Use the given values of nequals=93 and pequals=0.24 to find the maximum value that is significantly low, muμminus−2sigmaσ, and the minimum value that is significantly high, muμplus+2sigmaσ. Round your answer to the nearest hundredth unless otherwise noted.
Accepted Solution
A:
Answer:The maximum value that is significantly low is 14.0828The minimum value that is significantly high is 30.5572Step-by-step explanation:If we assume a binomial distribution, n is equal to 93 and p is equal to 0.24, the mean μ and the standard deviation σ are calculated as:μ = n*p = 93*0.24 = 22.32σ = [tex]\sqrt{n*p*(1-p)} =\sqrt{93*0.24*(1-0.24)} =4.1186[/tex]Then, the maximum value that is significantly low, μ−2σ, and the minimum value that is significantly high, μ+2σ, are equal to:μ − 2σ = 22.32 - 2(4.1186) = 14.0828μ + 2σ = 22.32 + 2(4.1186) = 30.5572