(IB18) A box contains 100 cards

[karush]

https://www.physicsforums.com/attachments/1141

(a) I presume the frequency row has to equal 100
so
\(\displaystyle k=100 - (26 + 10 + 20 + 29 + 11)= 4\)(b)(i) again presume the median is based on frequency and on ordered list
so
median of $4\ 10\ 11 \ 20\ 26\ 29 = \frac{31}{2}$ or $15.5$

(ii) interquartile range? isn't this data list 100 numbers long?
or is $Q_1=10$ and $Q_3=26$ so interquartile range$=26-10=13$

 

[MarkFL]

Re: (IB18) A box contains a 100 cards

For part b), you want to use the data, not the frequencies, in your calculations. To find the median (or $Q_2$) you observe that there is an even number of elements, so you take the arithmetic mean of the 50th and 51st elements.

Now, since there is an even number of elements in each half, you want to take the arithmetic mean of the 25th and 26th elements as $Q_1$ and the arithmetic mean of the 75th and 76th elements as $Q_3$. And then the inter-quartile range is given by:

\(\displaystyle IQR=Q_3-Q_1\)

 

[karush]

Re: (IB18) A box contains a 100 cards

For part b), you want to use the data, not the frequencies, in your calculations. To find the median (or $Q_2$) you observe that there is an even number of elements, so you take the arithmetic mean of the 50th and 51st elements.

Now, since there is an even number of elements in each half, you want to take the arithmetic mean of the 25th and 26th elements as $Q_1$ and the arithmetic mean of the 75th and 76th elements as $Q_3$. And then the inter-quartile range is given by:

\(\displaystyle IQR=Q_3-Q_1\)

I got 5-1=4 IQR

 

[MarkFL]

Re: (IB18) A box contains a 100 cards

I got 5-1=4 IQR

Yes, I got the same. :D

 

[CellCoree]

I would first clarify with the person providing the information to ensure that my understanding is correct. Assuming that the frequency row does indeed equal 100, I would then proceed to calculate the value of k as 4.

For part (b), I would also clarify if the data is indeed a list of 100 numbers. If it is not, and the median is based on the frequency and ordered list, then I would calculate the median as 15.5. However, if the data is a list of 100 numbers, then I would calculate the interquartile range as 13, based on the assumption that $Q_1=10$ and $Q_3=26$.

If there is any ambiguity or inconsistency in the information provided, I would communicate this to the person and ask for clarification. As a scientist, it is important to have clear and accurate data in order to make valid conclusions and interpretations.