4. - / 19 points consider a discrete probability distribution function: $p(x)=\frac{c(9,x)}{205}$ which is valid for the values of $x = 0, 3, 6, 7$. (a) complete the following probability distribution table. (enter your probabilities to 4 decimal places.) \begin{tabular}{|c|c|} \hline value of $x$ & probability \\ \hline $x = 0$ & \\ \hline $x = 3$ & \\ \hline $x = 6$ & \\ \hline $x = 7$ & \\ \hline \end{tabular} (b) determine $p(x = 4)$. (give your answer to 4 decimals.) (c) determine $p(x \geq 4)$. (give your answer to 4 decimals.) (d) determine the expected value of $x$. (give your answer to 2 decimals.) (e) fill in the blanks for the formula you would use to calculate the variance of $x$. $\sigma^2 = (\quad - \quad)^2 \cdot \quad + (\quad - \quad)^2 \cdot \quad + (\quad - \quad)^2 \cdot \quad + (\quad - \quad)^2 \cdot \quad$ 5. - / 1 points

Question

Accepted Answer

### Part (a)
# Explanation:
## Step1: Recall combination formula
The combination formula is $ C(n, x)=\frac{n!}{x!(n - x)!} $, where $ n = 9 $ here. We will calculate $ C(9,x) $ for $ x = 0,3,6,7 $ and then divide by 205 to get the probability.

## Step2: Calculate $ p(0) $
For $ x = 0 $: $ C(9,0)=\frac{9!}{0!(9 - 0)!}=\frac{9!}{9!}=1 $. So $ p(0)=\frac{1}{205}\approx0.0049 $

## Step3: Calculate $ p(3) $
For $ x = 3 $: $ C(9,3)=\frac{9!}{3!(9 - 3)!}=\frac{9\times8\times7}{3\times2\times1}=84 $. So $ p(3)=\frac{84}{205}\approx0.4098 $

## Step4: Calculate $ p(6) $
For $ x = 6 $: $ C(9,6)=C(9,3) = 84 $ (since $ C(n,x)=C(n,n - x) $). So $ p(6)=\frac{84}{205}\approx0.4098 $

## Step5: Calculate $ p(7) $
For $ x = 7 $: $ C(9,7)=C(9,2)=\frac{9!}{2!(9 - 2)!}=\frac{9\times8}{2\times1}=36 $. So $ p(7)=\frac{36}{205}\approx0.1756 $

# Answer (for part a table):
| Value of $ X $ | Probability |
|------------------|-------------|
| $ x = 0 $       | $ 0.0049 $ |
| $ x = 3 $       | $ 0.4098 $ |
| $ x = 6 $       | $ 0.4098 $ |
| $ x = 7 $       | $ 0.1756 $ |

### Part (b)
# Explanation:
## Step1: Check valid $ X $ values
The valid values of $ X $ are $ 0,3,6,7 $. $ X = 4 $ is not in this set.

## Step2: Determine probability
For a discrete probability distribution, if $ x $ is not in the support (valid values) of $ X $, then $ P(X = 4)=0 $

# Answer:
$ 0.0000 $

### Part (c)
# Explanation:
## Step1: Identify $ X\geq4 $ values
The valid values of $ X $ are $ 0,3,6,7 $. So $ X\geq4 $ corresponds to $ X = 6,7 $

## Step2: Calculate $ P(X\geq4) $
$ P(X\geq4)=p(6)+p(7) $. We know $ p(6)\approx0.4098 $ and $ p(7)\approx0.1756 $. So $ 0.4098 + 0.1756=0.5854 $

# Answer:
$ 0.5854 $

### Part (d)
# Explanation:
## Step1: Recall expected value formula
The expected value $ E(X)=\sum x\cdot p(x) $

## Step2: Calculate each term
- For $ x = 0 $: $ 0\times0.0049 = 0 $
- For $ x = 3 $: $ 3\times0.4098 = 1.2294 $
- For $ x = 6 $: $ 6\times0.4098 = 2.4588 $
- For $ x = 7 $: $ 7\times0.1756 = 1.2292 $

## Step3: Sum the terms
$ E(X)=0 + 1.2294+2.4588 + 1.2292=4.9174\approx4.92 $

# Answer:
$ 4.92 $

### Part (e)
# Explanation:
## Step1: Recall variance formula
The variance formula for a discrete random variable is $ \sigma^{2}=\sum(x - \mu)^{2}\cdot p(x) $, where $ \mu = E(X) \approx4.92 $

## Step2: Fill in the blanks
For each $ x $ (0, 3, 6, 7) and corresponding $ p(x) $:

- First term: $ (0 - 4.92)^{2}\cdot0.0049 $
- Second term: $ (3 - 4.92)^{2}\cdot0.4098 $
- Third term: $ (6 - 4.92)^{2}\cdot0.4098 $
- Fourth term: $ (7 - 4.92)^{2}\cdot0.1756 $

# Answer:
$ \sigma^{2}=(0 - 4.92)^{2}\cdot0.0049+(3 - 4.92)^{2}\cdot0.4098+(6 - 4.92)^{2}\cdot0.4098+(7 - 4.92)^{2}\cdot0.1756 $

Value of \( X \)	Probability
\( x = 3 \)	\( 0.4098 \)
\( x = 6 \)	\( 0.4098 \)
\( x = 7 \)	\( 0.1756 \)

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QUESTION IMAGE

Question

Explanation:

Part (a)

Step1: Recall combination formula

Step2: Calculate \( p(0) \)

Step3: Calculate \( p(3) \)

Step4: Calculate \( p(6) \)

Step5: Calculate \( p(7) \)

Step1: Check valid \( X \) values

Step2: Determine probability

Step1: Identify \( X\geq4 \) values

Step2: Calculate \( P(X\geq4) \)

Answer:

Part (b)