Normal distribution

Standard normal distribution takes a bell curve. It is also called as gaussian distribution. Values in nature are believed to take a normal distribution. The equation for normal distribution is

$$y = \frac{1}{\sigma\sqrt{2\pi}} e^{\frac{(x-\mu)^2}{2\sigma^2}}$$

where $\mu$ is mean

$\sigma$ is standard deviation

$\pi$ = 3.14159..

$e$ = 2.71828.. (natural log)

In [2]:
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline

In [10]:
vals = np.random.standard_normal(100000)


In [19]:
fig, ax = plt.subplots(1,1)
hist_vals = ax.hist(vals, bins=200, color='red', density=True)

The above is the standard normal distribution. Its mean is 0 and SD is 1. About 95% values fall within $\mu \pm 2 SD$ and 98% within $\mu \pm 3 SD$

The area under this curve is 1 which gives the probability of values falling within the range of standard normal.

A common use is to find the probability of a value falling at a particular range. For instance, find $p(-2 \le z \le 2)$ which is the probability of a value falling within $\mu \pm 2SD$. This calculated by summing the area under the curve between these bounds.

$$p(-2 \le z \le 2) = 0.9544$$

which is 95.44% probability. Its z score is 0.9544.

Similarly $$p(z \ge 5.1) = 0.00000029$$

Finding z score and p values using SciPy

The standard normal is useful as a z table to look up the probability of a z score (x axis). You can use Scipy to accomplish this.

Levels of significance

By rule of thumb, a z score greater than 0.005 is considered significant as such a value has a very low probability of occuring. Thus, there is less chance of it occurring randomly and hence, there is probably a force acting on it (significant force, not random chance).

Transformation to standard normal

If the distribution of a phenomena follows normal dist, then you can transform it to standard normal, so you can measure the z scores. To do so, $$std normal value = \frac{observed - \mu}{\sigma}$$ You subtract the mean and divide by SD of the distribution.

Example Let X be age of US presidents at inaugration. $X \in N(\mu = 54.8, \sigma=6.2)$. What is the probability of choosing a president at random that is less than 44 years of age.

We need to find $p(x<44)$. First we need to transform to standard normal.

$$p(z< \frac{44-54.8}{6.2})$$$$p(z<-1.741) = 0.0409 \approx 4\%$$

Finding z score and p values using SciPy

The standard normal is useful as a z table to look up the probability of a z score (x axis). You can use Scipy to accomplish this.

In [2]:
import scipy.stats as st

# compute the p value for a z score


Let us try for some common z scores:

In [4]:
[st.norm.cdf(-3), st.norm.cdf(-1), st.norm.cdf(0), st.norm.cdf(1), st.norm.cdf(2)]


As you noticed, the norm.cdf() function gives the cumulative probability (left tail) from -3 to 3 approx. If you need right tailed distribution, you simply subtract this value from 1.

Finding z score from a p value

Sometimes, you have the probability (p value), but want to find the z score or how many SD does this value fall from mean. You can do this inverse using ppf().

In [7]:
# Find Z score for a probability of 0.97 (2sd)


In [8]:
[st.norm.ppf(0.95), st.norm.ppf(0.97), st.norm.ppf(0.98), st.norm.ppf(0.99)]


As is the ppf() function gives only positive z scores, you need to apply $\pm$ to it.

Transformation to standard normal and machine learning

Transforming features to standard normal has applications in machine learning. As each feature has a different unit, their range, standard deviation vary. Hence we scale them all to standard normal distribution with mean=0 and SD=1. This way a learner finds those variables that are truly influencial and not simply because it has a larger range.

To accomplish this easily, we use scikit-learn's StandardScaler object as shown below:

In [3]:
demo_dist = 55 + np.random.randn(200) * 3.4
std_normal = np.random.randn(200)

In [6]:
[demo_dist.mean(), demo_dist.std(), demo_dist.min(), demo_dist.max()]

[55.11294611274521, 3.29635155084324, 46.56508960658229, 65.69499942017563]

In [7]:
[std_normal.mean(), std_normal.std(), std_normal.min(), std_normal.max()]


Now let us use scikit-learn to easily transform this dataset

In [8]:
from sklearn.preprocessing import StandardScaler
scaler = StandardScaler()

In [22]:
demo_dist = demo_dist.reshape(200,1)
demo_dist_scaled = scaler.fit_transform(demo_dist)

