Sure, let's solve this step-by-step.
We will use the properties of a binomial distribution to solve this problem. A binomial distribution is applicable here because:
1. We have a fixed number of trials (12 spins).
2. Each trial has two possible outcomes (either spinning a 1 or not).
3. The probability of success (spinning a 1) is constant on each trial.
4. The trials are independent of each other.
Step 1: Define the random variable
Let [tex]$X$[/tex] represent the number of times the spinner lands on 1. We are dealing with a binomial random variable because of the reasons stated above.
Step 2: Determine the parameters of the binomial distribution
The binomial distribution is characterized by two parameters [tex]\( n \)[/tex] and [tex]\( p \)[/tex]:
- [tex]\( n \)[/tex] is the number of trials, which is 12 (the number of times the spinner is spun).
- [tex]\( p \)[/tex] is the probability of success on each trial. Since the spinner is divided into six equal sectors, the probability of landing on any specific number (including 1) is [tex]\( \frac{1}{6} \)[/tex].
Step 3: Calculate the mean
The mean ([tex]\(\mu\)[/tex]) of a binomial distribution is given by the formula:
[tex]\[ \mu = n \cdot p \][/tex]
Substituting the values of [tex]\(n\)[/tex] and [tex]\(p\)[/tex]:
[tex]\[ \mu = 12 \cdot \frac{1}{6} = 2 \][/tex]
Step 4: Calculate the standard deviation
The standard deviation ([tex]\(\sigma\)[/tex]) of a binomial distribution is given by the formula:
[tex]\[ \sigma = \sqrt{n \cdot p \cdot (1 - p)} \][/tex]
Substituting the values of [tex]\(n\)[/tex] and [tex]\(p\)[/tex]:
[tex]\[ \sigma = \sqrt{12 \cdot \frac{1}{6} \cdot \left(1 - \frac{1}{6}\right)} \][/tex]
[tex]\[ \sigma = \sqrt{12 \cdot \frac{1}{6} \cdot \frac{5}{6}} \][/tex]
[tex]\[ \sigma = \sqrt{2 \cdot \frac{5}{6}} \][/tex]
[tex]\[ \sigma = \sqrt{\frac{10}{6}} \][/tex]
[tex]\[ \sigma = \sqrt{\frac{5}{3}} \approx 1.29 \][/tex]
Conclusion:
The mean ([tex]\(\mu_x\)[/tex]) of [tex]\(X\)[/tex] is 2, and the standard deviation ([tex]\(\sigma_x\)[/tex]) of [tex]\(X\)[/tex] is approximately 1.29.
Therefore, the answer is:
[tex]\[ \boxed{\mu_x=2, \sigma_x=1.29} \][/tex]
