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[SOLVED] -aux08.probability distribution find a and b

karush

Well-known member
Jan 31, 2012
3,071
A discrete random variable $X$ has a probability distribution as shown in the table below.
1103
(a) Find the value of $a+b$
if the sum of $E[X] = 1$ then $1-0.1-0.3 =0.6=a+b$
(b) Given $E[X]=1.5$ find the value of a and b
why would this be $1.5$
 
Last edited:

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,541
(b) Given \(\displaystyle E[X]=1.5\), find the value of \(\displaystyle a\) and \(\displaystyle b\)

why would this be \(\displaystyle 1.5\)
I am not sure I understand the question. The fact that $E[X]=1.5$ is given. You don't question why $P(X=0)=0.1$, do you?

Just write what $E[X]$ is by definition for this particular $X$, and you'll get a second linear equation in $a$ and $b$ in addition to $a+b=0.6$.
 

karush

Well-known member
Jan 31, 2012
3,071
I am not sure I understand the question. The fact that $E[X]=1.5$ is given. You don't question why $P(X=0)=0.1$, do you?

Just write what $E[X]$ is by definition for this particular $X$, and you'll get a second linear equation in $a$ and $b$ in addition to $a+b=0.6$.
well that gives \(\displaystyle a+b =1.1\) but still how do we get values for \(\displaystyle a\) and \(\displaystyle b\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
...
if the sum of \(\displaystyle E[X] = \)1 then \(\displaystyle 1-0.1-0.3 =0.6=a+b\)

(b) Given \(\displaystyle E[X]=1.5\), find the value of \(\displaystyle a\) and \(\displaystyle b\)

why would this be \(\displaystyle 1.5\)
It is instead:

\(\displaystyle P(0)+P(1)+P(2)+P(3)=1\)

and this gives, as you correctly found, \(\displaystyle a+b=0.6\)

Now, you also know:

\(\displaystyle 0.1\cdot0+a\cdot1+0.3\cdot2+b\cdot3=E[X]=1.5\)

or

\(\displaystyle a+3b=0.9\)

So, you have the linear system:

\(\displaystyle a+b=0.6\)

\(\displaystyle a+3b=0.9\)

I would suggest beginning by subtracting the first equation from the second to eliminate $a$...
 

karush

Well-known member
Jan 31, 2012
3,071
It is instead:

\(\displaystyle P(0)+P(1)+P(2)+P(3)=1\)

and this gives, as you correctly found, \(\displaystyle a+b=0.6\)

Now, you also know:

\(\displaystyle 0.1\cdot0+a\cdot1+0.3\cdot2+b\cdot3=E[X]=1.5\)

or

\(\displaystyle a+3b=0.9\)

So, you have the linear system:

\(\displaystyle a+b=0.6\)

\(\displaystyle a+3b=0.9\)

I would suggest beginning by subtracting the first equation from the second to eliminate $a$...
ok got.. \(\displaystyle a=0.45\) and \(\displaystyle b=0.15\)
 

karush

Well-known member
Jan 31, 2012
3,071
the original table is lost
I was wondering if is possible to reconstruct it from the comments
mahalo
SSCtw.png
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,541
the original table is lost
I was wondering if is possible to reconstruct it from the comments
From post #4 it seems that the distribution is the following.
\(\displaystyle
\begin{array}{c|c|c|c|c}
x & 0 & 1 & 2 & 3\\
\hline
P(X=x) & 0.1 & a & 0.3 & b
\end{array}
\)
 

karush

Well-known member
Jan 31, 2012
3,071
From post #4 it seems that the distribution is the following.
\(\displaystyle
\begin{array}{c|c|c|c|c}
x & 0 & 1 & 2 & 3\\
\hline
P(X=x) & 0.1 & a & 0.3 & b
\end{array}
\)
much mahalo