In [1]:
# Run this cell to set up packages for lecture.
from lec17_imports import *

Lecture 17 – Standardization and the Normal Distribution¶

DSC 10, Fall 2024¶

Announcements¶

  • Lab 4 is due Thursday and Homework 4 is due Sunday.
  • Monday is Veteran's Day, which is a university holiday. There will be no lecture, no discussion, and no office hours on Monday.

Agenda¶

  • Chebyshev's inequality.
  • Standardization.
  • The normal distribution.

Chebyshev's inequality¶

Recap: variance and standard deviation¶

$$\begin{align*}\text{variance} &= \text{average squared deviation from the mean}\\ &= \frac{(\text{value}_1 - \text{mean})^2 + ... + (\text{value}_n - \text{mean})^2}{n}\\ \text{standard deviation} &= \sqrt{\text{variance}} \end{align*}$$

where $n$ is the number of observations.

Standard deviation¶

  • The standard deviation (SD) measures something about how far the data values are from their average.
    • It is not directly interpretable because of the squaring and square rooting.
    • But generally, larger SD = more spread out.
  • The standard deviation has the same units as the original data.
  • numpy has a function, np.std, that calculates the standard deviation for us.
In [2]:
np.std([2, 3, 3, 9])
Out[2]:
2.7726341266023544

What can we do with the standard deviation?¶

It turns out, in any numerical distribution, the bulk of the data are in the range “mean ± a few SDs”.

Let's make this more precise.

Chebyshev’s inequality¶

Fact: In any numerical distribution, the proportion of values in the range “mean ± $z$ SDs” is at least

$$1 - \frac{1}{z^2} $$

Range Proportion
mean ± 2 SDs at least $1 - \frac{1}{4}$ (75%)
mean ± 3 SDs at least $1 - \frac{1}{9}$ (88.88..%)
mean ± 4 SDs at least $1 - \frac{1}{16}$ (93.75%)
mean ± 5 SDs at least $1 - \frac{1}{25}$ (96%)

Flight delays, revisited ✈️¶

In [3]:
delays = bpd.read_csv('data/united_summer2015.csv')
delays.plot(kind='hist', y='Delay', bins=np.arange(-20.5, 210, 5), density=True, ec='w', figsize=(10, 5), title='Flight Delays')
plt.xlabel('Delay (minutes)');
No description has been provided for this image
In [4]:
delay_mean = delays.get('Delay').mean()
delay_mean
Out[4]:
16.658155515370705
In [5]:
delay_std = np.std(delays.get('Delay')) # There is no .std() method in babypandas!
delay_std
Out[5]:
39.480199851609314

Mean and standard deviation¶

Chebyshev's inequality tells us that

  • At least 75% of delays are in the following interval:
In [6]:
delay_mean - 2 * delay_std, delay_mean + 2 * delay_std
Out[6]:
(-62.30224418784792, 95.61855521858934)
  • At least 88.88% of delays are in the following interval:
In [7]:
delay_mean - 3 * delay_std, delay_mean + 3 * delay_std
Out[7]:
(-101.78244403945723, 135.09875507019865)

Let's visualize these intervals!

In [8]:
delays.plot(kind='hist', y='Delay', bins=np.arange(-20.5, 210, 5), density=True, alpha=0.65, ec='w', figsize=(10, 5), title='Flight Delays')
plt.axvline(delay_mean - 2 * delay_std, color='maroon', label='± 2 SD')
plt.axvline(delay_mean + 2 * delay_std, color='maroon')

plt.axvline(delay_mean + 3 * delay_std, color='blue',  label='± 3 SD')
plt.axvline(delay_mean - 3 * delay_std, color='blue')

plt.axvline(delay_mean, color='green', label='Mean')
plt.scatter([delay_mean], [-0.0017], color='green', marker='^', s=250)
plt.ylim(-0.0038, 0.06)
plt.legend();
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Chebyshev's inequality provides lower bounds!¶

Remember, Chebyshev's inequality states that at least $1 - \frac{1}{z^2}$ of values are within $z$ SDs from the mean, for any numerical distribution.

