Substance X is an ingredient in over 400 commercially available food p

Question

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Substance X is an ingredient in over 400 commercially available food products. The graph shows the per-serving amount of Substance X, in milligrams (mg), for each of the 10 commercially available food products—Foods A through J—that have the greatest per-serving amount of Substance X.

Based on the information provided, select from each drop-down menu the option that creates the most accurate statement.

For the 10 foods in the graph, the median per-serving amount of Substance X is  GRAPH_1_PLACEHOLDER  1.1 mg.

For all commercially available foods that contain Substance X, the average (arithmetic mean) per-serving amount of Substance X is  GRAPH_2_PLACEHOLDER  0.8 mg.

ID: 700249

chetan2u · Accepted Answer

Kudos for correct solution

Substance X is an ingredient in over 400 commercially available food products. The graph shows the per-serving amount of Substance X, in milligrams (mg), for each of the 10 commercially available food products—Foods A through J—that have the greatest per-serving amount of Substance X.

Based on the information provided, select from each drop-down menu the option that creates the most accurate statement.

For the 10 foods in the graph, the median per-serving amount of Substance X is ____1___ 1.1 mg. 
Check the height of the bar for all 10 foods. The height decreases as we move from left to right.
As there are 10 bar, the median will be the average of 5th and 6th bar. So, (1.0+1.1)/2 = 1.05
Thus,  Less than

For all commercially available foods that contain Substance X, the average (arithmetic mean) per-serving amount of Substance X is ___2___ 0.8 mg.

If you miss out word  For all commercially available foods,  you will go wrong.
The bars are the largest for 10 foods with max average per serving amount, so maximum the remaining 390 foods can be is the least of all 10, so 0.71

Thus, these 390 will be short of (0.8-0.71)*390 or around 35 points, while these 10 bar would not give more than 9-10 pointe above 0.8
Clearly the average will be less than 0.8
You can add the values given + 390*0.71 and divide by 400 to get the average.
 Answer: Less than

VishalDamarla · Answer

1st: less than 1.1

1st- Median is avg. of E & F i.e 1.1 and ~1 = ~1.05 which is less than 1.1

2nd: less than 0.8

2nd - All 400 servings have to be considered {(1.6 + 1.4 + 1.38 .... + 0.71) + 390*0.7}/ 400 = ~0.71 (Use the calculator to save time)

arpit4k · Answer

Hey  chetan2u  , is the answer 1st: less than & 2nd: greater than?
Based on:
1st- when we find median of the set by taking avg. of E & F i.e 1.1 and 1 = 1.05 which is less than 1.1
2nd - Adding up all instances of servings i.e (1.6 + 1.4 + 1.38 .... + 0.71) / 8 = comes out to be more than 0.8

chetan2u · Answer

Hi,

You are missing out something in the wordings. The second option is not for just these 10 but for all 400

lalittandon · Answer

chetan2u   KarishmaB 
for the 2nd part we have to see if the mean of 400 items is less than 0.8. We are given that the 10 highest items from 400. But the 11th -400 item could be 0.0001 or 0.71. Why are we assuming the minimum value of the remaining 390 observations to be 0.71?
Sorry if I missed something obvious.
Thanks

chetan2u · Answer

lalittandon

We are looking for an answer to the question Ís mean greater than 0.8mg.

So, we take the maximum possible values for remaining, which is 0.71. Inspite of the largest possible individual values, the average is less than 0.8mg. Thus, we can say that whatever be the values of reamining 390, the answer will be NO.

ruis · Answer

Official answer:

1) Since the bars are arranged in order from greatest to least, the median will be the average of the heights of Bars E and F. The height of Bar E is 1.1 and the height of Bar F is less than 1.1, so the average of the heights of the two bars is less than 1.1.

2) First, since the mean, by definition, is the sum of all the data values in the data set divided by the number of data values, it would be useful to estimate the sum of the heights of the bars representing the amounts per serving of Substance X in the 10 commercially available food products containing the greatest per-serving amounts of Substance X. Observe that the heights of Bars B and C appear to be 1.4. Both Bar D and Bar E are shorter than 1.4 and while Bar A appears to be taller than 1.4 by 0.2, Bar E appears to be shorter than 1.4 by 0.3, so estimating the heights of Bars A–E as 1.4 ensures that the sum of their heights is less than 5(1.4) = 7.0. Estimating the heights of each of Bars F–J as 1.0 ensures that the sum of their heights is less than 5(1.0) = 5. Thus, the sum of the heights of the bars representing the amounts per serving of Substance X in the 10 commercially available food products containing the greatest per-serving amounts of Substance X is less than 7.0 + 5.0 = 12.0.

If N is the number of commercially available products that contain Substance X, then N > 400. Let n be the number in excess of 400 so N = 400 + n. The heights of Bars A–J, in order, represent the amounts per serving, in milligrams, of Substance X in the 10 commercially available food products containing the greatest per-serving amounts of Substance X and the sum of these heights is less than 12. The least of these heights is about 0.71, so the rest of the (400 + n) − 10 = 390 + n commercially available food products containing Substance X have at most 0.71 mg of Substance X each. In the worst case, the mean of all of the per-serving amounts of Substance X in commercially available products would be at most

then the mean is less than 0.8.

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KarishmaB · Answer

We have 400 food products with X. We are given the data on 10 of them that have maximum X.

The amounts of X are 1.6, 1.4, 1.38, ... etc

For the 10 foods in the graph, the median per-serving amount of Substance X is _____ 1.1 mg.

Median will be the average of the middle two values E (1.1) and F (1.02) approximately. So the median will certainly be less than 1.1 mg.
 ANSWER - less than

For all commercially available foods that contain Substance X, the average (arithmetic mean) per-serving amount of Substance X is ____ 0.8 mg.

Now we are talking about all 400 foods. We know about 10 top foods. The other 390 foods have X in lower quantity than J i.e. equal to less than 0.72 mg. EVEN IF all 390 foods have 0.72 mg of X, the average amount of X MUST BE less than 0.8 mg. There are just 9 food items with more than 0.8 mg of X and their excess is very little ('A' has maximum excess of 0.8 above 0.8 mg). The 391 items have a deficit of at least 0.08 each so the overall deficit will be far higher and the average will be lower than 0.8 mg for sure.

ANSWER - less than

The concept of Arithmetic Mean using deviations is very useful to visualise this and arrive at the answer quickly. 
Check it here:  https://anaprep.com/arithmetic-usefulness-of-deviations/

lalittandon  - I think Chetan has already addressed your issue so not repeating here.

Gemmie · Answer

For the 10 foods in the graph, the median per-serving amount of Substance X is   1.1 mg.

The median is avg of the 5th and 6th

The 10 foods in already in order from the highest amount of substance X to the lowest

The 5 and 6th is Food E (1.1 mg X) and F (1 mg X)
=> The median must be less than 1.1

For all commercially available foods that contain Substance X, the average (arithmetic mean) per-serving amount of Substance X is   0.8 mg.

The 10 foods shown in the graph have the greatest amount of X out of 400 foods

Food J (having lowest amount of X out of 10 Foods) already have the amount of X below 0.8mg

=> The other 390 foods must also have the amount of X below 0.8mg

The amount of X in 9 foods having the highest amount of X is not much higher than 0.8 (the highest is only 1.6)

==> The avg of 400 foods must be less than 0.8

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