Probability: 4 Coin Toss Cumulative Dist Table & Median

[chwala]

Homework Statement

(i) Construct the cumulative distribution table for the number of heads when the four coins are tossed. Coins are fair.
(ii) Find the Median.

Homework Equations

The Attempt at a Solution

(i)

x 0 1 2 3 4
f(x) 1/16 5/16 11/16 15/16 1

(ii)
how do we find the median? is it $1+2/2=1.5?$

 

[Ray Vickson]

Homework Statement

(i) Construct the cumulative distribution table for the number of heads when the four coins are tossed. Coins are fair.
(ii) Find the Median.

Homework Equations

The Attempt at a Solution

(i)

x 0 1 2 3 4
f(x) 1/16 5/16 11/16 15/16 1

(ii)
how do we find the median? is it $1+2/2=1.5?$

What is the meaning of "median"? How is it defined in your textbook or course notes (or whatever you use)?

 

[LCKurtz]

Also, $1 + 2/2\ne 1.5$.

 

[chwala]

What is the meaning of "median"? How is it defined in your textbook or course notes (or whatever you use)?

Find the Median of $x$ where $x$ denotes the random variable of the heads.

 

[chwala]

Also, $1 + 2/2\ne 1.5$.

seen the error, let me look at the working again, too much reading...

 

[Ray Vickson]

Find the Median of $x$ where $x$ denotes the random variable of the heads.

That is not what I asked you. Here you have just repeated the question in post #1, but that is not what I was seeking from you. I asked you what the word "median" actually means. If you have any kind of distribution (not necessarily the number of heads in 4 coin tosses), that distribution will have a median (or maybe, several "medians"), and I want you to tell me how they would be defined. Is there a formula? If I say that some distribution has a median of 4.84, what is that telling me about the distribution of probabillities?

 

[chwala]

That is not what I asked you. Here you have just repeated the question in post #1, but that is not what I was seeking from you. I asked you what the word "median" actually means. If you have any kind of distribution (not necessarily the number of heads in 4 coin tosses), that distribution will have a median (or maybe, several "medians"), and I want you to tell me how they would be defined. Is there a formula? If I say that some distribution has a median of 4.84, what is that telling me about the distribution of probabillities?

ok just give me a moment...

 

[Mark44]

how do we find the median? is it $1+2/2=1.5?$

seen the error, let me look at the working again, too much reading...

This is not complicated. As @LCKurtz wrote in post #3, $1 + 2/2 \ne 1.5$. Perhaps you meant $\frac {1 + 2} 2 = 1.5$, or without LaTeX, (1 + 2)/2 = 1.5. Without parentheses, 1 + 2/2 means 1 + 1 which equals 2.

 

[WWGD]

As a hint, do you know about percentiles in general, and maybe the median as percentile?

 

[chwala]

ok, my approach on this, on considering the probability distribution table for the above problem...
where, $F(X)$ is the cumulative distribution and $P(X)$ is the probability distribution,

$F(X) = P(X≤2) = 1/16 + 4/16 + 11/16 ≅0.7$
and
$P(X≥2) = 6/16 + 4/16 +1/16=11/16≅0.7$
therefore, the Median = 2.

 

[chwala]

As a hint, do you know about percentiles in general, and maybe the median as percentile?

I would be interested in seeing you use percentiles on this problem.

 

[Ray Vickson]

I would be interested in seeing you use percentiles on this problem.

Attached is a plot of the cumulative distribution function, together with the line at probability = 0.5. The median is the point at which the two graphs cross.

 

[chwala]

Attached is a plot of the cumulative distribution function, together with the line at probability = 0.5. The median is the point at which the two graphs cross.

Thanks Ray, is there another method that you can use apart from the graph. Thanks for your input.

 

[Ray Vickson]

Thanks Ray, is there another method that you can use apart from the graph. Thanks for your input.

Yes. The median is either at the 50th percentile (if there is one), or else is the point at which the CDF jumps from a value < 0.5 to a value > 0.5 as we go through that point. For example, in the above we have the cdf $F(x)$:
$$\begin{array}{cc}
x & F(x) \\
\hline
0 & 0.0625 \\
1 & 0.3125 \\
2 & 0.6875 \\
3 & 0.9375 \\
4 & 1
\end{array}$$
The function $F(x)$ jumps up from below 0.5 to above 0.5 as $x$ passes through $2$, so $x = 2$ is the median.

