Tim worker is doing his budget. he discovers that the average electric bill for the year was $206 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 = - .62.22 . the percent of area associated with $184.00 = 62.2 %. the z for $200 = - .622.22 . the percent of area associated with $200 = 48.6 %. subtracting the two percentages, the percent of expenses between $184.00 and $200.00 is 622.22

We use the z score formula:

z = (x – u) / s

Find z at x = 184 and x = 200 then use the standard tables to find for the P value using right tailed test.

at x1 = 184

z1 = (184 – 206) / 10 = -2.20

From the tables, P1 = 0.9861

at x2 = 200

z2 = (200 – 206) / 10 = -0.60

From the tables, P2 = 0.7257

So the probability it falls from 184 to 200 is:

P = 0.9861 – 0.7257

P = 0.2604 

twenty-six miles over two gallons because the proportion for the original 13 miles to one gallon would be written 13/1 and, doing the same thing to the top and bottom, i multiplied by 2 and got 26/2