Copyright 2016 Allen B. Downey

MIT License: https://opensource.org/licenses/MIT

```
In [1]:
```from __future__ import print_function, division
%matplotlib inline
import numpy as np
import nsfg
import first
import analytic
import thinkstats2
import thinkplot

```
In [2]:
```thinkplot.PrePlot(3)
for lam in [2.0, 1, 0.5]:
xs, ps = thinkstats2.RenderExpoCdf(lam, 0, 3.0, 50)
label = r'$\lambda=%g$' % lam
thinkplot.Plot(xs, ps, label=label)
thinkplot.Config(title='Exponential CDF', xlabel='x', ylabel='CDF',
loc='lower right')

```
```

Here's the distribution of interarrival times from a dataset of birth times.

```
In [3]:
```df = analytic.ReadBabyBoom()
diffs = df.minutes.diff()
cdf = thinkstats2.Cdf(diffs, label='actual')
thinkplot.Cdf(cdf)
thinkplot.Config(xlabel='Time between births (minutes)', ylabel='CDF')

```
```

```
In [4]:
```thinkplot.Cdf(cdf, complement=True)
thinkplot.Config(xlabel='Time between births (minutes)',
ylabel='CCDF', yscale='log', loc='upper right')

```
```

```
In [5]:
```thinkplot.PrePlot(3)
mus = [1.0, 2.0, 3.0]
sigmas = [0.5, 0.4, 0.3]
for mu, sigma in zip(mus, sigmas):
xs, ps = thinkstats2.RenderNormalCdf(mu=mu, sigma=sigma,
low=-1.0, high=4.0)
label = r'$\mu=%g$, $\sigma=%g$' % (mu, sigma)
thinkplot.Plot(xs, ps, label=label)
thinkplot.Config(title='Normal CDF', xlabel='x', ylabel='CDF',
loc='upper left')

```
```

I'll use a normal model to fit the distribution of birth weights from the NSFG.

```
In [6]:
```preg = nsfg.ReadFemPreg()
weights = preg.totalwgt_lb.dropna()

Here's the observed CDF and the model. The model fits the data well except in the left tail.

```
In [7]:
```# estimate parameters: trimming outliers yields a better fit
mu, var = thinkstats2.TrimmedMeanVar(weights, p=0.01)
print('Mean, Var', mu, var)
# plot the model
sigma = np.sqrt(var)
print('Sigma', sigma)
xs, ps = thinkstats2.RenderNormalCdf(mu, sigma, low=0, high=12.5)
thinkplot.Plot(xs, ps, label='model', color='0.6')
# plot the data
cdf = thinkstats2.Cdf(weights, label='data')
thinkplot.PrePlot(1)
thinkplot.Cdf(cdf)
thinkplot.Config(title='Birth weights',
xlabel='Birth weight (pounds)',
ylabel='CDF')

```
```

```
In [8]:
```n = 1000
thinkplot.PrePlot(3)
mus = [0, 1, 5]
sigmas = [1, 1, 2]
for mu, sigma in zip(mus, sigmas):
sample = np.random.normal(mu, sigma, n)
xs, ys = thinkstats2.NormalProbability(sample)
label = '$\mu=%d$, $\sigma=%d$' % (mu, sigma)
thinkplot.Plot(xs, ys, label=label)
thinkplot.Config(title='Normal probability plot',
xlabel='standard normal sample',
ylabel='sample values')

```
```

```
In [9]:
```mean, var = thinkstats2.TrimmedMeanVar(weights, p=0.01)
std = np.sqrt(var)
xs = [-4, 4]
fxs, fys = thinkstats2.FitLine(xs, mean, std)
thinkplot.Plot(fxs, fys, linewidth=4, color='0.8')
xs, ys = thinkstats2.NormalProbability(weights)
thinkplot.Plot(xs, ys, label='all live')
thinkplot.Config(title='Normal probability plot',
xlabel='Standard deviations from mean',
ylabel='Birth weight (lbs)')

```
```

```
In [10]:
```full_term = preg[preg.prglngth >= 37]
term_weights = full_term.totalwgt_lb.dropna()

