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 mu and standard deviation sigma. Also, use the range rule of thumb to find the minimum usual value mu minus 2 sigma and the maximum usual value mu plus 2 sigma. nequals50, pequals0.3
mu equals 15 (Do not round.)
sigma equals (Round to one decimal place as needed.)
Solution :
Given that,
p = 0.3
q = 1 - p = 1 - 0.3 = 0.7
n = 50
Using binomial distribution,
Mean = = n * p = 50 * 0.3 = 15
Standard deviation = = n * p * q = 50 * 0.3 * 0.7 = 3.2
minimum usual value = - 2
minimum usual value = 15 - 3.2
minimum usual value = 11.8
maximum usual value = + 2
maximum usual value = 15 + 3.2
maximum usual value = 18.2
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