-aux08.probability distribution find a and b

[karush]

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 $1.5$

 

[Evgeny.Makarov]

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

why would this be $$1.5$$

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]

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 $$a+b =1.1$$ but still how do we get values for $$a$$ and $$b$$

 

[MarkFL]

...
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 $$1.5$$

It is instead:

$$P(0)+P(1)+P(2)+P(3)=1$$

and this gives, as you correctly found, $$a+b=0.6$$

Now, you also know:

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

or

$$a+3b=0.9$$

So, you have the linear system:

$$a+b=0.6$$

$$a+3b=0.9$$

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

 

[karush]

It is instead:

$$P(0)+P(1)+P(2)+P(3)=1$$

and this gives, as you correctly found, $$a+b=0.6$$

Now, you also know:

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

or

$$a+3b=0.9$$

So, you have the linear system:

$$a+b=0.6$$

$$a+3b=0.9$$

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

ok got.. $$a=0.45$$ and $$b=0.15$$

 

[karush]

the original table is lost
I was wondering if is possible to reconstruct it from the comments
mahalo

 

[Evgeny.Makarov]

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.
[math]
\begin{array}{c|c|c|c|c}
x & 0 & 1 & 2 & 3\\
\hline
P(X=x) & 0.1 & a & 0.3 & b
\end{array}
[/math]

 

[karush]

From post #4 it seems that the distribution is the following.
[math]
\begin{array}{c|c|c|c|c}
x & 0 & 1 & 2 & 3\\
\hline
P(X=x) & 0.1 & a & 0.3 & b
\end{array}
[/math]

much mahalo