Mathematics

Question

normal distributed mean of 200pounds and a standard deviation of 20 pounds what percent would be over 250 pounds

1 Answer

  • [tex]\mathbb P(X>250)=\mathbb P\left(\dfrac{X-200}{20}>\dfrac{250-200}{20}\right)=\mathbb P(Z>2.5)\approx0.0062=0.62\%[/tex]
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