And what we want to do is write this right here just in terms of xi's.

我们要做就是在这里用 xi 形式写下来。

You can just say, OK, for each xi I just perform this operation.

你以，，对于每个 xi 我只执行这个操作。

Edit, Undo, minus 2 times mu times the sum from i is equal to 1 to n of xi.

编辑，撤消，减去 2 乘以 mu 乘以 i 和等于 xi 1 到 n。

So you take each of the data points in the population-- xi.

因此，您获取总体中每个数据点 - xi。

The songs can also be translated, but they communicate emotionally whether xi or not the words are understood.

歌曲也以翻译，但无论歌词是否被，们都会传达情感。

And it turns out my problem set up, the XIs are normal is too strong of a statement.

事实证明，我问题设置，XI 是正常，这太强烈了。

I plug in the actual data, the XIs.

我插入实际数据，即 XI。

If the xi's were normally distributed, this just happens.

如果 xi 是正态分布, 这种情况就会发生。

If the xi's are not normally distributed, this eventually happens.

如果 习 不是正态分布，这最终会发生。

So this is p to the sum of the xi's plus alpha minus 1.

所以这是 p 与 xi 总和加上 alpha 减去 1。

1 Minus p to the n minus sum of the xi's plus beta minus 1.

1减去p到n减去xi和加上β减去1。

We see some xi's from a Bernoulli distribution.

我们从伯努利分布中看到了一些xi。

So this looks like some of the xi's plus alpha over n plus alpha plus beta.

所以这看起来像一些 习 加 alpha 超过 n 加 alpha 加 beta。

For p, given all of my data, this is going to be proportional to n choose some of the xi's.

对于 p，给定我所有数据，这将与 n 选择一些 习 成正比。

What's the sampling distribution for all of the xi's?

所有xi抽样分布是什么?

Exactly, if all these xi's are independent.

确切地, 如果所有这些xi都是独立。

But keep in mind, the xi's are not independent.

但请记住, 习不是独立。

But it's more than just a general artifact that all I need to know is the sum of the xi's.

但这不仅仅是一个通用人工制品，我只需要知道 习 总和。

And all of the xi's is n-dimensional.

而所有xi都是n维。

This numerator is the sum of each of the xi minus the mean squared from i equals 1 to N. And that ended up to be 38.80.

这个分子是每个 xi 减去 i 等于 1 到 N 均方总和。最后是 38.80。