Question

Say you ate 29 hot dogs in a hot dog eating contest. If the average number of hots dogs eaten was 17 and the standard deviation was 6, what percentage of the contestants ate more hot dogs than you?

Answer 1

here from Chebychev distribution ;minimum % of values K std deviation above mean=(1/(2k²))*100

as 29 is 2 std deviation above mean;

therefore percentage of the contestants ate more hot dogs than you =(1/(2*2²))*100=12.5%

(Note: if we assume normal approximation ; 2 std deviation above values =2.5% ; please revert for any clarification required)