The problem provides a discrete random variable $X$ with its corresponding probabilities $P$. The probabilities are given in a table, with one unknown probability denoted by $c$. The goal is to determine the value of $c$, find the probability distribution function (PDF), draw its graph, calculate the expected value (mathematical expectation), variance, and standard deviation of $X$.

1. Problem Description

The problem provides a discrete random variable $X$ with its corresponding probabilities $P$. The probabilities are given in a table, with one unknown probability denoted by $c$. The goal is to determine the value of $c$, find the probability distribution function (PDF), draw its graph, calculate the expected value (mathematical expectation), variance, and standard deviation of $X$.

2. Solution Steps

First, we need to find the value of $c$. Since the sum of all probabilities must equal 1, we have:

$0.15 + 0.10 + 0.25 + 0.30 + c = 1$

$0.80 + c = 1$

$c = 1 - 0.80$

$c = 0.20$

Next, we find the probability distribution function (PDF). This is just the table provided, with the value of $c$ plugged in:

$P(X = -5) = 0.15$

$P(X = -3) = 0.10$

$P(X = -1) = 0.25$

$P(X = 1) = 0.30$

$P(X = 3) = 0.20$

Then we calculate the expected value (mathematical expectation) $E[X]$:

$E[X] = \sum_{i} x_i P(X = x_i)$

$E[X] = (-5)(0.15) + (-3)(0.10) + (-1)(0.25) + (1)(0.30) + (3)(0.20)$

$E[X] = -0.75 - 0.30 - 0.25 + 0.30 + 0.60$

$E[X] = -0.40$

Now, we calculate the variance $Var[X]$:

$Var[X] = E[X^2] - (E[X])^2$

First, find $E[X^2]$:

$E[X^2] = \sum_{i} x_i^2 P(X = x_i)$

$E[X^2] = (-5)^2(0.15) + (-3)^2(0.10) + (-1)^2(0.25) + (1)^2(0.30) + (3)^2(0.20)$

$E[X^2] = (25)(0.15) + (9)(0.10) + (1)(0.25) + (1)(0.30) + (9)(0.20)$

$E[X^2] = 3.75 + 0.90 + 0.25 + 0.30 + 1.80$

$E[X^2] = 7.00$

Now, we can compute the variance:

$Var[X] = E[X^2] - (E[X])^2$

$Var[X] = 7.00 - (-0.40)^2$

$Var[X] = 7.00 - 0.16$

$Var[X] = 6.84$

Finally, we calculate the standard deviation $\sigma_X$:

$\sigma_X = \sqrt{Var[X]}$

$\sigma_X = \sqrt{6.84}$

$\sigma_X \approx 2.615$

The graph of the PDF would have x-axis representing the values of X (-5, -3, -1, 1, 3) and y-axis representing the probabilities (0.15, 0.10, 0.25, 0.30, 0.20). It is a bar chart with bars at each x-value with heights equal to the corresponding probability.

3. Final Answer

$c = 0.20$

$E[X] = -0.40$

$Var[X] = 6.84$

$\sigma_X \approx 2.615$