F.5 Maths m1 distribution(2)

YTC

2015-05-11 1:51 am

question: http://postimg.org/image/ja9w8vz4n/
solution: http://postimg.org/image/52k7k8mfr/

I have some problems about part(ciii) 's solution.

The question asks for '' any psitive constant k ''.
Why in the solution, only put k=2√5 can already prove?
And why after adding 5 (blue words in the solution), the probabilities are still the
same?

Please help, thanks!!

回答 (1)

知足常樂

2015-05-11 3:14 am

✔ 最佳答案

你問了兩個問題：

(1)
The question asks for "any positive constant k".
Why in the solution, only put k=2√5 can already prove?

其實題目並不是 asks for "any positive constant k"。
題目是引用一個結果（那個結果叫 Chebyshev's inequality）。
這個結果是對於 any positive constant k，該句子都成立，那當然只看 k = 2√5 這個情況，句子都成立。

(2)
And why after adding 5 (blue words in the solution), the probabilities are still the same?

舉個例，{X = 2} 是一個 event，我們可以談論這個 event 的 probability。
{X = 2} 和 {X + k = 2 + k} 其實是同一個 event，故此有相同的 probability。

同理，P(-20 ≤ M - 5 ≤ 20) = P(-15 ≤ M ≤ 25) 是因為
{-20 ≤ M - 5 ≤ 20} 和 {-15 ≤ M ≤ 25} 是同一個 event。

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