Assume that a procedure yields a binomial distribution with n=1358
trials and the probability of success for one trial is p=.84
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.
The binomial distribution has the following mean (μ) and standard deviation (σ):
Mean, μ = n * p
Standard Deviation, σ = √(n * p * (1-p))
Where:
n = number of trials (1358 in this case)
p = probability of success for one trial (0.84 in this case)
First, let's calculate the mean (μ):
μ = 1358 * 0.84
Now, let's calculate the standard deviation (σ):
σ = √(1358 * 0.84 * (1-0.84))
Calculating these we get:
μ = 1358 * 0.84 = 1140.72
σ = √(1358 * 0.84 * 0.16) = √(183.168) ≈ 13.533
Using the range rule of thumb to find the usual values we find µ – 2σ and µ + 2σ:
Minimum usual value (μ – 2σ) = 1140.72 - 2 * 13.533 ≈ 1140.72 - 27.066 ≈ 1113.654
Maximum usual value (μ + 2σ) = 1140.72 + 2 * 13.533 ≈ 1140.72 + 27.066 ≈ 1167.786
These values mark the range for usual values for the binomial distribution. Rounding these to whole numbers, we get a minimum usual value of 1114 and a maximum usual value of 1168.
Therefore, the interval using square brackets is:
[1114, 1168]