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