Lesson: Boxplots

Comment on Boxplots

800 insects were weighed and resulting measurements are given in a box plot having mean 118, Q1 114 and Q3 126 with start point 105 and range 41. If the 80th percentile of the measurements is 130 mg, about how many measurements are between 126 mg and 130 mg?

a) 41 b)42 c) 43 d) 40

answer is 40, but i am not getting how calculate?

Brent, found the answer for the question i asked earlier.

it is important to remember that Q3 will be at 75 percentile, therefore 126 is at 75th percentile. we have 5 percentile difference.

5 percent of 800 will give us the number of measurements.
5/100 * 800 = 40 :)
greenlight-admin's picture

Good work!

Adding to this note of Yogasuhas we must note that percentile divides a group into hundred. So Q1= 25 Percentile , Q2= 50 percentile and Q3= 75 percentile.
Also can solve the question in this way as well. since percentile is similar to percentage, so 80% of 800 is 640, so 640th measurement is 130mg. Now 75% is 126mg we know 75% of 800 = 600. So 600th measurement is 126mg. Number of measurements between 640th and 600th are: 640-600 =40 measurements.
greenlight-admin's picture

Perfect!

Abdul Hannan's picture

Hi Mr Hanneson,

what is Interquartile range ?

Can you explain with example.
greenlight-admin's picture

The interquartile range = (3rd quartile) - (1st quartile)

For an example, let's use the boxplot that appears at 3:40 in the above video.
The 3rd quartile = 8
The 1st quartile = 3
So, the interquartile range = 8 - 3 = 5

Likewise, in the boxplot that appears at 6:15 in the video, we get:
Interquartile range = 6 - 2 = 4

Does that help?

Cheers,
Brent

Hello Brent,
Please correct me if i am wrong, If we have a even set of number we find Q2 which will be the part of finding Q1 and Q3. If set is odd then we find Q2 and exclude it from finding Q1 and Q3.
greenlight-admin's picture

Kind of.
If we have a even number of values, then Q2 = the average of the 2 middlemost numbers. So, Q2 is not part of the lesser and greater numbers.

Consider this set {1, 2, 3, 4, 5, 6, 7, 8}
In this case, Q2 = 4.5
So, the lesser values are 1, 2, 3, 4, and the greater values are 5, 6, 7, 8
This means Q1 = 2.5 and Q3 = 6.5

Now consider this set {1, 2, 3, 4, 5, 6, 7, 8, 9}
In this case, Q2 = 5
So, the lesser values are 1, 2, 3, 4, and the greater values are 6, 7, 8, 9
This means Q1 = 2.5 and Q3 = 7.5

Does that help?

Cheers,
Brent

I don't understand yogasuhas example problem. He said in his first post "Q2 [is] 126" and then in the next post "126 is at 75th percentile". I don't understand how he knows 126 is at the 75th percentile. Can you please explain.
greenlight-admin's picture

Good catch, Kevin!

I think yogasuha meant to say that Q3 (not Q2) is 126
I've edited yogasuha's original comment.

By the way, here's the original question: https://greprepclub.com/forum/eight-hundred-insects-were-weighed-and-the...

Cheers,
Brent

Hi Brent,

I always make a mistake for this type of questions.
https://greprepclub.com/forum/for-the-31-used-cars-sold-last-month-at-car-dealership-x-3265.html

"For the 31 used cars sold last month at Car Dealership X, which of the following could be the median price?"

I did wrongly by ascending the number of cars sold from a high number to low.
6,7,8 and 10 respectively. Then, I find the 16th of the card for the median which ranks in 8, the price is between 7000$-10,000$.
I am very confused now, or probably my solution will correct if the question asked "the median car"?
greenlight-admin's picture

Question link: https://greprepclub.com/forum/for-the-31-used-cars-sold-last-month-at-ca...

Even if you accidentally arrange the values from highest to lowest, the median will still be the same.

We're looking for the 16th value.
So, if we arrange the values from highest to lowest, the first 6 values are greater than $10,000.
The next 8 values are between $7500 and $10,000
At this point we have accounted for 14 numbers.
To find the 16th number, we'll start adding the next set of values, which are between $5000 and $7499
So, the 16th value (the median) will be between $5000 and $7499

Does that help?

Cheers,
Brent

The odd no. of values which you explained at the 5.18min, do you mean if the total no. of the set which is 11 (assume odd no.) or Q2 = 5?
your reply is highly appreciated
greenlight-admin's picture

Whenever we have a set with an ODD number of values, then we don't include the median among the lesser and greater numbers.

Consider the following set of values: {2,2,3,4,7,8,8,10,11}
Here we have an odd number of values.
The median = 7, so Q2 = 7
The lesser numbers are: 2,2,3,4
The median of the lesser numbers is 2.5. So Q1 = 2,5
The greater numbers are: 8,8,10,11
The median of the greater numbers is 9. So, Q3 = 9

Does that help?

Dear brent,
I am a little confused should we exclude Q2 = median from lesser no. and greater no. all the time or just when our set of no. is odd?

thanks
greenlight-admin's picture

We exclude the median from the lesser and greater numbers ONLY when the set has an ODD number of values.

When we have an EVEN number of values, the median is the average of the two middlemost values.
Since the median lies between the two middlemost values, the median divides the values into two equal sets.

Consider this set: {1,1,2,3,4,6,7,8,9,9}
In this case we have an EVEN number of values (10 values).
The median of the 10 values is 5. So, Q2 = 5
Notice that the median divides the set into two groups of five values.
The lesser numbers are 1,1,2,3,4
And the greater numbers are 6,7,8,9,9

Does that help?

How did you come up with the 800 value in the section b part of this question? https://greprepclub.com/forum/eight-hundred-insects-were-weighed-and-the-resulting-measur-13743.html
greenlight-admin's picture

Solution link: https://greprepclub.com/forum/eight-hundred-insects-were-weighed-and-the...

The question tells us that "Eight hundred insects were weighed..."

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