For instance, it tells us that at least 75% of delays are in the following interval:

In [9]:
delay_mean - 2 * delay_std, delay_mean + 2 * delay_std
Out[9]:
(-62.30224418784792, 95.61855521858934)

However, in this case, a much larger fraction of delays are in that interval.

In [10]:
within_2_sds = delays[(delays.get('Delay') >= delay_mean - 2 * delay_std) & 
                      (delays.get('Delay') <= delay_mean + 2 * delay_std)]

within_2_sds.shape[0] / delays.shape[0]
Out[10]:
0.9560940325497288

If we know more about the shape of the distribution, we can provide better guarantees for the proportion of values within $z$ SDs of the mean.

Activity¶

For a particular set of data points, Chebyshev's inequality states that at least $\frac{8}{9}$ of the data points are between $-20$ and $40$. What is the standard deviation of the data?

✅ Click here to see the answer after you've tried it yourself.
  • Chebyshev's inequality states that at least $1 - \frac{1}{z^2}$ of values are within $z$ standard deviations of the mean.
  • When $z = 3$, $1 - \frac{1}{z^2} = \frac{8}{9}$.
  • So, $-20$ is $3$ standard deviations below the mean, and $40$ is $3$ standard deviations above the mean.
  • $10$ is in the middle of $-20$ and $40$, so the mean is $10$.
  • $3$ standard deviations are between $10$ and $40$, so $1$ standard deviation is $\frac{30}{3} = 10$.

Standardization¶

Heights and weights 📏¶

We'll work with a data set containing the heights and weights of 5000 adult males.

In [11]:
height_and_weight = bpd.read_csv('data/height_and_weight.csv')
height_and_weight
Out[11]:
Height Weight
0 73.85 241.89
1 68.78 162.31
2 74.11 212.74
... ... ...
4997 67.01 199.20
4998 71.56 185.91
4999 70.35 198.90

5000 rows × 2 columns

Distributions of height and weight¶

Let's look at the distributions of both numerical variables.

In [12]:
height_and_weight.plot(kind='hist', y='Height', density=True, ec='w', bins=30, alpha=0.8, figsize=(10, 5));
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In [13]:
height_and_weight.plot(kind='hist', y='Weight', density=True, ec='w', bins=30, alpha=0.8, color='C1', figsize=(10, 5));
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In [14]:
height_and_weight.plot(kind='hist', density=True, ec='w', bins=60, alpha=0.8, figsize=(10, 5));
No description has been provided for this image

Observation: The two distributions look like shifted and stretched versions of the same basic shape, called a bell curve 🔔. Distributions shaped like this are called normal distributions.

Many normal distributions¶

  • There are many normal distributions, with different means and different standard deviations.
  • All normal distributions are shaped like bell curves, but they vary in center and spread.
  • The mean and standard deviation uniquely define a normal distribution. There is only one normal distribution with a given mean and SD.
In [15]:
show_many_normal_distributions()
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  • Note that the area underneath each curve is 1. Therefore, the taller curves are narrower, and the shorter curves are wider.
  • Any normal distribution can be shifted and scaled to look like any other normal distribution. Let's see how with height and weight!

Standard units¶

Suppose $x$ is a numerical variable, and $x_i$ is one value of that variable. Then, $$x_{i \: \text{(su)}} = \frac{x_i - \text{mean of $x$}}{\text{SD of $x$}}$$

represents $x_i$ in standard units – the number of standard deviations $x_i$ is above the mean.

Example: Suppose someone weighs 225 pounds. What is their weight in standard units?

In [16]:
weights = height_and_weight.get('Weight')
(225 - weights.mean()) / np.std(weights)
Out[16]:
1.9201699181580782
  • Interpretation: 225 is 1.92 standard deviations above the mean weight.
  • 225 becomes 1.92 in standard units.

