Assume that a procedure yields a binomial distribution with n=297
trials and the probability of success for one trial is p=37%
Use the range rule of thumb to find the minimum usual value μ–2σ and the maximum usual value μ+2σ.
Enter answer as an interval using square-brackets only with whole numbers.
To find the minimum and maximum usual values within a binomial distribution using the range rule of thumb, we first need to calculate the mean (μ) and standard deviation (σ) of the distribution.
The mean (μ) of a binomial distribution is given by:
μ = n * p
where n is the number of trials and p is the probability of success on a single trial. Plugging in the values we have:
μ = 297 * 0.37
μ = 109.89
Since we usually round to whole numbers when dealing with the binomial distribution, we'll take μ = 110 for practical purposes.
The standard deviation (σ) of a binomial distribution is given by:
σ = sqrt(n * p * q)
where q is the probability of failure (q = 1 - p). Plugging in the values:
q = 1 - p
q = 1 - 0.37
q = 0.63
σ = sqrt(297 * 0.37 * 0.63)
σ = sqrt(297 * 0.23271)
σ = sqrt(69.11457)
σ ≈ 8.31
The range rule of thumb suggests that most values should fall within the range of μ ± 2σ. Therefore, we can calculate the minimum usual value and maximum usual value as:
Minimum usual value (μ – 2σ):
Minimum = 110 - 2 * 8.31
Minimum ≈ 110 - 16.62
Minimum ≈ 93.38
We round to the nearest whole number for practical purposes:
Minimum ≈ 93
Maximum usual value (μ + 2σ):
Maximum = 110 + 2 * 8.31
Maximum ≈ 110 + 16.62
Maximum ≈ 126.62
We round to the nearest whole number for practical purposes:
Maximum ≈ 127
Therefore, the interval for the minimum usual value to the maximum usual value within the binomial distribution using the range rule of thumb is [93, 127].