Now the deviation in the left tail is almost gone, but the heaviest babies are still heavy.

```
In [11]:
```mean, var = thinkstats2.TrimmedMeanVar(weights, p=0.01)
std = np.sqrt(var)
xs = [-4, 4]
fxs, fys = thinkstats2.FitLine(xs, mean, std)
thinkplot.Plot(fxs, fys, linewidth=4, color='0.8')
thinkplot.PrePlot(2)
xs, ys = thinkstats2.NormalProbability(weights)
thinkplot.Plot(xs, ys, label='all live')
xs, ys = thinkstats2.NormalProbability(term_weights)
thinkplot.Plot(xs, ys, label='full term')
thinkplot.Config(title='Normal probability plot',
xlabel='Standard deviations from mean',
ylabel='Birth weight (lbs)')

```
```

```
In [14]:
```import brfss
df = brfss.ReadBrfss()
weights = df.wtkg2.dropna()

```
In [15]:
```def MakeNormalModel(weights):
"""Plots a CDF with a Normal model.
weights: sequence
"""
cdf = thinkstats2.Cdf(weights, label='weights')
mean, var = thinkstats2.TrimmedMeanVar(weights)
std = np.sqrt(var)
print('n, mean, std', len(weights), mean, std)
xmin = mean - 4 * std
xmax = mean + 4 * std
xs, ps = thinkstats2.RenderNormalCdf(mean, std, xmin, xmax)
thinkplot.Plot(xs, ps, label='model', linewidth=4, color='0.8')
thinkplot.Cdf(cdf)

Here's the distribution of adult weights and a normal model, which is not a very good fit.

```
In [16]:
```MakeNormalModel(weights)
thinkplot.Config(title='Adult weight, linear scale', xlabel='Weight (kg)',
ylabel='CDF', loc='upper right')

```
```

```
In [17]:
```log_weights = np.log10(weights)
MakeNormalModel(log_weights)
thinkplot.Config(title='Adult weight, log scale', xlabel='Weight (log10 kg)',
ylabel='CDF', loc='upper right')

```
```

The following function generates a normal probability plot.

```
In [18]:
```def MakeNormalPlot(weights):
"""Generates a normal probability plot of birth weights.
weights: sequence
"""
mean, var = thinkstats2.TrimmedMeanVar(weights, p=0.01)
std = np.sqrt(var)
xs = [-5, 5]
xs, ys = thinkstats2.FitLine(xs, mean, std)
thinkplot.Plot(xs, ys, color='0.8', label='model')
xs, ys = thinkstats2.NormalProbability(weights)
thinkplot.Plot(xs, ys, label='weights')

```
In [19]:
```MakeNormalPlot(weights)
thinkplot.Config(title='Adult weight, normal plot', xlabel='Weight (kg)',
ylabel='CDF', loc='upper left')

```
```

```
In [20]:
```MakeNormalPlot(log_weights)
thinkplot.Config(title='Adult weight, lognormal plot', xlabel='Weight (log10 kg)',
ylabel='CDF', loc='upper left')

```
```

```
In [21]:
```xmin = 0.5
thinkplot.PrePlot(3)
for alpha in [2.0, 1.0, 0.5]:
xs, ps = thinkstats2.RenderParetoCdf(xmin, alpha, 0, 10.0, n=100)
thinkplot.Plot(xs, ps, label=r'$\alpha=%g$' % alpha)
thinkplot.Config(title='Pareto CDF', xlabel='x',
ylabel='CDF', loc='lower right')

```
```

The distribution of populations for cities and towns is sometimes said to be Pareto-like.

```
In [24]:
```import populations
pops = populations.ReadData()
print('Number of cities/towns', len(pops))

```
```

```
In [25]:
```log_pops = np.log10(pops)
cdf = thinkstats2.Cdf(pops, label='data')
cdf_log = thinkstats2.Cdf(log_pops, label='data')
# pareto plot
xs, ys = thinkstats2.RenderParetoCdf(xmin=5000, alpha=1.4, low=0, high=1e7)
thinkplot.Plot(np.log10(xs), 1-ys, label='model', color='0.8')
thinkplot.Cdf(cdf_log, complement=True)
thinkplot.Config(xlabel='log10 population',
ylabel='CCDF',
yscale='log', loc='lower left')

```
```

```
In [26]:
```thinkplot.PrePlot(cols=2)
mu, sigma = log_pops.mean(), log_pops.std()
xs, ps = thinkstats2.RenderNormalCdf(mu, sigma, low=0, high=8)
thinkplot.Plot(xs, ps, label='model', color='0.8')
thinkplot.Cdf(cdf_log)
thinkplot.Config(xlabel='log10 population',
ylabel='CDF', loc='lower right')

```
```

```
In [27]:
```thinkstats2.NormalProbabilityPlot(log_pops, label='data')
thinkplot.Config(xlabel='Random variate',
ylabel='log10 population',
xlim=[-5, 5])

```
```

```
In [28]:
```import random
def expovariate(lam):
p = random.random()
x = -np.log(1-p) / lam
return x

We can test it by generating a sample.

```
In [29]:
```t = [expovariate(lam=2) for _ in range(1000)]

And plotting the CCDF on a log-y scale.

```
In [30]:
```cdf = thinkstats2.Cdf(t)
thinkplot.Cdf(cdf, complement=True)
thinkplot.Config(xlabel='Exponential variate', ylabel='CCDF', yscale='log')

```
```

A straight line is consistent with an exponential distribution.

