For a given city, the graph represents the daily deviation, in degrees

Question

https://gmatclub.com/forum/download/file.php?id=73580 For a given city, the graph represents the daily deviation, in degrees Fahrenheit (°F), of the high temperature from the expected high temperature for each day in a 100-day period. Data is grouped into disjoint classes of deviations: for each value of T marked on the horizontal axis, the class centered at T includes all observed deviations greater than or equal to (T-2)°F but less than (T+2)°F. The height of each bar represents the number of deviations in the corresponding class. A given day's high temperature is x°F less than seasonal if it is x°F less than the left endpoint of the class centered at 0, and the high temperature is x°F greater than seasonal if it is x°F greater than the right endpoint of the class centered at 0. From each drop-down menu, select the option that creates the most accurate statement based on the information provided. For a randomly selected day in this 100-day period, the probability that the high temperature was more than 4°F less than seasonal is GRAPH_1_PLACEHOLDER and the probability that the high temperature was more than 8°F less than seasonal is GRAPH_2_PLACEHOLDER https://gmatclub.com/forum/download/file.php?id=81464 GMAT-Club-Forum-puu3xcxg.png 12.jpg

GMATCoachBen · Accepted Answer

Video solution here (3:38):

https://www.youtube.com/watch?v=O7-KVz8HX3w

It's essential to read carefully, to fully understand the graph and what it's asking for. We must find the specific definition of "4°F less than seasonal" in the passage.

KarishmaB · Answer

How to read the graph:

- for each value of T marked on the horizontal axis, the class centered at T includes all observed deviations greater than or equal to (T-2)°F but less than (T+2)°F.

This means that the bar centred at 0 goes from -2 to 2 (obvious from the graph) but the values it includes are -2 <= x < 2

- The height of each bar represents the number of deviations in the corresponding class.

The height of this bar centred at 0 is 34 so there were 34 days when the daily deviation was -2 to 2.

- A given day's high temperature is x°F less than seasonal if it is x°F less than the left endpoint of the class centered at 0, and the high temperature is x°F greater than seasonal if it is x°F greater than the right endpoint of the class centered at 0.

The left endpoint of class centred at 0 is -2 and right endpoint is 2.
A given day's high temperature is x°F less than seasonal if it is x°F less than -2 
A given day's high temperature is x°F greater than seasonal if it is x°F greater than 2.

Example, for a day, the high temperature was more than 4°F less than seasonal when it was less than -6.
and for a day, the high temperature was more than 4°F greater than seasonal when it was more than 6.

Question: For a randomly selected day in this 100-day period, the probability that the high temperature was more than 4°F less than seasonal is ___ and the probability that the high temperature was more than 8°F less than seasonal is ___

There are 100 days. 
4°F less than seasonal is values to the left of -6. There are 12 + 2 = 14 such values.
Probability = 14/100 = 0.14

8°F less than seasonal is values to the left of -10. There are 2 such values.
Probability = 2/100 = 0.02

ANSWER: 0.14, 0.02

Sixcarbs · Answer

I watched your video. GMAC is giving different answers.

They say it is 0.18 and .03.

I think it is easier than you make it out to be.

4 to the left is a bar with a height of 18, and 8 to the right is a bar with a height of 3. (You counted left for the second one, the second one is actually more not less.)

GMATCoachBen · Answer

Where is GMAC giving different answers? Do you have a screenshot?

Here's a screenshot of the question on exam 1 (taken on 6/16/23), with the same answers as above.

The reason we are counting to the left is that the stem says: "more than 8°F LESS than seasonal". And the passage gives the definition of "LESS than seasonal": "A given day's high temperature is x°F less than seasonal if it is x°F less than the left endpoint of the class centered at 0"

@ Sajjad1994  One little detail I notice in the screenshot below is that "less than seasonal" and "greater than seasonal" are italicized; this could be nice to add to the question above, since it's helpful in finding the definition efficiently. Italics are reasonably common on DI and can help save time.

egmat · Answer

There are at least 2 different versions of this question. I believe the datasets for these versions are the same. The difference stems from the question statement.

So, when you are solving questions and watching solutions, please ensure that you are watching the solution of the appropriate question stem.
Secondly, should different versions be posted as separate threads that are linked, or should they all be in the same thread since the dataset analysis will be common for all? @Sajjad1994.

Version 1: Question in the main post in this thread and in the solution posted by GMATCoachBen

Version 2: Question in Mock 4 that I believe is being referenced by Sixcarbs

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MS61 · Answer

This statement " "more than 8°F LESS than seasonal". And the passage gives the definition of "LESS than seasonal": "A given day's high temperature is x°F less than seasonal if it is x°F less than the left endpoint of the class centered at 0""
is very difficult to comprehend in such less time. I almost guessed this que because I found it very time consuming to comprehend. Any opinion on this ques wrt time and same experience?

egmat · Answer

This is a difficult GI question – from multiple aspects.
 
 The chart – a histogram – not your typical line or bar chart. Since we don’t handle histograms as much as we handle line or bar charts, the chart itself adds the layer of complexity. 
 The translation required to understand the histogram – For this one needs to visualize one of the datapoints. 
 Two definitions – faily complex in the dataset. If one does not use strategic pausing to understand these definitions, then solving this question will be very challenging. 
 The question statement – that uses the definition is also complex to understand.

https://www.youtube.com/watch?v=bVpz7yJuuXw

The only thing that is not complex in this question is the calculation of probability. That is the simplest aspect of the solution.

So, as you watch this solution, pay close attention to each of the above pointers and work towards building your skill set in ‘owning the dataset’.

