Monthly water bills are $40, $42, $40, $38, $48, $50, $58, $62, $56, $46, $44, $44 for January to December. Using sigma notation, find the mean bill for the whole year, for the first six months, and for April through November.

Step-by-step solution

1. 1

   Write the general template. A mean over a block of terms is

   $\bar{x} = \frac{1}{n}\sum_{i=a}^{b} x_i, \qquad n = b - a + 1$

   All three parts use this same expression; only the limits $a$, $b$ — and therefore $n$ — change. The count must always be recomputed from the limits, never reused from the previous part.

2. 2

   (a) Whole year: $i$ from 1 to 12.

   $\bar{x} = \frac{1}{12}\sum_{i=1}^{12} x_i = \frac{40+42+40+38+48+50+58+62+56+46+44+44}{12} = \frac{568}{12}$

   $= 47.3333\ldots = \$47.33$

3. 3

   (b) First six months: $i$ from 1 to 6.

   $\bar{x} = \frac{1}{6}\sum_{i=1}^{6} x_i = \frac{40+42+40+38+48+50}{6} = \frac{258}{6} = 43 = \$43.00$

4. 4

   (c) April through November: $i$ from 4 to 11. April is $x_4$ and November is $x_{11}$, so $n = 11 - 4 + 1 = 8$ — eight months, not seven. Off-by-one on this count is the classic error:

   $\bar{x} = \frac{1}{8}\sum_{i=4}^{11} x_i = \frac{38+48+50+58+62+56+46+44}{8} = \frac{402}{8} = 50.25 = \$50.25$

5. 5

   Check the three totals against each other. The first six months total $258$ and months $4$–$11$ total $402$; the full year is $568$. Consistency test: months $1$–$3$ total $40+42+40 = 122$ and month $12$ is $44$, so $122 + 402 + 44 = 568$ ✓ — the three blocks partition the year exactly.

6. 6

   Interpret the pattern. The summer stretch (April–November) averages \50.25$, well above both the yearly$$47.33$and the winter-heavy first half at$$43.00$, which matches bills that peak in July and August.