Seven students report school-year discretionary expenses of 2800, 1990, 2005, 2400, 1860, 2200 and 2000 dollars, against an estimate of 2110 dollars. Find the mean, say whether it beats the estimate, and find what an eighth student's expenses would have to be for the eight values to average 2110.

Step-by-step solution

1. 1

   Add the seven values. Running total:

   $2800+1990=4790,\ +2005=6795,\ +2400=9195,$
   $+1860=11055,\ +2200=13255,\ +2000=15255.$

   The sum is \15{,}255$.

2. 2

   Divide by seven for the mean.

   $\bar{x}=\frac{15255}{7}=2179.2857\ldots\ \Longrightarrow\ \$2179.29.$

3. 3

   Compare with the estimate. \2179.29>$2110$, so these seven students spend **more** than the estimate — by about$$69$ each on average. Note this is a sample of friends, not a random sample, so it supports the suspicion but does not settle it statistically.

4. 4

   Set up part (c) using the total, not the mean. The key identity is $\text{mean}=\dfrac{\text{sum}}{\text{count}}$, so a target mean fixes a target sum. With eight values averaging \2110$:

   $\text{required total}=8\times 2110=\$16{,}880.$

5. 5

   Subtract the seven known values.

   $x=16880-15255=\$1625.$

6. 6

   Check the answer. Adding \1625$to the list gives a total of$$16{,}880$, and$16880\div 8=2110$✓. The value has to fall well below the others precisely because the seven friends already average$$69$ above the target, and one value must absorb all seven of those overshoots.