 

[StoneTemplePython]

since the coins are fair, this problem is equivalent to looking for the midpoint of Pascal's Triangle for n= 4. By symmetry you can eyeball it and see the midpoint at $x = 2$

In fact any symmetric distribution will have its midpoint as a median.

(Footnotes: if the first moment exists, then we also have midpoint = mean for symmetric distribution; medians in general are not unique so there can be a range of values associated with the median though.)

 

[Ray Vickson]

since the coins are fair, this problem is equivalent to looking for the midpoint of Pascal's Triangle for n= 4. By symmetry you can eyeball it and see the midpoint at $x = 2$

In fact any symmetric distribution will have its midpoint as a median.

(Footnotes: if the first moment exists, then we also have midpoint = mean for symmetric distribution; medians in general are not unique so there can be a range of values associated with the median though.)

I agree, of course, but the OP seemed to segue off into a question of finding the median in general, presumably also in cases where an obvious symmetry argument won't work anymore.

 

[chwala]

I agree, of course, but the OP seemed to segue off into a question of finding the median in general, presumably also in cases where an obvious symmetry argument won't work anymore.

so are we saying that there may be other values for the median the textbook answer gives $1.5$
that is $({1+2})/{2}$= $1.5$
could this be a solution?

 

[chwala]

As a hint, do you know about percentiles in general, and maybe the median as percentile?

I am waiting for your method in using percentiles as you indicated.

 

[chwala]

since the coins are fair, this problem is equivalent to looking for the midpoint of Pascal's Triangle for n= 4. By symmetry you can eyeball it and see the midpoint at $x = 2$

In fact any symmetric distribution will have its midpoint as a median.

(Footnotes: if the first moment exists, then we also have midpoint = mean for symmetric distribution; medians in general are not unique so there can be a range of values associated with the median though.)

I thought for symmetric distribution the values of the mean = mode = median. Is this statement correct?

 

[StoneTemplePython]

I thought for symmetric distribution the values of the mean = mode = median. Is this statement correct?

No. See my post above that you quoted re: "Footnotes". Thinking like this will get you in trouble with e.g. the Cauchy distribution.

Also there is no reason for the mode to coincide with median. It simply is not true in general. For a basic counter example that comes up in randomwalks see: https://en.wikipedia.org/wiki/Arcsine_distribution

 

[WWGD]

I am waiting for your method in using percentiles as you indicated.

Basically, the median is the 50-th percentile, so that half the data lies below it and the other half lies above it. The simplest example would be, for 1,2,3, 2 is the median.

 

[Klystron]

Basically, the median is the 50-th percentile, so that half the data lies below it and the other half lies above it. The simplest example would be, for 1,2,3, 2 is the median.

To expand on this method the following excerpt from Wiki introduces more terms associated with percentile measures. Note how the author like @WWGD defines median as the 50th percentile:

----- wikipedia -----------
The term percentile and the related term percentile rank are often used in the reporting of scores from norm-referenced tests. For example, if a score is at the 86th percentile, where 86 is the percentile rank, it is equal to the value below which 86% of the observations may be found (carefully contrast with in the 86th percentile, which means the score is at or below the value below which 86% of the observations may be found - every score is in the 100th percentile). The 25th percentile is also known as the first quartile (Q1), the 50th percentile as the median or second quartile (Q2), and the 75th percentile as the third quartile (Q3). In general, percentiles and quartiles are specific types of quantiles.
--------- end excerpt from https://en.wikipedia.org/wiki/Percentile

 

[Ray Vickson]

I thought for symmetric distribution the values of the mean = mode = median. Is this statement correct?

No. Consider the "discrete" example with $P(X=1)=P(X=5)= 0.4, P(X=2) = P(X = 4) = 0.1.$ The median is at $x = 3$, and there are two modes ($x=1$ and $x=5$). In this example, the median is not at the 50th percentile, because there is no 50th percentile at all (as happens also in your example that started this thread). In this case it happens to be true that the median equals the mean, but it is possible to devise examples where there is no mean at all, but there still modes and medians.

 

[chwala]

Basically, the median is the 50-th percentile, so that half the data lies below it and the other half lies above it. The simplest example would be, for 1,2,3, 2 is the median.

so this may not be correct statement, going with what Ray Vickson has mentioned...