As an exercise, write a function that generates a Pareto variate.

**Exercise:** In the BRFSS (see Section 5.4), the distribution of heights is roughly normal with parameters µ = 178 cm and σ = 7.7 cm for men, and µ = 163 cm and σ = 7.3 cm for women.

In order to join Blue Man Group, you have to be male between 5’10” and 6’1” (see http://bluemancasting.com). What percentage of the U.S. male population is in this range? Hint: use `scipy.stats.norm.cdf`

.

`scipy.stats`

contains objects that represent analytic distributions

```
In [31]:
```import scipy.stats

For example `scipy.stats.norm` represents a normal distribution.

```
In [32]:
```mu = 178
sigma = 7.7
dist = scipy.stats.norm(loc=mu, scale=sigma)
type(dist)

```
Out[32]:
```

A "frozen random variable" can compute its mean and standard deviation.

```
In [33]:
```dist.mean(), dist.std()

```
Out[33]:
```

```
In [34]:
```dist.cdf(mu-sigma)

```
Out[34]:
```

How many people are between 5'10" and 6'1"?

```
In [35]:
```# Solution
low = dist.cdf(177.8) # 5'10"
high = dist.cdf(185.4) # 6'1"
low, high, high-low

```
Out[35]:
```

**Exercise:** To get a feel for the Pareto distribution, let’s see how different the world would be if the distribution of human height were Pareto. With the parameters xm = 1 m and α = 1.7, we get a distribution with a reasonable minimum, 1 m, and median, 1.5 m.

Plot this distribution. What is the mean human height in Pareto world? What fraction of the population is shorter than the mean? If there are 7 billion people in Pareto world, how many do we expect to be taller than 1 km? How tall do we expect the tallest person to be?

`scipy.stats.pareto`

represents a pareto distribution. In Pareto world, the distribution of human heights has parameters alpha=1.7 and xmin=1 meter. So the shortest person is 100 cm and the median is 150.

```
In [36]:
```alpha = 1.7
xmin = 1 # meter
dist = scipy.stats.pareto(b=alpha, scale=xmin)
dist.median()

```
Out[36]:
```

What is the mean height in Pareto world?

```
In [37]:
```# Solution
dist.mean()

```
Out[37]:
```

What fraction of people are shorter than the mean?

```
In [38]:
```# Solution
dist.cdf(dist.mean())

```
Out[38]:
```

`dist.cdf` or `dist.sf`.

```
In [39]:
```# Solution
(1 - dist.cdf(1000)) * 7e9, dist.sf(1000) * 7e9

```
Out[39]:
```

How tall do we expect the tallest person to be?

```
In [40]:
```# Solution
# One way to solve this is to search for a height that we
# expect one person out of 7 billion to exceed.
# It comes in at roughly 600 kilometers.
dist.sf(600000) * 7e9

```
Out[40]:
```

```
In [41]:
```# Solution
# Another way is to use `ppf`, which evaluates the "percent point function", which
# is the inverse CDF. So we can compute the height in meters that corresponds to
# the probability (1 - 1/7e9).
dist.ppf(1 - 1/7e9)

```
Out[41]:
```

**Exercise:** The Weibull distribution is a generalization of the exponential distribution that comes up in failure analysis (see http://wikipedia.org/wiki/Weibull_distribution). Its CDF is

$\mathrm{CDF}(x) = 1 − \exp[−(x / λ)^k]$

Can you find a transformation that makes a Weibull distribution look like a straight line? What do the slope and intercept of the line indicate?

Use `random.weibullvariate`

to generate a sample from a Weibull distribution and use it to test your transformation.

Generate a sample from a Weibull distribution and plot it using a transform that makes a Weibull distribution look like a straight line.

`thinkplot.Cdf`

, which provides a transform that makes the CDF of a Weibull distribution look like a straight line. Here's an example that shows how it's used.

```
In [42]:
```sample = [random.weibullvariate(2, 1) for _ in range(1000)]
cdf = thinkstats2.Cdf(sample)
thinkplot.Cdf(cdf, transform='weibull')
thinkplot.Config(xlabel='Weibull variate', ylabel='CCDF')

```
```

**Exercise:** For small values of `n`

, we don’t expect an empirical distribution to fit an analytic distribution exactly. One way to evaluate the quality of fit is to generate a sample from an analytic distribution and see how well it matches the data.

