the standard deviation is 35.
What percent of the data is between 515 and 585?
1 answer
1 answer