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# In the course of an experiment, 95 measurements were recorde

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In the course of an experiment, 95 measurements were recorde [#permalink]  23 Dec 2015, 18:53
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#GREpracticequestion In the course of an experiment, 95 measurements were recorded.jpg [ 22.67 KiB | Viewed 8961 times ]

In the course of an experiment, 95 measurements were recorded, and all of the measurements were integers.
The 95 measurements were then grouped into 7 measurement intervals.
The graph above shows the frequency distribution of the 95 measurements by measurement interval.

 Quantity A Quantity B The average (arithmetic mean) of the 95 measurements The median of the 95 measurements

A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.

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Question: 2
Page: 155
Difficulty: hard
[Reveal] Spoiler: OA

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Re: course of an experiment, 95 measurements were recorded [#permalink]  23 Dec 2015, 19:16
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Now here we have a frequency plot for 95 measurement. If we observe the first 3 columns of the graph we have 15, 35 and 15 in the ranges 1-5, 6-10, 11-15.

Now median of 95 is the 48th number so it must lie in 6 - 10 range.

Now the way to find average is to assume that all the numbers in a single column is the same i.e the mid point. So all the number of the column 1-5 is 3. So now we take these and multiply by frequency.

$$\frac{(3*15 + 35*8 + 15*13)}{(15 + 35+ 15)}$$ = 8 considering t just the first 3 columns now as we add more measurements of higher values the average increases further. Hence Average > Median for this experiment.

PS: A useful tip if the distribution is balanced i.e symmetric about the midpoint. The average is close to median. If it is skewed to left (like in the figure above) the average is greater than median and if its skewed to right the average is less than the median.
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Re: course of an experiment, 95 measurements were recorded [#permalink]  29 Mar 2016, 09:07
Now the way to find average is to assume that all the numbers in a single column is the same i.e the mid point. So all the number of the column 1-5 is 3. So now we take these and multiply by frequency.

^^^^ I don't get this part I thought 1-5 is about 13 according to graph, also where does 3*15 + 35 * 18 + 15 * 13 come from? where does the 15+35 + 15 come from?
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Re: course of an experiment, 95 measurements were recorded [#permalink]  20 Apr 2017, 05:46
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sandy wrote:

In the course of an experiment, 95 measurements were recorded, and all of the measurements were integers. The 95 measurements were then grouped into 7 measurement intervals. The graph above shows the frequency distribution of the 95 measurements by measurement interval.

 Quantity A Quantity B The average (arithmetic mean) of the 95 measurements The median of the 95 measurements

A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.

Tough/time-consuming question.

QUANTITY A
To find the mean, we must add all 95 values.
Notice that, that each interval, there are 5 possible values for each data point.
For example, the first entry is 1-5, which means the each of the 15 data points can be 1, 2, 3, 4, or 5 (NOTE: the data points are all INTEGERS)
So, which value do we use?
If all 15 data points are 1, then the mean will be different from the case where are 15 data points are 5.
So, let's see what happens if each data point is the MINIMUM POSSIBLE VALUE

1-5 interval: there are 15 data points. So if each data point is 1, then the SUM = (15)(1) = 15
6-10 interval: there are 35 data points. So if each data point is 6, then the SUM = (35)(6) = 210
11-15 interval: there are 15 data points. So if each data point is 11, then the SUM = (15)(11) = 165
16-20 interval: there are 12 data points. So if each data point is 16, then the SUM = (12)(16) = 192
21-25 interval: there are 10 data points. So if each data point is 21, then the SUM = (10)(21) = 210
26-30 interval: there are 5 data points. So if each data point is 26, then the SUM = (5)(26) = 130
31-35 interval: there are 3 data points. So if each data point is 31, then the SUM = (3)(31) = 93

So, the SUM of all 95 data points = 15 + 210 + 165 + 192 + 210 + 130 + 93
= 1015
So, MEAN = 1015/95 ≈ 10.6
Since we used the smallest possible value for each interval, we can conclude that the SMALLEST possible value of Quantity A is 10.6

QUANTITY B
The median of the 95 data points will be the MIDDLEMOST value when all of the values are arranged in ASCENDING ORDER
So, the median will equal the 48th value when all of the values are arranged in ASCENDING ORDER
In the 1-5 interval, there are 15 data points.
In the 6-10 interval, there are 35 data points.
At this point, we've already accounted for 50 data points (15 + 35 = 50).
This means the 48th value will be in the 6-10 interval.
So, the median of the set must be 6, 7, 8, 9, or 10

So, the largest possible value of Quantity B is 10
And, the smallest possible value of Quantity A is 10.6

So, we can be certain that Quantity A is greater.

[Reveal] Spoiler:
A

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Re: course of an experiment, 95 measurements were recorded [#permalink]  14 Nov 2018, 09:02
sagnik242 wrote:
Now the way to find average is to assume that all the numbers in a single column is the same i.e the mid point. So all the number of the column 1-5 is 3. So now we take these and multiply by frequency.

^^^^ I don't get this part I thought 1-5 is about 13 according to graph, also where does 3*15 + 35 * 18 + 15 * 13 come from? where does the 15+35 + 15 come from?

