algebra one i dont know what to type here lol

Question

Quinn mailed four postcards that each cost $0.25, five letters that each cost $0.44, two large envelopes that each cost $0.80, and one small package for $5.10. Which cost on the list of mailing fees on Quinn’s receipt would skew the mean?

the cost of a postcard

the cost of a letter

the cost of a large envelope

the cost of the small package

Answer 1

4P +5L + 2M +1B= 12 items

4P = 4(.25) = 1

5L= 5(.44) =2.2

2M =2(.8) = 1.76

1B = 5.1

total cost 1+2.2+1.76+5.1 = 10.06

mean cost 10.06/12 = $0.838333...= about 84 cents

median cost is 44 cents

another small package would skew the mean further to the right of the median

another postcard, letter or large envelope would move the mean closer to the median

5.1>.84

.8<.84

..25<.84

.44< .84

Answer 2

First, calculate the mean:

( 4(0.25)+5(0.44)+2(0.80)+1(5.10) ) / (4+5+2+1) = 9.9 / 12 = 0.825

Next, consider the effect of removal on each of the cost of a postcard, a letter, a large envelope, or a small package.

If a postcard was removed from the receipt, then the new mean would be

( 3(0.25)+5(0.44)+2(0.80)+1(5.10) ) / (3+5+2+1) = 9.65 / 11 = 0.88

If a letter was removed from the receipt, then the new mean would be

( 4(0.25)+4(0.44)+2(0.80)+1(5.10) ) / (4+4+2+1) = 9.46 / 11 = 0.86

If a large envelope was removed from the receipt, then the new mean would be

( 4(0.25)+5(0.44)+1(0.80)+1(5.10) ) / (4+5+1+1) = 9.1 / 11 = 0.83

If a small package was removed from the receipt, then the new mean would be

( 4(0.25)+5(0.44)+2(0.80)+0(5.10) ) / (4+5+2+0) = 4.8 / 11 = 0.44

As you can see, the item which caused the largest change in the mean (or skew) was the small package. That's because the small package was much more expensive than any of the rest of the items. For future problems like this one, try examining the values of your dataset and focusing on the outlier. Hope this helps!