Question

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

Answer 1

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

$z_{1} = \frac{184-206}{10} = -2.20$

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

$z_{2} = \frac{200-206}{10} = -0.60$

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

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