Python 统计学入门
Maggie Matsui
Content Developer, DataCamp




期望值:概率分布的均值
公平骰子的期望值 = $(1 \times \frac{1}{6}) + (2 \times \frac{1}{6}) +(3 \times \frac{1}{6}) +(4 \times \frac{1}{6}) +(5 \times \frac{1}{6}) +(6 \times \frac{1}{6}) = 3.5$

$$P(\text{掷骰子}) \le 2 = ~?$$

$$P(\text{掷骰子}) \le 2 = 1/3$$


不均匀骰子的期望值 = $(1 \times \frac{1}{6}) +(2 \times 0) +(3 \times \frac{1}{3}) +(4 \times \frac{1}{6}) +(5 \times \frac{1}{6}) +(6 \times \frac{1}{6}) = 3.67$

$$P(\text{不均匀骰}) \le 2 = ~?$$

$$P(\text{不均匀骰}) \le 2 = 1/6$$

描述离散结果的概率

离散均匀分布

print(die)
number prob
0 1 0.166667
1 2 0.166667
2 3 0.166667
3 4 0.166667
4 5 0.166667
5 6 0.166667
np.mean(die['number'])
3.5
rolls_10 = die.sample(10, replace = True)
rolls_10
number prob
0 1 0.166667
0 1 0.166667
4 5 0.166667
1 2 0.166667
0 1 0.166667
0 1 0.166667
5 6 0.166667
5 6 0.166667
...
rolls_10['number'].hist(bins=np.linspace(1,7,7))
plt.show()


np.mean(rolls_10['number']) = 3.0

mean(die['number']) = 3.5

np.mean(rolls_100['number']) = 3.4

mean(die['number']) = 3.5

np.mean(rolls_1000['number']) = 3.48

mean(die['number']) = 3.5
随着样本量增大,样本均值将趋近期望值。
| 样本量 | 均值 |
|---|---|
| 10 | 3.00 |
| 100 | 3.40 |
| 1000 | 3.48 |
Python 统计学入门