document.write( "Question 1196439: The salary of assistant professors at Kent State University is normally distributed with a mean of $45,000 and a standard deviation of $1,500.If I chose an assistant professor at random, what is the probability that he/she makes more than 1.3 standard deviations above the mean? \n" ); document.write( "
Algebra.Com's Answer #829281 by Theo(13342)\"\" \"About 
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1.3 standard deviations above the mean gives you a z-score of 1.3
\n" ); document.write( "area to the right of a z-score of 1.3 = .0968005494.
\n" ); document.write( "i used the ti-84 plus to get that number.
\n" ); document.write( "the z-score table would give you 1 - .90320 = .09680.
\n" ); document.write( "if you round to 4 decimal places, the answer will be the same as .0968.
\n" ); document.write( "i confirmed using david lane calculator.
\n" ); document.write( "1.3 * 1500 = 1950
\n" ); document.write( "add that to 45000 to get 46950
\n" ); document.write( "results from david lane calculator are shown below.
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\n" ); document.write( "results from z-score table are shown below.
\n" ); document.write( "
\n" ); document.write( "david lane calculator at https://davidmlane.com/hyperstat/z_table.html
\n" ); document.write( "z-score table at https://www.rit.edu/academicsuccesscenter/sites/rit.edu.academicsuccesscenter/files/documents/math-handouts/Standard%20Normal%20Distribution%20Table.pdf
\n" ); document.write( "note that the z-score table gives you area to the left of the z-score.
\n" ); document.write( "area to the right is 1 minus that.
\n" ); document.write( "note that if you used the z-score formula to get the z-score, you would have gotten z-score of 1.3 as shown below.
\n" ); document.write( "z = (x - m) / s
\n" ); document.write( "z is the z-score
\n" ); document.write( "x is the raw score
\n" ); document.write( "m is the raw mean
\n" ); document.write( "s is the standard deviation.
\n" ); document.write( "formula becomes z = (46950 - 45000) / 1500 = = 1950 / 1500 = 1.3.
\n" ); document.write( "46950 is 45000 + 1.3 * 1500 which is 1.3 standard deviations above the mean.
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