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tetexe · 2020年09月20日

问一道题:NO.PZ2020010303000009

问题如下:

Either using a Z table or the Excel function NORM.S.DIST, compute

a. Pr(-1.5 < Z < 0), where Z ∼ N(0, 1)

b. Pr(Z < -1.5), where Z ∼ N(0, 1)

c. Pr(-1.5 < X < 0), where X ∼ N(1, 2)

d. Pr(X > 2), where X ∼ N(1, 2)

e. Pr(W > 12), where W ∼ N(3, 9)

选项:

解释:

a. 43.3%. In Excel, the command to compute this value is NORM.S.DIST(0,TRUE) – NORM.S.DIST(-1.5,TRUE).

b. 6.7%. In Excel, the command to compute this value is NORM.S.DIST(-1.5,TRUE).

c. 20.1%. In Excel, the command to compute this value is NORM.S.DIST((0 – 1)/SQRT(2),TRUE) – NORM.S.DIST((-1.5-1)/SQRT(2),TRUE).

d. 24.0%. In Excel, the command to compute this value is 1 – NORM.S.DIST((2 – 1)/SQRT(2),TRUE).

e. 0.13%. In Excel, the command to compute this value is 1 – NORM.S.DIST((12 – 3)/3,TRUE).

第三小问详细计算能不能列一下啊?还是没看明白。谢谢

2 个答案

袁园_品职助教 · 2023年04月26日

嗨,从没放弃的小努力你好:


你算的没错。这是由于这个答案是用excel拉的,算的比较精确,小数点保留的比较多而已。

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就算太阳没有迎着我们而来,我们正在朝着它而去,加油!

袁园_品职助教 · 2020年09月20日

同学你好!

第三问的分布是服从(1,2)的正态分布,所以0 到均值1的距离就是(0-1)/根号2 倍的标准差(后面公式的前半部分就是求负无穷到这个点的累计概率),-1.5到均值1的距离就是(-1.5-1)/根号2 倍的标准差(后面公式的后半部分就是求负无穷到这个点的累计概率)。

负无穷到0的累计概率减去负无穷到-1.5的累计概率,就是得到他俩之间的累计概率了。

大山 · 2023年04月24日

老师,为什么我第三问算出来是20.28%?

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