Question

before the last school year began, it was estimated that the average discretionary personal expenses each school year for a student attending a 4 - year public college were $2,110. this past summer ashley decided to poll seven of her friends attending a 4 - year public college because she thought that estimate was low. she made a list of their actual school - year expenses: $2,800, $1,990, $2,005, $2,400, $1,860, $2,200, $2,000. a. what is the mean of her friends personal expenses? round your answer to the nearest cent. b. is the average higher or lower than the estimate? c. what would ashleys actual personal expenses for that school year have to be so that her amount and her friends amounts together would have an average of $2,110?

Explanation:

Step1: Calculate the sum of her friends' expenses

$2800 + 1990+2005 + 2400+1860+2200+2000=15255$

Step2: Calculate the mean of her friends' expenses

The number of friends is $n = 7$. Mean $\bar{x}=\frac{15255}{7}\approx2179.29$

Step3: Compare with the estimate

The estimate was $2110$. Since $2179.29>2110$, it is higher.

Step4: Let Ashley's expense be $x$.

We know that the new - average with Ashley included is $2110$ and the number of data - points is $n = 8$. The sum of her friends' expenses is $15255$. Using the formula for the mean $\bar{x}=\frac{\text{Sum of all values}}{\text{Number of values}}$, we have $2110=\frac{15255 + x}{8}$.

Step5: Solve for $x$

Multiply both sides of the equation $2110=\frac{15255 + x}{8}$ by $8$: $2110\times8=15255 + x$. So $16880=15255 + x$. Then $x=16880 - 15255=1625$

Answer:

a. $\$2179.29$
b. higher
c. $\$1625$