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=1405, p= 2 / 5
p = 2 / 5 = 0.4
q = 1 - p = 1 - 0.4 = 0.6
n = 1405
Using binomial distribution,
Mean = = n * p = 1405 * 0.4 = 562
Standard deviation = = n * p * q = 1405 * 0.4 * 0.6 = 9.0
minimum usual value = μ−2σ
minimum usual value = 562 - 2 * 9
minimum usual value = 544
maximum usual value = μ+2σ
maximum usual value = 562 + 2 * 9
maximum usual value = 580
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