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Sajjad1994
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Sajjad1994
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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?
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For question 3 I get 75%.
Please check if this is correct

3) Days on which the high temperature was less than 70 degrees accounted for approximately _______ of the days for which sales failed to exceed $1500.

- #days below 70f: 9
- #days sales<$1500: 12

- % of #days below 70f over #days sales<$1500: 9/12*100 = 75%.

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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 than

2) 
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
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Sajjad1994
Seems to me like an equal split for the second question. Where did I go wrong?
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for ii) one must note that although the split between the points seems to be equal, on higher temperature days the avg. revenue will be much higher than the average revenue for colder days, thus the answer would be more than half of the revenue will come from days with temperatures higher than 75 degrees.
Sajjad1994
The attached graph displays the daily sales figures for a beachfront ice cream shop for the month of June, with each data point representing the sales totals and the high temperature for the day.
Using the graph, please answer the questions that follow.

1. The median sales figure for the month of June was $1600.

2. half of the shop's sales revenue came on days when the high temperature was greater than 75 degrees.



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