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

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)

­Hi,

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

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

 

­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.

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.­

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.
­

It seems like the graph is missing for me