6. In a genetics investigation, cat litters with ten kittens are studied which of three are male. The scientist selects three kittens randomly. Let $$ X $$ as the number of female kittens that being selected and construct probability distribution for $$ X $$ (you may use tree diagram to represent the above event). Based on the probability distribution obtained, find the: a) mean. b) standard deviation. 7. A box holds 5 whites and 3 black marbles. If two marbles are drawn randomly without replacement and $$ X $$ denoted the number of white marbles, $$ \begin{array}{ll} ext { a) Find the probability distribution of } X ext {. } \ ext { b) } & ext { Plot the cumulative frequency distribution (CFD) of } X ext {. }\end{array} $$

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6. In a genetics investigation, cat litters with ten kittens are studied which of three are male. The scientist selects three kittens randomly. Let $$ X $$ as the number of female kittens that being selected and construct probability distribution for $$ X $$ (you may use tree diagram to represent the above event). Based on the probability distribution obtained, find the: a) mean. b) standard deviation. 7. A box holds 5 whites and 3 black marbles. If two marbles are drawn randomly without replacement and $$ X $$ denoted the number of white marbles, $$ \begin{array}{ll}	ext { a) Find the probability distribution of } X 	ext {. } \ 	ext { b) } & 	ext { Plot the cumulative frequency distribution (CFD) of } X 	ext {. }\end{array} $$

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## **Problem 6**

**Context:**
- **Total Kittens:** 10 (3 males and 7 females)
- **Selection:** 3 kittens are selected randomly.
- **Variable $$ X $$:** Number of female kittens selected.

### **6.a) Calculating the Mean ($$ \mu $$)**

The mean of a hypergeometric distribution can be calculated using the formula:

$$
\mu = n \times \frac{K}{N}
$$

Where:
- $$ n = 3 $$ (number of trials/selections)
- $$ K = 7 $$ (number of favorable outcomes, i.e., female kittens)
- $$ N = 10 $$ (total number of kittens)

Plugging in the values:

$$
\mu = 3 \times \frac{7}{10} = 2.1
$$

**Answer:**  
The **mean** number of female kittens selected, $$ \mu $$, is **2.1**.

---

### **6.b) Calculating the Standard Deviation ($$ \sigma $$)**

The standard deviation for a hypergeometric distribution is derived from its variance ($$ \sigma^2 $$):

$$
\sigma^2 = n \times \frac{K}{N} \times \frac{N - K}{N} \times \frac{N - n}{N - 1}
$$

Substituting the known values:

$$
\sigma^2 = 3 \times \frac{7}{10} \times \frac{3}{10} \times \frac{7}{9} = \frac{441}{900} = 0.49
$$

Taking the square root to find the standard deviation:

$$
\sigma = \sqrt{0.49} = 0.7
$$

**Answer:**  
The **standard deviation** of the number of female kittens selected, $$ \sigma $$, is **0.7**.

---

## **Problem 7**

**Context:**
- **Box Contents:** 5 white marbles and 3 black marbles (Total = 8 marbles)
- **Selection:** 2 marbles are drawn randomly without replacement.
- **Variable $$ X $$:** Number of white marbles drawn.

### **7.a) Probability Distribution of $$ X $$**

Possible values of $$ X $$: 0, 1, 2

1. **$$ X = 0 $$** (No white marbles; both are black):
   $$
   P(X = 0) = \frac{\binom{3}{2}}{\binom{8}{2}} = \frac{3}{28} \approx 0.1071
   $$

2. **$$ X = 1 $$** (One white and one black marble):
   $$
   P(X = 1) = \frac{\binom{5}{1} \times \binom{3}{1}}{\binom{8}{2}} = \frac{15}{28} \approx 0.5357
   $$

3. **$$ X = 2 $$** (Both marbles are white):
   $$
   P(X = 2) = \frac{\binom{5}{2}}{\binom{8}{2}} = \frac{10}{28} \approx 0.3571
   $$

**Probability Distribution Table:**

| $$ X $$ | $$ P(X) $$   |
|--------|--------------|
| 0      | $$ \frac{3}{28} \approx 0.1071 $$ |
| 1      | $$ \frac{15}{28} \approx 0.5357 $$ |
| 2      | $$ \frac{10}{28} \approx 0.3571 $$ |

---

### **7.b) Cumulative Frequency Distribution (CFD) of $$ X $$**

The cumulative frequency distribution $$ F(X) $$ represents the probability that $$ X $$ is less than or equal to a specific value.

1. **$$ F(0) = P(X \leq 0) = P(X = 0) = \frac{3}{28} \approx 0.1071 $$**
2. **$$ F(1) = P(X \leq 1) = P(X = 0) + P(X = 1) = \frac{3}{28} + \frac{15}{28} = \frac{18}{28} \approx 0.6429 $$**
3. **$$ F(2) = P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = \frac{3}{28} + \frac{15}{28} + \frac{10}{28} = \frac{28}{28} = 1 $$**

**Cumulative Frequency Distribution Table:**

| $$ X $$ | $$ F(X) $$     |
|--------|----------------|
| 0      | $$ 0.1071 $$   |
| 1      | $$ 0.6429 $$   |
| 2      | $$ 1.0000 $$   |

**Graphical Representation:**

While I cannot draw the graph here, the CFD can be visualized as a step function:
- Start at $$ X = 0 $$ with $$ F(X) = 0.1071 $$
- Jump to $$ F(X) = 0.6429 $$ at $$ X = 1 $$
- Jump to $$ F(X) = 1.0000 $$ at $$ X = 2 $$

---

**Summary of Answers:**

- **6.a)** Mean ($$ \mu $$) = **2.1**
- **6.b)** Standard Deviation ($$ \sigma $$) = **0.7**
- **7.a)** Probability Distribution of $$ X $$:
  - $$ P(X=0) = \frac{3}{28} \approx 0.1071 $$
  - $$ P(X=1) = \frac{15}{28} \approx 0.5357 $$
  - $$ P(X=2) = \frac{10}{28} \approx 0.3571 $$
- **7.b)** Cumulative Frequency Distribution (CFD) of $$ X $$:
  - $$ F(0) = 0.1071 $$
  - $$ F(1) = 0.6429 $$
  - $$ F(2) = 1.0000 $$

---

If you need further clarification or assistance with constructing the tree diagram or graphical plots, feel free to ask!
**Problem 6:** - **Mean ($$ \mu $$) = 2.1** - **Standard Deviation ($$ \sigma $$) = 0.7** **Problem 7:** - **Probability Distribution of $$ X $$:** - $$ P(X=0) = 0.1071 $$ - $$ P(X=1) = 0.5357 $$ - $$ P(X=2) = 0.3571 $$ - **Cumulative Frequency Distribution (CFD) of $$ X $$:** - $$ F(0) = 0.1071 $$ - $$ F(1) = 0.6429 $$ - $$ F(2) = 1.0000 $$

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