What is the least number of rolls should a fair standard 6-sided die be done so that the recorded numbers that have been shown approximates a binomial distribution of standard deviation of 10 and a success occurs when a 4 is tossed?
To determine the least number of rolls needed for a fair 6-sided die to approximate a binomial distribution with a standard deviation of 10, we need to consider the properties of a binomial distribution and calculate the required parameters.
In a binomial distribution, the standard deviation (σ) is given by the formula:
σ = √(n * p * (1 - p))
Where:
- n is the number of trials
- p is the probability of success in each trial
In this case, a success occurs when a 4 is tossed, so the probability of success (p) is 1/6 (since there is one favorable outcome out of six possible outcomes).
We are given that the standard deviation (σ) is 10. Substituting these values into the formula, we get:
10 = √(n * (1/6) * (1 - 1/6))
To solve for n, we can square both sides of the equation:
100 = n * (1/6) * (1 - 1/6)
Simplifying further:
100 = n * (1/6) * (5/6)
Multiply both sides by 6/5 to isolate n:
n = (100 * (6/5))/(1/6)
Simplifying this expression:
n = (100 * 6 * 6)/(5 * 1)
n = 720
Therefore, the least number of rolls required for a 6-sided die to approximately follow a binomial distribution with a standard deviation of 10 is 720 rolls.