Probability Question
2014-11-21 6:27 am
Find k such that P(|Z| ≤ k) = 0.97 where Z follows the standard normal distribution.
回答 (3)
2014-11-21 6:24 pm
✔ 最佳答案
Find k such that P(｜Z｜<=k)=0.97 where Z follows the standard normal distributionSol
P(｜Z｜<=k)=0.97
P(－k<=Z<=k)=0.97
P(k)－P(－k)=0.97
P(k)－[1－P(k)]=0.97
2P(k)－1=0.97
2P(k)=1.97
P(k)=0.985
k=2.17
2014-11-21 7:35 am
Yin 網友，應該不對。
考慮
P(|Z| ≤ k) = P(-k ≤ Z ≤ k) = 0.97
P(0 ≤ Z ≤ k) = 0.97/2 = 0.485
再查表。
2014-11-20 23:35:52 補充：
Yin 網友，應該不對。
考慮
P(|Z| ≤ k) = P(-k ≤ Z ≤ k) = 0.97
P(0 ≤ Z ≤ k) = 0.97/2 = 0.485
再查表。
考慮
P(|Z| ≤ k) = P(-k ≤ Z ≤ k) = 0.97
P(0 ≤ Z ≤ k) = 0.97/2 = 0.485
再查表。
2014-11-20 23:35:52 補充：
Yin 網友，應該不對。
考慮
P(|Z| ≤ k) = P(-k ≤ Z ≤ k) = 0.97
P(0 ≤ Z ≤ k) = 0.97/2 = 0.485
再查表。
2014-11-21 7:24 am
P(|Z| ≤ k) = 0.97
P(0≤Z ≤ k)+P(-k≤Z ≤ 0) = 0.97
by symmetry
P(-k≤Z ≤ 0) = 0.485
1 - P(Z≤-k) = 0.485
P(Z≤-k) = 0.515
k = -0.04
P(0≤Z ≤ k)+P(-k≤Z ≤ 0) = 0.97
by symmetry
P(-k≤Z ≤ 0) = 0.485
1 - P(Z≤-k) = 0.485
P(Z≤-k) = 0.515
k = -0.04
收錄日期: 2021-04-15 17:18:22
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20141120000051KK00118