Three sigma rule

Stomach weights are normally distributed, with a mean of 1314g and a standard deviation of 113g.

State the probability that a randomly selected stomach weighs more than 1118g.

(Report the probabilities using at least four decimal places. )

Correct answer:

p =  0.9586

Step-by-step explanation:

m=1314 g σ=113 g  p = N(x>1118)  m1=mσ=1314113=1201 m2=m1σ=1201113=1088 m3=m2σ=1088113=975 m2 < 1118 < m1  689599.7 rule  p0=0.5+0.95/2=4039=0.975  p=0.9586

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