# Answer on Statistics and Probability Question for Andrew

Question #475

Let 2S be the total weight of a set of bobs. Number k is called a mean if it is possible to choose k bobs with the total weight S. What is the greatest possible amount of means in a set of 100 bobs?

Expert's answer

Notice, if k is a mean, then 100 − k is also a mean. Thus, if k = 1 is not a mean, then k = 99 is also not a mean, and the maximum number of means is 97 (k ≠ 100). If k = 1 is a mean, then the weight of one bob is S and hence only k = 99 is also a mean. Thus, the maximum number of means is not greater than 97.

Here is an example of 100 bobs with the weights a1, a2, … , a100 for which all numbers from 2 to 98 (97 numbers in total) are means. Let a1 = a2 = 1, a(n+2) = an + a(n+1), n = 1, 2, … , 97, – serial numbers of Fibonacci and S = a1 + a2 + ⋯ + a98. Select a100 = S − a99. Then the total weight of all bobs is 2S and also a100 + a99 = a100 + a98 + a97 = a100 + a98 + a96 + a95 = ⋯ = a100 + a98 + a96 + a94 + a92 + ⋯ + a6 + a4 + a2 + a1 = S

Therefore, the numbers 2, 3, 4, ..., 51 are means. But then the numbers 100–2=98, 100–3 =97, … , 100–48=52 are also means, i.e. all numbers from 2 to 98 are means.

Answer. 97 means

Here is an example of 100 bobs with the weights a1, a2, … , a100 for which all numbers from 2 to 98 (97 numbers in total) are means. Let a1 = a2 = 1, a(n+2) = an + a(n+1), n = 1, 2, … , 97, – serial numbers of Fibonacci and S = a1 + a2 + ⋯ + a98. Select a100 = S − a99. Then the total weight of all bobs is 2S and also a100 + a99 = a100 + a98 + a97 = a100 + a98 + a96 + a95 = ⋯ = a100 + a98 + a96 + a94 + a92 + ⋯ + a6 + a4 + a2 + a1 = S

Therefore, the numbers 2, 3, 4, ..., 51 are means. But then the numbers 100–2=98, 100–3 =97, … , 100–48=52 are also means, i.e. all numbers from 2 to 98 are means.

Answer. 97 means

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