中央極限定理

Python 統計學入門

Maggie Matsui

Content Developer, DataCamp

擲骰 5 次

die = pd.Series([1, 2, 3, 4, 5, 6])

# Roll 5 times
samp_5 = die.sample(5, replace=True)
print(samp_5)

array([3, 1, 4, 1, 1])

np.mean(samp_5)

2.0

六面骰

# Roll 5 times and take mean
samp_5 = die.sample(5, replace=True)
np.mean(samp_5)

4.4

samp_5 = die.sample(5, replace=True)
np.mean(samp_5)

3.8

重複 10 次：

sample_means = []

for i in range(10):

  samp_5 = die.sample(5, replace=True)
  sample_means.append(np.mean(samp_5))

print(sample_means)

[3.8, 4.0, 3.8, 3.6, 3.2, 4.8, 2.6,
3.0, 2.6, 2.0]

樣本平均的抽樣分配

10 個樣本平均的長條圖

sample_means = []
for i in range(100):
    sample_means.append(np.mean(die.sample(5, replace=True)))

100 個樣本平均的長條圖

sample_means = []
for i in range(1000):
    sample_means.append(np.mean(die.sample(5, replace=True)))

1000 個樣本平均的長條圖

某統計量的抽樣分配，隨試驗次數增加，會越接近常態分配。

10、100、1000 個樣本平均的長條圖；數量越多，分配越像鐘形曲線

* 樣本需隨機且相互獨立

sample_sds = []
for i in range(1000):
  sample_sds.append(np.std(die.sample(5, replace=True)))

擲 5 次骰子之樣本標準差的 1000 次分配

sales_team = pd.Series(["Amir", "Brian", "Claire", "Damian"])

sales_team.sample(10, replace=True)

array(['Claire', 'Damian', 'Brian', 'Damian', 'Damian', 'Amir', 'Amir', 'Amir', 
      'Amir', 'Damian'], dtype=object)

sales_team.sample(10, replace=True)

array(['Brian', 'Amir', 'Brian', 'Claire', 'Brian', 'Damian', 'Claire', 'Brian', 
      'Claire', 'Claire'], dtype=object)

樣本比例的分配也近似常態

# Estimate expected value of die
np.mean(sample_means)

3.48

# Estimate proportion of "Claire"s
np.mean(sample_props)

0.26

樣本平均的抽樣分配，中線為虛線

Python 統計學入門