In [23]:
[round(demo_dist_scaled.mean(),3), demo_dist_scaled.std(), demo_dist_scaled.min(), demo_dist_scaled.max()]

[0.0, 1.0, -2.5931264837260137, 3.2102320229538095]

In [49]:
fig, axs = plt.subplots(2,2, figsize=(15,8))
p1 = axs[0][0].scatter(sorted(demo_dist), sorted(std_normal))
axs[0][0].set_title("Scatter of original dataset against standard normal")

p1 = axs[0][1].scatter(sorted(demo_dist_scaled), sorted(std_normal))
axs[0][1].set_title("Scatter of scaled dataset against standard normal")

p2 = axs[1][0].hist(demo_dist, bins=50)
axs[1][0].set_title("Histogram of original dataset against standard normal")

p3 = axs[1][1].hist(demo_dist_scaled, bins=50)
axs[1][1].set_title("Histogram of scaled dataset against standard normal")

Text(0.5,1,'Histogram of scaled dataset against standard normal')

As you see above, the shape of distribution is the same, just the values are scaled.

Assessing normality of a distribution

To assess how normal a distribution of values is, we sort the values, then plot them against sorted values of standard normal distribution. If the values fall on a straight line, then they are normally distributed, else they exhibit skewness and or kurtosis.

In [20]:
demo_dist = 55 + np.random.randn(200) * 3.4
std_normal = np.random.randn(200)

In [21]:
demo_dist = sorted(demo_dist)
std_normal = sorted(std_normal)

In [22]:
plt.scatter(demo_dist, std_normal)

<matplotlib.collections.PathCollection at 0x111e56dd8>

For the most part, the values fall on a straight line, except in the fringes. Thus, the demo distribution is fairly normal.

Standard error

As can be seen, we use statistics to estimate population mean $\mu$ from sample mean $\bar x$. Standard error of $\bar x$ represents on average, how far will it be from $\mu$.

As you suspect, the quality of $\bar x$, or its standard error will depend on the sample size. In addition, it also depends on population standard deviation. Thus for a tighter population, it is much easy to estimate mean from a small sample as there are fewer outliers.

Nonetheless, $$SE(\bar x) = \frac{\sigma}{\sqrt{n}}$$ where $\sigma$ is population SD and $n$ is sample size.

Empirically, $SE(\bar x)$ is same as the SD of a distribution of sample means. If you were to collect a number of samples, find their means to form a distribution, the SD of this distribution represents the standard error of that estimate (mean in this case).

Confidence intervals

From a population, many samples of size > 30 is drawn and their means are computed and plotted, then with $\bar x$ or $\bar y$ -> mean of a sample and $n$ -> size of 1 sample, $\sigma_{\bar x}$ or $\sigma_{\bar y}$ is SD of distribution of samples, you can observe that

  • $\mu_{\bar x} = \mu$ (mean of distribution of sample means equals population mean)
  • $\frac{\sigma}{\sqrt n} = \sigma_{\bar x}$ (SD of population over sqrt of sample size equals SD of sampling distribution)
  • relationship between population mean and mean of a single sample is
  • $$ \mu = \bar y \pm z(\frac{\sigma}{\sqrt n})$$

The z table and normal distribution are used to derive confidence intervals. Popular intervals and their corresponding z scores are

interval z-value
99% $\pm 2.576$
98% $\pm 2.326$
95% $\pm 1.96$
90% $\pm 1.645$

As you imagine, these are the values of z on X axis of the standard normal distribution and the area they cover.

For a normal distribution, confidence intervals for an estimate (such as mean) can be given as $$CI = \bar x \pm z\frac{s}{\sqrt{n}}$$ where $s$ is sample SD that is substituted in place of population SD, if sample size is larger than 30.

Example The average TV viewing times of 40 adults sampled in Iowa is 7.75 hours per week. The SD of this sample is 12.5. Find the 95% CI population's average TV viewing times.

$\bar x = 40$, $s=12.5$, $n=40$, $Z=1.96$ for 95% CI. Thus $$95\%CI = 7.75 \pm 1.96\frac{12.5}{\sqrt{40}}$$ $$95\%CI = (3.877 | 11.623)$$

Thus the 95% CI is pretty wide. Intuitively, if SD of sample is smaller, then so is the CI.

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