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

First of all, let's fill the gaps. The z score for 184 is -0.62.22. The percent associated with this is 62.22%. The z score for 200 is -0.622.22. The percent associated with this is 48.6%. The percent of the expenses is 622.22

We have to apply z-score formula in order to solve this problem. The formula is

z = (x-u) / s

Plugging z-scores at 184 and 200, we find

The corresponding probability with this number is P(1) = 0.9861

The corresponding probability with this number is P(2) = 0.7257

Probability between 184 and 200 is P(1) - P(2) = 0.2604

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