The problem consists of several parts related to statistics and probability. We need to: A. Define statistical measures (CV, $\delta$, $\bar{x}$, R) - this part is omitted as it asks for definitions. B. Complete a table and represent the data as a pie chart - this part is omitted as the table is already completed and pie chart is drawn. C. Calculate the sample standard deviation of the observations: 6, 8, 9, 10, 12. D. (i) Calculate the probability that all 3 children in a family are girls, and write down the sample space $\Omega$. (ii) Calculate the probability that all 3 children in a family are girls, given that the first child is a girl. Write down the sample space $\Omega$. E. Define independent and mutually exclusive events and give examples - this part is omitted as it asks for definitions and examples. F. Given $P(A) = \frac{1}{2}$, $P(B) = \frac{1}{3}$ and $P(A \cap B) = \frac{1}{5}$, find $P(A^c)$, $P(B^c)$, $P(A \cup B)$, $P(A \cap B^c)$ and $P(A^c \cup B^c)$.

1. Problem Description

The problem consists of several parts related to statistics and probability. We need to:

A. Define statistical measures (CV, $\delta$, $\bar{x}$, R) - this part is omitted as it asks for definitions.

B. Complete a table and represent the data as a pie chart - this part is omitted as the table is already completed and pie chart is drawn.

C. Calculate the sample standard deviation of the observations: 6, 8, 9, 10,

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2. D. (i) Calculate the probability that all 3 children in a family are girls, and write down the sample space $\Omega$.

(ii) Calculate the probability that all 3 children in a family are girls, given that the first child is a girl. Write down the sample space $\Omega$.

E. Define independent and mutually exclusive events and give examples - this part is omitted as it asks for definitions and examples.

F. Given $P(A) = \frac{1}{2}$, $P(B) = \frac{1}{3}$ and $P(A \cap B) = \frac{1}{5}$, find $P(A^c)$, $P(B^c)$, $P(A \cup B)$, $P(A \cap B^c)$ and $P(A^c \cup B^c)$.

2. Solution Steps

C. Calculate the sample standard deviation of the observations: 6, 8, 9, 10,

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2. First, calculate the mean $\bar{x}$:

$\bar{x} = \frac{6 + 8 + 9 + 10 + 12}{5} = \frac{45}{5} = 9$

Next, calculate the sample variance $s^2$:

$s^2 = \frac{\sum_{i=1}^n (x_i - \bar{x})^2}{n-1} = \frac{(6-9)^2 + (8-9)^2 + (9-9)^2 + (10-9)^2 + (12-9)^2}{5-1}$

$s^2 = \frac{(-3)^2 + (-1)^2 + (0)^2 + (1)^2 + (3)^2}{4} = \frac{9 + 1 + 0 + 1 + 9}{4} = \frac{20}{4} = 5$

Finally, calculate the sample standard deviation $s$:

$s = \sqrt{s^2} = \sqrt{5} \approx 2.236$

D. (i) Calculate the probability that all 3 children in a family are girls.

Assume the probability of having a girl is $\frac{1}{2}$ and the probability of having a boy is $\frac{1}{2}$.

The sample space $\Omega = \{BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG\}$. There are 8 possible outcomes.

The event of interest is $GGG$.

$P(GGG) = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}$

(ii) Calculate the probability that all 3 children in a family are girls, given that the first child is a girl.

Let A be the event that all 3 children are girls.

Let B be the event that the first child is a girl.

We want to find $P(A|B) = \frac{P(A \cap B)}{P(B)}$.

The sample space $\Omega$ given the first child is a girl is $\{GBB, GBG, GGB, GGG\}$. There are 4 possible outcomes.

$P(A \cap B) = P(GGG) = \frac{1}{8}$

$P(B) = \frac{1}{2}$

$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{P(GGG)}{P(\text{first is girl})} = \frac{\frac{1}{8}}{\frac{1}{2}} = \frac{1}{8} \cdot 2 = \frac{1}{4}$.

Alternatively, since we know the first child is a girl, we want the probability that the next two children are girls. The probability is $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$.

New sample space $\Omega' = \{G BB, G BG, G GB, G GG\}$

F. Given $P(A) = \frac{1}{2}$, $P(B) = \frac{1}{3}$ and $P(A \cap B) = \frac{1}{5}$, find $P(A^c)$, $P(B^c)$, $P(A \cup B)$, $P(A \cap B^c)$ and $P(A^c \cup B^c)$.

$P(A^c) = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2}$

$P(B^c) = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3}$

$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{5} = \frac{15 + 10 - 6}{30} = \frac{19}{30}$

$P(A \cap B^c) = P(A) - P(A \cap B) = \frac{1}{2} - \frac{1}{5} = \frac{5 - 2}{10} = \frac{3}{10}$

$P(A^c \cup B^c) = P((A \cap B)^c) = 1 - P(A \cap B) = 1 - \frac{1}{5} = \frac{4}{5}$

3. Final Answer

C. Sample standard deviation: $\sqrt{5} \approx 2.236$

D. (i) $P(GGG) = \frac{1}{8}$, $\Omega = \{BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG\}$

(ii) $P(\text{3 girls } | \text{ first is girl}) = \frac{1}{4}$, $\Omega' = \{GBB, GBG, GGB, GGG\}$

F. $P(A^c) = \frac{1}{2}$, $P(B^c) = \frac{2}{3}$, $P(A \cup B) = \frac{19}{30}$, $P(A \cap B^c) = \frac{3}{10}$, $P(A^c \cup B^c) = \frac{4}{5}$