To determine how many times rolling a 6 would make you question your model, we can analyze the expected number of 6s and calculate the variability around this expectation.
1. Expected Value:
- The probability of rolling a 6 on a single roll is [tex]\(\frac{1}{6}\)[/tex].
- If you roll the cube 60 times, the expected number of times you should roll a 6 is given by:
[tex]\[
\text{Expected number of 6s} = \frac{1}{6} \times 60 = 10
\][/tex]
2. Variance and Standard Deviation:
- The variance ([tex]\(\sigma^2\)[/tex]) for a binomial distribution can be calculated using the formula:
[tex]\[
\sigma^2 = n \cdot p \cdot (1 - p)
\][/tex]
where [tex]\(n\)[/tex] is the number of trials, and [tex]\(p\)[/tex] is the probability of the event occurring.
- Plugging in the values, we get:
[tex]\[
\sigma^2 = 60 \times \frac{1}{6} \times \left(1 - \frac{1}{6}\right) = 60 \times \frac{1}{6} \times \frac{5}{6} \approx 8.33
\][/tex]
3. Standard Deviation:
- The standard deviation ([tex]\(\sigma\)[/tex]) is the square root of the variance:
[tex]\[
\sigma \approx \sqrt{8.33} \approx 2.89
\][/tex]
4. Unusual Values:
- A common rule is to consider any value more than 2 standard deviations away from the mean as unusual.
- The mean (expected number of 6s) is 10.
5. Lower and Upper Bounds:
- Lower bound:
[tex]\[
\text{Mean} - 2 \times \text{Standard Deviation} \approx 10 - 2 \times 2.89 \approx 4.23
\][/tex]
- Upper bound:
[tex]\[
\text{Mean} + 2 \times \text{Standard Deviation} \approx 10 + 2 \times 2.89 \approx 15.77
\][/tex]
Therefore, if you rolled the cube 60 times and observed the number of 6s, any count falling outside the range of approximately 4 to 16 (rounded from 4.23 to 15.77) would be considered unusual. Thus, getting fewer than 4 or more than 16 sixes would make you question your model.
