F.5 Maths m1 distribution(2)

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)

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。


收錄日期: 2021-04-15 20:08:17
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