A set of data is normally distributed with a mean of 16 and a standard deviation of 0.3. What percent of the data is between 15.2 and 16?
I had: 16-t(0.3)=15.2 and 16-t(0.3)=15.7
which came to 0.497-.341=0.156. rounded off I came up with 16%. What did I do wrong??
Pre-cal - Reiny, Saturday, January 3, 2009 at 9:33am
I use the following z-score conversion
z = (mean - given data)/standard deviation
Your data of 15.2 translates into a z-score of -2.66666
and of course your data of 16 gives a z-score of 0
(I don't know where the 15.7 in your second calculation comes from, is it a typo?)
so you want the region between a z-score of 0 and -2.6666 which would be
.5 - .00383 = .19617
I use the following applet instead of tables, in this one you don't even have to convert to z-scores.
typo correction Pre-cal - Reiny, Saturday, January 3, 2009 at 9:39am
of course .5 - .00383 = .49617 and not .19617
so 49.6% of the data falls in your given range.