Seven students reported school-year expenses of $2,800, $1,990, $2,005, $2,400, $1,860, $2,200 and $2,000. Find the mean, say whether it beats the $2,110 estimate, and find the eighth value that would make the mean of all eight exactly $2,110.

Step-by-step solution

1. 1

   (a) Add the seven values. Grouping to keep the running total manageable: $2{,}800+1{,}990 = 4{,}790$; $2{,}005+2{,}400 = 4{,}405$; $1{,}860+2{,}200 = 4{,}060$; then $4{,}790+4{,}405+4{,}060+2{,}000 = 15{,}255$.

   $\sum X = 15{,}255$

2. 2

   Divide by 7 and round to the cent.

   $\bar{X} = \frac{15{,}255}{7} = 2{,}179.285714\ldots$

   The third decimal is $5$ followed by more digits, so rounding to two places gives

   $\bar{X} = \$2{,}179.29$

3. 3

   (b) Compare with the estimate. $2{,}179.29 > 2{,}110$, so the friends' average is higher than the published estimate — by \69.29$. Ashley's suspicion that the estimate was low is supported by her sample (though seven friends is far too small to settle the question statistically).

4. 4

   (c) Convert the target mean into a target total. For eight students to average \2{,}110$:

   $\text{required total} = 2{,}110 \times 8 = 16{,}880$

5. 5

   Subtract what the seven friends already contribute. Let $x$ be Ashley's expenses:

   $15{,}255 + x = 16{,}880 \;\Longrightarrow\; x = 1{,}625$

6. 6

   Check the direction and the arithmetic. Because the seven friends already average above the target, the eighth value must sit below it — and \1{,}625 < $2{,}110$✓. Verifying:$15{,}255 + 1{,}625 = 16{,}880$and$16{,}880 \div 8 = 2{,}110$ exactly ✓.