Tim worker is doing his budget. he discovers that the average electric bill for the year was $206.00 with a standard deviation of $10.00. what percent of his expenses in this category would he expect to fall between $184.00 and $200.00? the z for $184.00 = - the percent of area associated with $184.00 = % the z for $200.00 = - the percent of area associated with $200.00 = % subtracting the two percentages, the percent of expenses between $184.00 and $200.00 is %

The z-value is computed from ...

... z = (x -µ)/σ

... z = (184 -206)/10 = -2.2 . . . . for $184

... z = (200 -206)/10 = -0.6 . . . . for $200

You can look up these values in a normal distribution table, or you can use an appropriate calculator to find the corresponding percentiles.

... -2.2 corresponds to the 1.390 percentile. (That amount of area is below -2.2 standard deviations from the mean.)

... -0.6 corresponds to the 27.425 percentile.

Subtracting the two percentages gives the percentage of expenses between $184 and $200. That number is 26.035% ≈ 26%.

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Comment on the calculator display

The difference that got cut off from the display in the attachment is ...

... 0.2603496703

The normalcdf( ) function requires a lower limit. Using -8 standard deviations is effectively equivalent to -∞ for this purpose, as any lower number has no effect on the least-significant digits of the result.