Question

Let be a random variable with possible values of 1,2,3,4,5 and pdf P(x)x^2/55. Find the cumulative distribution function for X.

Answer 1

A probability and statistic problem

Answer 2

If \(p(x)\) denotes the pdf, \(P(x)\) denotes the cdf, and \(\mathbb{P}(x=k)\) the probability that \(x\) takes value \(k\), then
\[\large P(x)=\mathbb{P}(X\le k)=\sum_{x=-\infty}^kp(x)\]
You'll have 7 cases to consider.

For all \(x<1\), you have a cumulative probability because (presumably) the pdf is defined to be \(\dfrac{x^2}{55}\) for \(x=1,...,5\), and 0 elsewhere:
\[\large \begin{align*}P(x)&=\begin{cases}\displaystyle\sum_{x=-\infty}^00&\text{for }x<1\end{cases}\\\\
&=
\begin{cases}
0&\text{for }x<1
\end{cases}\end{align*}\]
For \(x=1\), you have a cumulative probability of
\[\large \begin{align*}P(x)&=\begin{cases}\displaystyle\sum_{x=-\infty}^00&\text{for }x<1\\\\
\displaystyle\sum_{x=-\infty}^1\frac{x^2}{55}&\text{for }x=1\end{cases}\\\\
&=
\begin{cases}
0&\text{for }x<1\\\\
\dfrac{1}{55}&\text{for }x=1
\end{cases}\end{align*}\]
For \(x=2\), you have
\[\large \begin{align*}P(x)&=\begin{cases}\displaystyle\sum_{x=-\infty}^00&\text{for }x<1\\\\
\displaystyle\sum_{x=-\infty}^1\frac{x^2}{55}&\text{for }x=1\\\\
\displaystyle\sum_{x=-\infty}^2\frac{x^2}{55}&\text{for }x=2\end{cases}\\\\
&=
\begin{cases}
0&\text{for }x<1\\\\
\dfrac{1}{55}&\text{for }x=1\\\\
\dfrac{1}{55}+\dfrac{4}{55}&\text{for }x=2
\end{cases}\end{align*}\]
and so on, up to the case where \(x>5\).

Answer 3

thank you much i was so loss about that problem lol