Median of seven random numbers above 0.6

Question

Draw 7 independent numbers uniformly from [0,1]. The sample median is the 4th smallest (the middle one). What is the probability that this median exceeds 0.6?

Accepted Answer

The median (4th of 7) exceeds 0.6 exactly when at least 4 of the seven numbers land above 0.6. Each lands above 0.6 with probability 0.4, so the count above 0.6 is Binomial(7, 0.4). We need 4, 5, 6, or 7: P = \binom{7}{4}(0.4)^4(0.6)^3 + \binom{7}{5}(0.4)^5(0.6)^2 + \binom{7}{6}(0.4)^6(0.6) + \binom{7}{7}(0.4)^7. Over the denominator 5^7 = 78125, the numerators are 15120, 6048, 1344, and 128, summing to 22640: P = \frac{22640}{78125} = \frac{4528}{15625} \approx 0.290. Takeaway. For an odd sample of size 2m+1, the median beats t iff at least m+1 values do; the workload is just a longer binomi…

Median of seven random numbers above 0.6

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