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

因此，您获取总体中的每个数据点 - xi。

Exactly, if all these xi's are independent.

确切地说, 如果所有这些xi都是独立的。

The sum of the xi's is sufficient.

xi 的总就足够了。

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

如果 xi 是正态分布的, 这种情况就会发生。

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

辑，撤销，减去2倍均值乘以从i等于1到n的xi之。

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

歌曲也可以翻译，无论歌词是否被理解，它们都会传达情感。

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

所以这是 p 与 xi 的总 alpha 减去 1。

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

你可以直接说，好吧，对于每个xi我只需要执行这个操作。

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的β减去1。

Each of these, just so you can relate to the formula, each of that is xi minus the mean squared.

每个数据点，为了让你能与公式对应起来，都是xi减去均值的平方。

Some of the XIs is sufficient, and so I could just model some of the XIs.

一些 XI 就足够了，因此我可以只对一些 XI 进行建模。

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

对于 p，给定我的所有数据，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的抽样分布是什么?

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

而所有的xi都是n维的。

We see some xi's from a Bernoulli distribution.

我们从伯努利分布中看到了一些xi。

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.

这个分子是从i等于1到N的每个xi减去均值后的平方。最终结果是38.80。

Jan, what do you make of Stevie leaving Florian Wirtz on the bench saying he has no place for him in his XI if he were Arne Slott?

一月，你怎么看史蒂夫把弗洛里安·维尔茨放在替补席的决定？他说如果他是阿尔内·斯洛特，他的首发阵容里就没有维尔茨的位置。

I goes from 1 to n. 1 over, I'll just write down the sampling distribution, square root 2 pi sigma e to the minus 1 half xi minus x bar squared over sigma squared.

我从 1 到 n. 1， 我只写下采样分布，平方根 2 pi sigma e 到负 1 半 习 减去 x bar 的平方超过 sigma 的平方。

With parameter p, then we would have f, sum of the xi's, i goes from 1 to n, given p. This is going to be n choose sum of the xi's.

使用参数 p， 那么我们将得到 f，习 之，i 从 1 到 n，给定 p。 这将是 习 的 n 选择。

But basically all I need to know is X bar, some of the XIs over N. And I'll put an N in here to make you remember that's a function of N. It has something to do with N. N is important here.

基本我需要知道的是 X 条，一些 XI 超过 N。 我在这里放一个 N， 让你记住这是 N 的函数。 它与 N 有关。