Standardization¶

The process of converting all values of a variable (i.e. a column) to standard units is known as standardization, and the resulting values are considered to be standardized.

In [17]:
def standard_units(col):
    return (col - col.mean()) / np.std(col)
In [18]:
standardized_height = standard_units(height_and_weight.get('Height'))
standardized_height
Out[18]:
0       1.68
1      -0.09
2       1.78
        ... 
4997   -0.70
4998    0.88
4999    0.46
Name: Height, Length: 5000, dtype: float64
In [19]:
standardized_weight = standard_units(height_and_weight.get('Weight'))
standardized_weight
Out[19]:
0       2.77
1      -1.25
2       1.30
        ... 
4997    0.62
4998   -0.06
4999    0.60
Name: Weight, Length: 5000, dtype: float64

The effect of standardization¶

Standardized variables have:

  • A mean of 0.
  • An SD of 1.

We often standardize variables to bring them to the same scale.

In [20]:
# e-15 means 10^(-15), which is a very small number, effectively zero.
standardized_height.describe()
Out[20]:
count    5.00e+03
mean     1.49e-15
std      1.00e+00
           ...   
50%      4.76e-04
75%      6.85e-01
max      3.48e+00
Name: Height, Length: 8, dtype: float64
In [21]:
standardized_weight.describe()
Out[21]:
count    5.00e+03
mean     5.98e-16
std      1.00e+00
           ...   
50%      6.53e-04
75%      6.74e-01
max      4.19e+00
Name: Weight, Length: 8, dtype: float64

Let's look at how the process of standardization works visually.

In [22]:
HTML('data/height_anim.html')
Out[22]:
Your browser does not support the video tag.
In [23]:
HTML('data/weight_anim.html')
Out[23]:
Your browser does not support the video tag.

Standardized histograms¶

Now that we've standardized the distributions of height and weight, let's see how they look on the same set of axes.

In [24]:
standardized_height_and_weight = bpd.DataFrame().assign(
    Height=standardized_height,
    Weight=standardized_weight
)
standardized_height_and_weight.plot(kind='hist', density=True, ec='w',bins=30, alpha=0.8, figsize=(10, 5));
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These both look very similar!

The standard normal distribution¶

The standard normal distribution¶

  • The distributions we've seen look essentially the same once standardized.
  • This distribution is called the standard normal distribution. It is defined by its mean of 0 and its standard deviation of 1. The shape of such a distribution is called the standard normal curve.

$$ \phi(z) = \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2}z^2} $$

  • You don't need to know the formula – just the shape!
    • We'll just use the formula today to make plots.

The standard normal curve¶

In [25]:
def normal_curve(z):
    return 1 / np.sqrt(2 * np.pi) * np.exp((-z**2)/2)

x = np.linspace(-4, 4, 1000)
y = normal_curve(x)

plt.figure(figsize=(10, 5))
plt.plot(x, y, color='black');
plt.xlabel('$z$');
plt.title(r'$\phi(z) = \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2}z^2}$');
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Heights/weights are roughly normal¶

If a distribution follows this shape, we say it is roughly normal.

In [26]:
standardized_height_and_weight.plot(kind='hist', density=True, ec='w', bins=120, alpha=0.8, figsize=(10, 5));
plt.plot(x, y, color='black', linestyle='--', label='Normal', linewidth=5)
plt.legend(loc='upper right');
No description has been provided for this image

The standard normal distribution¶

  • Think of the normal distribution as a "continuous histogram".
  • Its mean and median are both 0 – it is symmetric.
  • It has inflection points at $\pm 1$.
    • More on this later.
  • Like a histogram:
    • The area between $a$ and $b$ is the proportion of values between $a$ and $b$.
    • The total area underneath the normal curve is is 1.
In [27]:
sliders()

HBox(children=(FloatSlider(value=0.0, description='a', max=3.0, min=-4.0, step=0.25), FloatSlider(value=1.0, d…
Output()

Cumulative density functions¶

  • The cumulative density function (CDF) of a distribution is a function that takes in a value $z$ and returns the proportion of values in the distribution that are less than or equal to $z$, i.e. the area under the curve to the left of $z$.
  • To find areas under curves, we typically use integration (calculus). However, the standard normal curve has no closed-form integral.
  • Often, people refer to tables that contain approximations of the CDF of the standard normal distribution.
  • We'll use an approximation built into the scipy module in Python. The function scipy.stats.norm.cdf(z) computes the area under the standard normal curve to the left of z.

