We are given the probability distribution of a discrete random variable $X$. The possible values of $X$ are 1, 2, 4, and 8, with corresponding probabilities 0.3, 0.4, $a$, and $2a$. We need to find the value of $a$, the cumulative distribution function $F(x)$, the mode, the expected value $E[X]$ (denoted $M(X)$), the variance $Var(X)$ (denoted $D(X)$), and the standard deviation $\sigma(X)$.

1. Problem Description

We are given the probability distribution of a discrete random variable $X$. The possible values of $X$ are 1, 2, 4, and 8, with corresponding probabilities 0.3, 0.4, $a$, and $2a$. We need to find the value of $a$, the cumulative distribution function $F(x)$, the mode, the expected value $E[X]$ (denoted $M(X)$), the variance $Var(X)$ (denoted $D(X)$), and the standard deviation $\sigma(X)$.

2. Solution Steps

(1) Finding $a$:

Since the probabilities must sum to 1, we have:

$0.3 + 0.4 + a + 2a = 1$

$0.7 + 3a = 1$

$3a = 1 - 0.7$

$3a = 0.3$

$a = 0.1$

(2) Finding the cumulative distribution function $F(x)$:

$F(x) = P(X \le x)$

For $x < 1$, $F(x) = 0$.

For $1 \le x < 2$, $F(x) = P(X = 1) = 0.3$.

For $2 \le x < 4$, $F(x) = P(X = 1) + P(X = 2) = 0.3 + 0.4 = 0.7$.

For $4 \le x < 8$, $F(x) = P(X = 1) + P(X = 2) + P(X = 4) = 0.3 + 0.4 + 0.1 = 0.8$.

For $x \ge 8$, $F(x) = P(X = 1) + P(X = 2) + P(X = 4) + P(X = 8) = 0.3 + 0.4 + 0.1 + 2(0.1) = 0.3 + 0.4 + 0.1 + 0.2 = 1$.

So, the cumulative distribution function is:

$F(x) = \begin{cases}

0, & x < 1 \\

0.3, & 1 \le x < 2 \\

0.7, & 2 \le x < 4 \\

0.8, & 4 \le x < 8 \\

1, & x \ge 8

\end{cases}$

(3) Finding the mode:

The mode is the value of $X$ with the highest probability.

$P(X=1) = 0.3$

$P(X=2) = 0.4$

$P(X=4) = a = 0.1$

$P(X=8) = 2a = 0.2$

The highest probability is 0.4, which corresponds to $X = 2$. So, the mode is

2.

(4) Finding $M(X), D(X), \sigma(X)$:

$E[X] = M(X) = \sum x P(X=x) = 1(0.3) + 2(0.4) + 4(0.1) + 8(0.2) = 0.3 + 0.8 + 0.4 + 1.6 = 3.1$

$E[X^2] = \sum x^2 P(X=x) = 1^2(0.3) + 2^2(0.4) + 4^2(0.1) + 8^2(0.2) = 1(0.3) + 4(0.4) + 16(0.1) + 64(0.2) = 0.3 + 1.6 + 1.6 + 12.8 = 16.3$

$Var(X) = D(X) = E[X^2] - (E[X])^2 = 16.3 - (3.1)^2 = 16.3 - 9.61 = 6.69$

$\sigma(X) = \sqrt{Var(X)} = \sqrt{6.69} \approx 2.5865$

3. Final Answer

$a = 0.1$

$F(x) = \begin{cases}

0, & x < 1 \\

0.3, & 1 \le x < 2 \\

0.7, & 2 \le x < 4 \\

0.8, & 4 \le x < 8 \\

1, & x \ge 8

\end{cases}$

Mode $= 2$

$M(X) = 3.1$

$D(X) = 6.69$

$\sigma(X) \approx 2.5865$