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, \( \begin{array}{ll}\text { a) Find the probability distribution of } X \text {. } \\ \text { b) } & \text { Plot the cumulative frequency distribution (CFD) of } X \text {. }\end{array} \)

Let's tackle each part of your questions systematically. --- ## **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!