$$P(A|B) = \frac{P(B|A)P(A)}{P(B)}$$
$$\sigma^2 = \frac{\sum (x_i - \mu)^2}{N}$$
$$\bar{x} = \frac{\sum x_i}{N}$$
$$Z = \frac{X - \mu}{\sigma}$$
$$R^2 = 1 - \frac{SS_{res}}{SS_{tot}}$$
$$y = \beta_0 + \beta_1 x_1 + \epsilon$$
$$f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}$$
$$n=2025$$
$$p < 0.05$$
$$\beta_1 = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}$$
$$IQR = Q_3 - Q_1$$
$$E[\bar{X}] = \mu$$
$$\log(x)$$
$$2\pi r$$
$$N \sim (0,1)$$
$$H_0: \mu = 0$$
$$0.999$$
$$\text{Cov}(X,Y)$$
$$\text{Var}(X)$$
$$df = n-1$$
$$RMSE$$
$$MAE$$
$$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$
$$\lambda = 0.7$$
$$\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$$
$$ANOVA$$
$$t = \frac{\bar{x} - \mu}{s/\sqrt{n}}$$
$$F = \frac{\text{Variance between groups}}{\text{Variance within groups}}$$
$$e^{i\pi} + 1 = 0$$
$$A \cap B$$
$$\sqrt{n}$$
$$Q_1, Q_2, Q_3$$
$$0.12345$$
$$\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$$
$$\sum_{i=1}^{N} x_i$$
$$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad-bc$$
$$\mathbb{E}[X]$$
$$\text{Binomial}(n,p)$$
$$\text{Normal}(\mu, \sigma^2)$$
$$n!$$
$$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$$
$$1.618$$
$$10^{5}$$
$$\pm \text{SE}$$
$$\alpha = 0.01$$
$$s = \sqrt{\frac{1}{N-1} \sum_{i=1}^N (x_i - \bar{x})^2}$$
$$R = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum (x_i - \bar{x})^2 \sum (y_i - \bar{y})^2}}$$
$$A \cup B$$
$$X \sim \text{Poisson}(\lambda)$$
$$E = \sum x_i P(x_i)$$
$$N(p, q)$$
$$\Phi(z)$$
$$\delta^2$$
$$F-test$$
$$\text{Mode}$$
$$\text{Median}$$
$$s_p^2$$
أحمد ستار جبار
محلل بيانات وإحصائي
أطمح لصناعة تجارب رقمية رائعة، أحب التحدي والتطوير المستمر، وأسعى لصنع فرق في عالم البيانات.