Gemmie · Answer

For a randomly selected day in this 100-day period, the probability that the high temperature was more than 4°F less than seasonal is   and the probability that the high temperature was more than 8°F less than seasonal is

4°F less => - 4
more than 4°F less than seasonal => first two columns in the left: 2 + 12 = 14 => 0.14

8°F less => -8
more than 8°F less than seasonal => first column in the left: 2
=> 0.02

pranjalshah · Answer

This is a silly question but can someone clarify why we cannot perceive the statement "the high temperature was more than 4°F less than seasonal" as (the high temperature was more than) (4°F  less than seasonal ), instead of (the high temperature was more than 4°F) ( less than seasonal ).

Like for example, if I say "the temperature is more than 10°C less than room temperature".
If the Room Temp. is 25°C, then the temp is more than (10°C less than room temperature)
So, the temp is more than 15°C.

GMATCoachBen · Answer

pranjalshah  Interesting question. For the meaning you wrote above, "the temp is more than (10°C less than room temperature)", we would normally use the word HIGHER or ABOVE instead of "more".

Then, the meaning becomes: "the temp is higher than 15°C", or "the temp is above 15°C".

Oppenheimer1945 · Answer

I really have hard time understanding the meaning of more than less in mathematical terms. Otherwise this is a fairly simple ques of 605 level.

a smart way would be : Try to see both sides of histogram and calculate the probability pair for both cases as the ques is asked for same side of graph (left or right of histogram)

Pair would be either (0.14, 0.02) or (0.18 , 0.03).

First ones goes right .

callingTardis · Answer

Q1:
high temperature was more than 4°F less than seasonal
i.e deviation was more than -4°F
i.e it could be -5°F or -7°F
so it should be  2 days + 12 days + some more days (perhaps 18/2) 
so I marked 0.24

Q2:
high temperature was more than 8°F less than seasonal
i.e deviation was more than -8°F
i.e it could be -9°F or -10°F
so it should be  2 days + some more days (perhaps 12/2) 
so I marked 0.05

What am I missing?  KarishmaB

cheshire · Answer

When you actually go to compute those probabilities, you don’t “add some more days” – you just take the full counts shown in the bars and divide by 100.

For “more than 4 °F below seasonal”  you want every day with deviation < −4. That is exactly the two left-most bars (centered at −8 and −12) with counts 12 and 2. So:
P(deviation < −4) =  \frac{12+2}{100}  = 0.14

For “more than 8 °F below seasonal”  you want every day with deviation < −8. That’s only the bar at −12, count = 2, so:
P(deviation < −8) =  \frac{2}{100}  = .02

Dream009 · Answer

Here is my approach:

This is a tricky one but let me try to explain this to you - first trick is to read the question stem t-2 and t+2 = which means the x axis is showing us a mid point of the range - eg if the bar in axis is showing 4 - the bar is effectively for range t-2 and t+2 => the bar is looking at high temp in the range 2<x<6 => when 4 is mentioned. Second thing that you need to catch is the negative and positive sides from 0 - as per definition for less than seasonal - we should only look at left side of the graph - so right side is not in your scope both blanks speak about left. If you are with me till here now things will be easy

4 less than seasonal means - I need range of high temp in bars flagged as -8 and- 12. I cannot add the -4 bar because effectively the bar is for range -4-2<x<-4+2 =.> -6 < high temp < -2 and the bar 8 would be -8-2 < x< -8 +2 => -10< high temp <-6 see in the first case -3 will also be added which will extrapolate answer

thus I will add only 2 bars and get 14/100 = .14 - I hope it helps

Apply same logic to the second blank

parkhydel

liberoasperiores · Answer

More than 4 means 5 6 7 etc.. so subtract these values from -2 which is the left end of zero. you get -7 -8 and so on. These are part of the last two boxes.

Frost777 · Answer

Hi, in the question it asks "At least 4 degrees greater" so will the bar with 16 not be counted, same for the other part of the question as well?

Bunuel · Answer

No. The bar labeled 16 includes deviations below 6°F. Since “at least 4°F greater than seasonal” starts at 6°F, that bar is not counted.

P.S. I merged the previous discussion with the more detailed one above. You can find detailed solutions to this question in the earlier posts in this thread.

egmat · Answer

Hi Frost777,

Good instinct. Yes, the bar with count  16  (the class centered at T =  4 ) is  not  counted for the "at least 4°F greater than seasonal" part. Let me show you why, because it comes straight from the definition.

"Greater than seasonal" is measured from +2, not from 0

The stimulus says a day is  x°F greater than seasonal  when its deviation is x°F above the  right endpoint of the class centered at 0 . That class covers deviations from  -2  up to (but not including)  +2 , so its right endpoint is  +2°F .

So "at least  4 °F greater than seasonal" means:

- deviation ≥  2  +  4  =  6°F

Now look at where each bar sits:

- Class centered at  4  covers  2 ,  6 ) - every value here is  below 6 , so it's excluded. That's your bar of  16 .
- Class centered at  8  covers  6 ,  10 ) -  15  days
- Class centered at  12  covers  10 ,  14 ) -  3  days

Total =  15  +  3  =  18  -  18 / 100  =  0.18 .

Same logic for the second dropdown

"At least  8 °F greater than seasonal" - deviation ≥  2  +  8  =  10°F .

- Class centered at  8  covers  6 ,  10 ) - all below  10 , so  excluded  (its  15  days drop out).
- Only the class centered at  12 , covering  10 ,  14 ) -  3  days, qualifies.

So the probability is  3 / 100  =  0.03 .

Yes - the exclusion idea you spotted applies both times. The key is to first convert "x°F greater than seasonal" into an actual deviation cutoff (add x to  +2 ), then keep only the classes whose entire range clears that cutoff.

Answer: 0.18

For a given city, the graph represents the daily deviation, in degrees

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For a given city, the graph represents the daily deviation, in degrees

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