Areas under the standard normal curve¶

What does scipy.stats.norm.cdf(0) evaluate to? Why?

In [28]:
normal_area(-np.inf, 0)
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In [29]:
from scipy import stats
stats.norm.cdf(0)
Out[29]:
0.5

Areas under the standard normal curve¶

Suppose we want to find the area to the right of 2 under the standard normal curve.

In [30]:
normal_area(2, np.inf)
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The following expression gives us the area to the left of 2.

In [31]:
stats.norm.cdf(2)
Out[31]:
0.9772498680518208
In [32]:
normal_area(-np.inf, 2)
No description has been provided for this image

However, since the total area under the standard normal curve is 1:

$$\text{area right of $2$} = 1 - (\text{area left of $2$})$$

In [33]:
1 - stats.norm.cdf(2)
Out[33]:
0.02275013194817921

Areas under the standard normal curve¶

How might we use stats.norm.cdf to compute the area between -1 and 0?

In [34]:
normal_area(-1, 0)
No description has been provided for this image

Strategy:

$$\text{area from $-1$ to $0$} = (\text{area left of $0$}) - (\text{area left of $-1$})$$

In [35]:
stats.norm.cdf(0) - stats.norm.cdf(-1)
Out[35]:
0.3413447460685429

General strategy for finding area¶

The area under a standard normal curve in the interval $[a, b]$ is

stats.norm.cdf(b) - stats.norm.cdf(a)

What can we do with this? We're about to see!

Using the normal distribution¶

Let's return to our data set of heights and weights.

In [36]:
height_and_weight
Out[36]:
Height Weight
0 73.85 241.89
1 68.78 162.31
2 74.11 212.74
... ... ...
4997 67.01 199.20
4998 71.56 185.91
4999 70.35 198.90

5000 rows × 2 columns

As we saw before, both variables are roughly normal. What benefit is there to knowing that the two distributions are roughly normal?

Standard units and the normal distribution¶

  • Key idea: The $x$-axis in a plot of the standard normal distribution is in standard units.
    • For instance, the area between -1 and 1 is the proportion of values within 1 standard deviation of the mean.
  • Suppose a distribution is (roughly) normal. Then, these are two are approximately equal:
    • The proportion of values in the distribution between $a$ and $b$.
    • The area between $a_{\: \text{(su)}}$ and $b_{\: \text{(su)}}$ under the standard normal curve.
      • Recall, $x_{i \: \text{(su)}} = \frac{x_i - \text{mean of $x$}}{\text{SD of $x$}}$.

Example: Proportion of weights between 200 and 225 pounds¶

Let's suppose, as is often the case, that we don't have access to the entire distribution of weights, but just the mean and SD.

In [37]:
weight_mean = weights.mean()
weight_mean
Out[37]:
187.0206206581932
In [38]:
weight_std = np.std(weights)
weight_std
Out[38]:
19.779176302396458

Using just this information, we can estimate the proportion of weights between 200 and 225 pounds:

  1. Convert 200 to standard units.
  2. Convert 225 to standard units.
  3. Use stats.norm.cdf to find the area between (1) and (2).
In [39]:
left = (200 - weight_mean) / weight_std
left
Out[39]:
0.656214351061435
In [40]:
right = (225 - weight_mean) / weight_std
right
Out[40]:
1.9201699181580782
In [41]:
normal_area(left, right)