Yes, it is confusing to me too.
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Re: course of an experiment, 95 measurements were recorded [#permalink]  14 Nov 2018, 13:31
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AE wrote:
sagnik242 wrote:
Now the way to find average is to assume that all the numbers in a single column is the same i.e the mid point. So all the number of the column 1-5 is 3. So now we take these and multiply by frequency.

^^^^ I don't get this part I thought 1-5 is about 13 according to graph, also where does 3*15 + 35 * 18 + 15 * 13 come from? where does the 15+35 + 15 come from?

Yes, it is confusing to me too.

If you want to calculate the exact average of 95 values you need all 95 values explicitly. Now we do not have that but we know that 13 values of the 95 values lie between 1 and 5 (inclusive).

It could be 13 1s' or 13 5s' we dont know. Hence we take the average value from the range to calculate the values so $$13 *3$$, 3 being the average

$$13*3=39$$

Now in worst case the sum is $$13*1=13$$.

In best case the sum is $$13*5=65$$.

39 is exactly in the middle. It is not accurate but a close enough approximation.

It is important to note that in statistics we will always have 'losses'.

If we had all 13 numbers we could tell exactly all of their properties however we want to represent the whole data set with minimum numbers.

In this case 13 is the frequency (i.e. number of times) 1 is the lower bound and 5 is the upper bound. We described 13 number using only 3 numbers.
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Re: course of an experiment, 95 measurements were recorded [#permalink]  28 Apr 2019, 07:01
PS: A useful tip if the distribution is balanced i.e symmetric about the midpoint. The average is close to median. If it is skewed to left (like in the figure above) the average is greater than median and if its skewed to right the average is less than the median.[/quote]

Hello
Correct me if I'm wrong but the graph above is skewed TO RIGHT (not the left as you said)
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Re: course of an experiment, 95 measurements were recorded [#permalink]  28 Apr 2019, 08:38
The graph is skewed to the right, and the skew pulls the mean along with it, but generally leaves the median alone. So in this right-skewed case, the mean is greater than the median. Not hard, not time-consuming.

Hope this helps.
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Re: course of an experiment, 95 measurements were recorded [#permalink]  22 Sep 2019, 00:35
sandy wrote:
Now here we have a frequency plot for 95 measurement. If we observe the first 3 columns of the graph we have 15, 35 and 15 in the ranges 1-5, 6-10, 11-15.

Now median of 95 is the 48th number so it must lie in 6 - 10 range.

Now the way to find average is to assume that all the numbers in a single column is the same i.e the mid point. So all the number of the column 1-5 is 3. So now we take these and multiply by frequency.

$$\frac{(3*15 + 35*8 + 15*13)}{(15 + 35+ 15)}$$ = 8 considering t just the first 3 columns now as we add more measurements of higher values the average increases further. Hence Average > Median for this experiment.

PS: A useful tip if the distribution is balanced i.e symmetric about the midpoint. The average is close to median. If it is skewed to left (like in the figure above) the average is greater than median and if its skewed to right the average is less than the median.

The graph is skewed to the left or right? the below discussion confused me. What I see it is to the left because more number are clustered in 3 intervals
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Re: course of an experiment, 95 measurements were recorded [#permalink]  22 Sep 2019, 15:37
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Asmakan wrote:

The graph is skewed to the left or right? the below discussion confused me. What I see it is to the left because more number are clustered in 3 intervals

It's skewed to the right. You can think of it as the tail that's been pulled out. The pulling-out drags the mean with it, but doesn't affect the median.

HTH
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Re: In the course of an experiment, 95 measurements were recorde [#permalink]  01 Oct 2019, 08:35
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Some on this page have said that the median is not affected in a skewed distribution. It seems to me that when comparing a normal distribution to a skewed distribution the median is pulled as well as the mean. The mean is simply pulled more. https://www.youtube.com/watch?v=s6N_l3Bu-Mc&t=88s
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Re: In the course of an experiment, 95 measurements were recorde [#permalink]  01 Oct 2019, 09:36
arc601 wrote:
Some on this page have said that the median is not affected in a skewed distribution. It seems to me that when comparing a normal distribution to a skewed distribution the median is pulled as well as the mean. The mean is simply pulled more. https://www.youtube.com/watch?v=s6N_l3Bu-Mc&t=88s

In order to affect the median, you'd have to change the values of at least 50% of the entries. So not impossible, but quite unlikely.
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Re: In the course of an experiment, 95 measurements were recorde [#permalink]  01 Oct 2019, 10:09
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Kahn Academy has a good series on the effect of adding values to a data set or changing values on a data set and the effect on the mean and median: https://www.khanacademy.org/math/ap-sta ... hest-value

My takeaway was that
1) If a value on the end of a set is changed, then the mean changes but the median is unchanged.
2) If a value on the end of a set is added, then the mean changes AND the median changes.

The 2nd and 3rd attachments are from a video about skewed distributions in general.
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Screen Shot 2019-10-02 at 1.47.08 PM.jpg [ 100.49 KiB | Viewed 5592 times ]

pulledmedian.jpg [ 91.79 KiB | Viewed 5638 times ]

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Re: In the course of an experiment, 95 measurements were recorde [#permalink]  15 Oct 2019, 15:25
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Re: In the course of an experiment, 95 measurements were recorde   [#permalink] 15 Oct 2019, 15:25
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