Assume that a procedure yields a binomial distribution with n=297

Question

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.

Answer 1

For a binomial distribution, the mean (μ) and standard deviation (σ) are calculated using the formulas:

μ = n * p
σ = √(n * p * (1 - p))

For the given binomial distribution, with n = 297 trials and probability of success p = 0.37, we calculate the mean and standard deviation first:

μ = 297 * 0.37
μ = 109.89 (We will keep the decimal place for increased accuracy in further calculations)

σ = √(297 * 0.37 * (1 - 0.37))
σ = √(297 * 0.37 * 0.63)
σ = √(69.471)
σ = 8.334 (keeping decimals for accuracy)

Now, using the range rule of thumb to find the minimum (μ - 2σ) and the maximum (μ + 2σ) usual values:

Minimum: μ - 2σ = 109.89 - 2 * 8.334
Minimum: ≈ 109.89 - 16.668
Minimum: ≈ 93.222 (rounded to the nearest whole number: 93)

Maximum: μ + 2σ = 109.89 + 2 * 8.334
Maximum: ≈ 109.89 + 16.668
Maximum: ≈ 126.558 (rounded to the nearest whole number: 127)

Therefore, the interval of the minimum usual value to the maximum usual value, using whole numbers, is:

[93, 127]