chebyshev theorem

Question

Americans spend an average of 3 hours per day online. If the standard deviation is 31 minutes, find the range in which at least 88.89% of the data will lie. Use Chebyshev’s theorem. Round your k to the nearest whole number.

Answer 1

By Chebyshevs's inequality, for a distribution with mean m and std deviation σ,

P[|X -m| > kσ ] ≤ (1/k²)

Or P[|X -m] <k(sigma)] = 1 - P[|X -m| > kσ] ≥ 1 - (1/k²)

1 - (1/k²) = 0.8889 => k = 1/√(1 -0.8889) = 3

The range is 3hr±3x31min = 3 hr ± 93 min i.e [1hr 27min, 4hr 33min]