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楼主 |
发表于 31-7-2010 05:05 PM
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发表于 31-7-2010 05:10 PM
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回复 201# harry_lim
有吗?!据我所知全都是essay。如果有的话应该不多吧~ |
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楼主 |
发表于 31-7-2010 05:16 PM
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发表于 31-7-2010 06:18 PM
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回复 203# harry_lim
,原来你也不清楚...哈哈 |
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楼主 |
发表于 31-7-2010 06:22 PM
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发表于 31-7-2010 06:28 PM
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回复 205# harry_lim
不好意思哦,。。。
未老先衰啦 |
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楼主 |
发表于 31-7-2010 06:29 PM
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回复 harry_lim
不好意思哦,。。。
未老先衰啦
佩琪 发表于 31-7-2010 06:28 PM 
你是明年的考生??? |
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发表于 31-7-2010 06:35 PM
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回复 207# harry_lim
是的~不过学校小考也要来了...怕~
因为没试过那么多科一起不及格~哎 |
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楼主 |
发表于 31-7-2010 06:36 PM
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发表于 31-7-2010 06:40 PM
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回复 209# harry_lim
尽量吧~我要加油!你也是哦。。。 |
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楼主 |
发表于 31-7-2010 06:43 PM
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发表于 31-7-2010 06:45 PM
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回复 211# harry_lim
这我知道~你考过UPSR,PMR,SPM 和STPM... |
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楼主 |
发表于 31-7-2010 09:37 PM
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发表于 31-7-2010 09:52 PM
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当然当然 ,可否请教楼主数学2的参考书那本较好?Q&A还是练习题? |
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楼主 |
发表于 31-7-2010 10:17 PM
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发表于 1-8-2010 12:02 AM
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回复 215# harry_lim
哦~谢谢咯。。。 |
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楼主 |
发表于 1-8-2010 12:04 AM
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发表于 2-8-2010 12:17 PM
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怎么开帖没人问问题啊???
小弟有个问题不会做,想请各位大大帮忙...
Sampling and Estimation
Greyfriars School is very proud of the athletic achievements of its students. The times(in seconds) taken by a random sample of 100 students aged 15 years, to run a mile are given by sigma x = 8000, sigma x^2 = 2569000. Get an unbiased estimate for the variance of this population. By assuming that the true value of the variance of this population is 400s^2 and the times can be taken to be independent observations from a normal distribution, give the 94% symmetric confidence interval for the mean of this distribution. |
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发表于 2-8-2010 02:38 PM
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回复 218# kelfaru
a) The unbiased estimator for population variance is S^2 = sigma (X - X_bar)^2 / (n-1) = (sigma X^2 - nX_bar^2) / (n-1), where X_bar is sample mean.
Thus, the unbiased estimate for population variance is S^2 = (2569000 - 100(80^2)) / (100-1) = ???
b) Suppose that the population is a normal distribution with variance = 400s^2. Then we can write as N(u, 400), where u is the population mean. It follows that the sampling distribution of the sample mean X_bar is N_sampling(u, 400/100 = 4).
Now, P(-z_0.06/2 <= Z <= z_0.06/2) = 0.94 implies that z_0.06/2 = ??. The 94% symmetric confidence interval is
-z_0.06/2 <= (X_bar - u)/2 <= z_0.06/2 implies that 2(-z_0.06/2) <= X_bar - u <= 2(z_0.06/2) implies that
X_bar - 2(-z_0.06/2) <= u <= X_bar + 2(z_0.06/2) |
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发表于 2-8-2010 02:45 PM
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回复 218# kelfaru
计算的部分,就留个你吧 :p |
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