Answer :
Sure, let's go through the solution step-by-step:
Given:
- [tex]\( n = 15 \)[/tex] (sample size)
- [tex]\( P = 0.6 \)[/tex] (proportion of students who like to visit the beach)
Here, [tex]\( x \)[/tex] follows a binomial distribution, [tex]\( x \sim \text{Binomial}(n=15, P=0.6) \)[/tex].
### Part (i): Mean, Variance, and Standard Deviation
1. Mean of [tex]\( x \)[/tex]:
[tex]\[ \text{Mean} = \mu = n \times P = 15 \times 0.6 = 9.0 \][/tex]
2. Variance of [tex]\( x \)[/tex]:
[tex]\[ \text{Variance} = \sigma^2 = n \times P \times (1 - P) = 15 \times 0.6 \times 0.4 = 3.6 \][/tex]
3. Standard Deviation of [tex]\( x \)[/tex]:
[tex]\[ \text{Standard Deviation} = \sigma = \sqrt{\sigma^2} = \sqrt{3.6} \approx 1.8973665961010275 \][/tex]
### Part (ii): Probability Calculations
1. (a) Probability [tex]\( P(x = 7) \)[/tex]:
[tex]\[ P(x = 7) \approx 0.11805577445376002 \][/tex]
2. (b) Probability [tex]\( P(x \geq 10) \)[/tex]:
[tex]\[ P(x \geq 10) \approx 0.40321555041484797 \][/tex]
3. (c) Probability [tex]\( P(x = 0) \)[/tex]:
[tex]\[ P(x = 0) \approx 1.0737418240000026 \times 10^{-6} \][/tex]
### Summary of Results
- Mean ([tex]\(\mu\)[/tex]): [tex]\( 9.0 \)[/tex]
- Variance ([tex]\(\sigma^2\)[/tex]): [tex]\( 3.6 \)[/tex]
- Standard Deviation ([tex]\(\sigma\)[/tex]): [tex]\( \approx 1.8973665961010275 \)[/tex]
For the probability calculations:
- [tex]\( P(x = 7) \approx 0.11805577445376002 \)[/tex]
- [tex]\( P(x \geq 10) \approx 0.40321555041484797 \)[/tex]
- [tex]\( P(x = 0) \approx 1.0737418240000026 \times 10^{-6} \)[/tex]
These steps and results provide a comprehensive breakdown of the problem's solution.
Given:
- [tex]\( n = 15 \)[/tex] (sample size)
- [tex]\( P = 0.6 \)[/tex] (proportion of students who like to visit the beach)
Here, [tex]\( x \)[/tex] follows a binomial distribution, [tex]\( x \sim \text{Binomial}(n=15, P=0.6) \)[/tex].
### Part (i): Mean, Variance, and Standard Deviation
1. Mean of [tex]\( x \)[/tex]:
[tex]\[ \text{Mean} = \mu = n \times P = 15 \times 0.6 = 9.0 \][/tex]
2. Variance of [tex]\( x \)[/tex]:
[tex]\[ \text{Variance} = \sigma^2 = n \times P \times (1 - P) = 15 \times 0.6 \times 0.4 = 3.6 \][/tex]
3. Standard Deviation of [tex]\( x \)[/tex]:
[tex]\[ \text{Standard Deviation} = \sigma = \sqrt{\sigma^2} = \sqrt{3.6} \approx 1.8973665961010275 \][/tex]
### Part (ii): Probability Calculations
1. (a) Probability [tex]\( P(x = 7) \)[/tex]:
[tex]\[ P(x = 7) \approx 0.11805577445376002 \][/tex]
2. (b) Probability [tex]\( P(x \geq 10) \)[/tex]:
[tex]\[ P(x \geq 10) \approx 0.40321555041484797 \][/tex]
3. (c) Probability [tex]\( P(x = 0) \)[/tex]:
[tex]\[ P(x = 0) \approx 1.0737418240000026 \times 10^{-6} \][/tex]
### Summary of Results
- Mean ([tex]\(\mu\)[/tex]): [tex]\( 9.0 \)[/tex]
- Variance ([tex]\(\sigma^2\)[/tex]): [tex]\( 3.6 \)[/tex]
- Standard Deviation ([tex]\(\sigma\)[/tex]): [tex]\( \approx 1.8973665961010275 \)[/tex]
For the probability calculations:
- [tex]\( P(x = 7) \approx 0.11805577445376002 \)[/tex]
- [tex]\( P(x \geq 10) \approx 0.40321555041484797 \)[/tex]
- [tex]\( P(x = 0) \approx 1.0737418240000026 \times 10^{-6} \)[/tex]
These steps and results provide a comprehensive breakdown of the problem's solution.
