Let's analyze the given data and solve the problem step-by-step.
### Data Given:
[tex]\[
\begin{array}{|l|c|c|}
\hline
& \text{Under 35} & \text{35 and over} \\
\hline
\text{Male} & 345 & 380 \\
\hline
\text{Female} & 362 & 472 \\
\hline
\end{array}
\][/tex]
From this data, we can determine the necessary totals.
- Number of males under 35: [tex]\( 345 \)[/tex]
- Number of males 35 and over: [tex]\( 380 \)[/tex]
- Number of females under 35: [tex]\( 362 \)[/tex]
- Number of females 35 and over: [tex]\( 472 \)[/tex]
#### a) Total Population
First, let's find the total population:
[tex]\[
\text{Total population} = 345 + 380 + 362 + 472 = 1559
\][/tex]
#### b) Total Number of Males
Next, we find the total number of males:
[tex]\[
\text{Total males} = 345 + 380 = 725
\][/tex]
### i) Calculate [tex]\( P(M) \)[/tex]
The probability that a randomly chosen person is male, [tex]\( P(M) \)[/tex], is given by the ratio of the total number of males to the total population:
[tex]\[
P(M) = \frac{\text{Total males}}{\text{Total population}} = \frac{725}{1559} \approx 0.465
\][/tex]
### ii) Calculate [tex]\( P(M \text{ and } Y) \)[/tex]
The probability that a randomly chosen person is both male and under 35, [tex]\( P(M \text{ and } Y) \)[/tex], is given by the ratio of the number of males under 35 to the total population:
[tex]\[
P(M \text{ and } Y) = \frac{\text{Number of males under 35}}{\text{Total population}} = \frac{345}{1559} \approx 0.221
\][/tex]
### iii) Check for Independence
Events [tex]\( M \)[/tex] and [tex]\( Y \)[/tex] are independent if [tex]\( P(M \text{ and } Y) = P(M) \cdot P(Y) \)[/tex].
First, let's calculate [tex]\( P(Y) \)[/tex], the probability that a randomly chosen person is under 35:
[tex]\[
\text{Total under 35} = 345 + 362 = 707
\][/tex]
[tex]\[
P(Y) = \frac{\text{Total under 35}}{\text{Total population}} = \frac{707}{1559} \approx 0.454
\][/tex]
We have:
- [tex]\( P(M) \approx 0.465 \)[/tex]
- [tex]\( P(Y) \approx 0.454 \)[/tex]
- [tex]\( P(M \text{ and } Y) \approx 0.221 \)[/tex]
Now compute [tex]\( P(M) \cdot P(Y) \)[/tex]:
[tex]\[
P(M) \cdot P(Y) = 0.465 \times 0.454 \approx 0.211
\][/tex]
Since [tex]\( P(M \text{ and } Y) \neq P(M) \cdot P(Y) \)[/tex]:
[tex]\[
0.221 \neq 0.211
\][/tex]
The events [tex]\( M \)[/tex] and [tex]\( Y \)[/tex] are not independent. This means that the occurrence of one event affects the probability of the occurrence of the other.
### Summary of Answers:
i) [tex]\( P(M) \approx 0.465 \)[/tex]
ii) [tex]\( P(M \text{ and } Y) \approx 0.221 \)[/tex]
iii) The events [tex]\( M \)[/tex] and [tex]\( Y \)[/tex] are not independent because [tex]\( P(M \text{ and } Y) \neq P(M) \cdot P(Y) \)[/tex].
