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Using Chebysheff's theorm, you have concluded that at least 65.57% of the 2,803 runners took between...

Using Chebysheff's theorm, you have concluded that at least 65.57% of the 2,803 runners took between 56.7 and 87.1 minutes to complete the 10km race. What was the standard deviation of these 2,803 runners? (Please explain work, I have seen other solutions but they don't make sense)

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Answer #1

mean =(56.7+87.1)/2=71.9

here as we know that k std deviaton away values from mean =(1-1/k2)*100%

hence (1-1/k2)*100=65.57

k=sqrt(1/(1-0.6557))=1.7042

therefore standard deviation =(71.9-56.7)/1.7042=8.92

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