Pre-cal

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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??

Thanks

  • Pre-cal -

    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.

    http://davidmlane.com/hyperstat/z_table.html

  • typo correction Pre-cal -

    of course .5 - .00383 = .49617 and not .19617

    so 49.6% of the data falls in your given range.

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