minimum and max usual value

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Use the given values of n and p to find the minimum usual value μ - 2σ and the maximum usual value μ + 2σ. Round your answer to the nearest hundredth unless otherwise noted.

 n = 103, p = 0.26  
Jun 17th, 2015

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the mean=n*p=103*0.26=26.78

standard deviation=sqrt(n*p*(1-p))

σ=sqrt(103*0.26*0.74)=sqrt(19.8172)=4.45165

minimum usual value is 26.78-2(4.45165)=17.88

maximum usual value i 26.78+2(4.45165)=35.68

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 17th, 2015

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