Assume that a procedure yields a binomial distribution with n trials and the probability of success for one trial is p. Use the given values of n and p to find the mean μ and standard deviation σ. Also, use the range rule of thumb to find the minimum usual value μ - 2σ and
the maximum usual value μ + 2σ.
n = 1445, p = 2/5
I could use some help in figuring this out!
To find the mean (μ) and standard deviation (σ) of a binomial distribution with given values of n and p, we can use the following formulas:
μ = n * p
σ = sqrt(n * p * (1 - p))
Given:
n = 1445
p = 2/5
Using the formulas above, we can calculate the values:
μ = 1445 * (2/5)
= 289/5
= 57.8
σ = sqrt(1445 * (2/5) * (1 - 2/5))
= sqrt(1445 * 2/5 * 3/5)
= sqrt(17.16)
≈ 4.141
So, the mean (μ) is approximately 57.8 and the standard deviation (σ) is approximately 4.141.
To find the minimum usual value (μ - 2σ) and maximum usual value (μ + 2σ) using the range rule of thumb, we need to subtract and add two standard deviations from the mean, respectively.
Minimum usual value (μ - 2σ):
57.8 - 2 * 4.141
≈ 57.8 - 8.282
≈ 49.518
Maximum usual value (μ + 2σ):
57.8 + 2 * 4.141
≈ 57.8 + 8.282
≈ 66.082
So, the minimum usual value (μ - 2σ) is approximately 49.518 and the maximum usual value (μ + 2σ) is approximately 66.082.