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sample distribution of 10 (20,12,8,10,7,11,13,6,4,16) miles What percent of the commuters travel more than 8 miles to class?

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    $\begingroup$ Hint: Why do you need z-scores to calculate a percent here? If one person is male and one is female how would you calculate percent male? $\endgroup$ – Nick Cox Nov 1 '15 at 23:34
  • $\begingroup$ Not only do you not need them, you shouldn't use them unless you can assume the data are normal. With a normal distribution, z scores can be used to determine percentages, but this won't work with other distributions (e.g. a uniform distribution). $\endgroup$ – zbicyclist Nov 1 '15 at 23:42
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    $\begingroup$ @zbicyclist Not so, in the sense that if you knew you had say a uniform distribution and you knew a z score you could do calculations using the density function. It would be perhaps unusual to be in that situation, but it would be feasible. Any distribution for which mean and SD can be related to density, distribution, or quantile function would work. $\endgroup$ – Nick Cox Nov 2 '15 at 8:48
  • $\begingroup$ @NickCox, thanks for the correction -- as you note, this would be unusual. I commonly see analysts calculating z scores, and then over-assuming normality. $\endgroup$ – zbicyclist Nov 3 '15 at 1:25
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Using the Stabilized Normal Probability Plot technique, the data tests as normal:

   Obs    Xprime    Yprime      AbsDif alpha=.10 alpha=.05 alpha=.01
1    4 0.1435663 0.1728315 0.029265254 0.1324046 0.1453979 0.1724372
2    6 0.2531833 0.2554858 0.002302456 0.1324046 0.1453979 0.1724372
3    7 0.3333333 0.3028740 0.030459327 0.1324046 0.1453979 0.1724372
4    8 0.4030133 0.3534983 0.049515038 0.1324046 0.1453979 0.1724372
5   10 0.4681157 0.4612687 0.006847004 0.1324046 0.1453979 0.1724372
6   11 0.5318843 0.5166183 0.015266013 0.1324046 0.1453979 0.1724372
7   12 0.5969867 0.5716814 0.025305320 0.1324046 0.1453979 0.1724372
8   13 0.6666667 0.6255154 0.041151249 0.1324046 0.1453979 0.1724372
9   16 0.7468167 0.7711199 0.024303246 0.1324046 0.1453979 0.1724372
10  20 0.8564337 0.9069053 0.050471623 0.1324046 0.1453979 0.1724372

(JQT v21 i3 Technical Aids; Jul 1989. Lloyd S. Nelson. The power of this test was shown to compare favorably with the Shapiro-Wilk test and the Anderson-Darling test.)

Accepting the null hypothesis of normality, percentages can be calculated based on z-scores in the following manner:

Using R:

commute <- c(20,12,8,10,7,11,13,6,4,16)
pnorm((mean(commute)-8)/sd(commute))

Using Excel:

=NORM.S.DIST((AVERAGE(A1:A10)-8)/STDEV.S(A1:A10),TRUE)

where your sample is in the range A1:A10.

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