CEdward
kntombat
1) The median sales figure for the month of June was more than $1600.
There are a total of 28 points and the median has to be the \(\frac{14th point+15th point}{2}\) which is \(\frac{~1500$ + ~1750$}{2} = 1625$\). Therefore the OA is A.
2) More than half of the shop's sales revenue came on days when the high temperature was greater than 75 degrees.
Well after completing the set I did calculate just to be sure of what I did was correct, but the OA has to be A, you can see on the graph that as we cross 75 degrees, barring one point most of the other points are above $1700.
3) Days on which the high temperature was less than 70 degrees accounted for approximately 90% of the days for which sales failed to exceed $1500.
If you check the points before 70 degrees, you will find that all the 9 points are below 1500$. Therefore, the OA is C.
Hmmm for 2, I counted a 14-14 split? What did I miss?
It doest matter if is 14-14 split or 13-15 (see below my explanation for question 2)
1) To get whether the median is greater/lower/equal 1600, when you see the graph, there are only one point in the middle of it that has 1600 as ice cream sales. so mentally draw a line and count how any points are above and how many are below. after find out that 12 below and 15 above, therefore the median is higher than this.
Answer: greter than2)
The question asks whether more/less/equal half of the shop's sales revenue came on days when the high temperature was greater than 75 degrees. you don't need to make any calculation here too. there are about the same (it doesnt matter it is EXACTLY the same) amount of sales on the left of 75F and on the right of 75F. Given that there is a very positive trend, higher avg temperature => higher avg Sales. Since "avg revenue = # sales * avg sale price", the
answer: More than