For example, in Section 5.1 we plotted the distribution of time between births and saw that it is approximately exponential. But the distribution is based on only 44 data points. To see whether the data might have come from an exponential distribution, generate 44 values from an exponential distribution with the same mean as the data, about 33 minutes between births.

Plot the distribution of the random values and compare it to the actual distribution. You can use random.expovariate to generate the values.

```
In [43]:
```import analytic
df = analytic.ReadBabyBoom()
diffs = df.minutes.diff()
cdf = thinkstats2.Cdf(diffs, label='actual')
n = len(diffs)
lam = 44.0 / 24 / 60
sample = [random.expovariate(lam) for _ in range(n)]
1/lam, np.mean(sample)

```
Out[43]:
```

```
In [44]:
```# Solution
model = thinkstats2.Cdf(sample, label='model')
thinkplot.PrePlot(2)
thinkplot.Cdfs([cdf, model], complement=True)
thinkplot.Config(xlabel='Time between births (minutes)',
ylabel='CCDF',
yscale='log')

```
```

```
In [45]:
```# Solution
# If you plot distributions for a large number of samples, you get a sense
# of how much random variation to expect. In this case, the data fall within
# the range we expect, so there is no compelling reason to think it is
# not exponential.
for i in range(100):
sample = [random.expovariate(lam) for _ in range(n)]
thinkplot.Cdf(thinkstats2.Cdf(sample), complement=True, color='0.9')
thinkplot.Cdf(cdf, complement=True)
thinkplot.Config(xlabel='Time between births (minutes)',
ylabel='CCDF',
yscale='log')

```
```

**Worked Example:** The distributions of wealth and income are sometimes modeled using lognormal and Pareto distributions. To see which is better, let’s look at some data.

The Current Population Survey (CPS) is a joint effort of the Bureau of Labor Statistics and the Census Bureau to study income and related variables. Data collected in 2013 is available from http://www.census.gov/hhes/www/cpstables/032013/hhinc/toc.htm. I downloaded `hinc06.xls`

, which is an Excel spreadsheet with information about household income, and converted it to `hinc06.csv`

, a CSV file you will find in the repository for this book. You will also find `hinc.py`

, which reads this file.

Extract the distribution of incomes from this dataset. Are any of the analytic distributions in this chapter a good model of the data?

```
In [48]:
```import hinc
df = hinc.ReadData()
df

```
Out[48]:
```

Here's what the CDF looks like on a linear scale.

```
In [49]:
```xs, ps = df.income.values, df.ps.values
cdf = thinkstats2.Cdf(xs, ps, label='data')
cdf_log = thinkstats2.Cdf(np.log10(xs), ps, label='data')
# linear plot
thinkplot.Cdf(cdf)
thinkplot.Config(xlabel='household income',
ylabel='CDF')

```
```

To check whether a Pareto model describes the data well, I plot the CCDF on a log-log scale.

I found parameters for the Pareto model that match the tail of the distribution.

```
In [50]:
```xs, ys = thinkstats2.RenderParetoCdf(xmin=55000, alpha=2.5,
low=0, high=250000)
thinkplot.Plot(xs, 1-ys, label='model', color='0.8')
thinkplot.Cdf(cdf, complement=True)
thinkplot.Config(xlabel='log10 household income',
ylabel='CCDF',
xscale='log',
yscale='log',
loc='lower left')

```
```

For the lognormal model I estimate mu and sigma using percentile-based statistics (median and IQR).

```
In [51]:
```median = cdf_log.Percentile(50)
iqr = cdf_log.Percentile(75) - cdf_log.Percentile(25)
std = iqr / 1.349
# choose std to match the upper tail
std = 0.35
print(median, std)

```
```

Here's what the distribution, and fitted model, look like on a log-x scale.

```
In [52]:
```xs, ps = thinkstats2.RenderNormalCdf(median, std, low=3.5, high=5.5)
thinkplot.Plot(xs, ps, label='model', color='0.8')
thinkplot.Cdf(cdf_log)
thinkplot.Config(xlabel='log10 household income',
ylabel='CDF')

```
```

My conclusions based on these figures are:

1) The Pareto model might be a reasonable choice for the top 10-20% of incomes.

2) The lognormal model captures the shape of the distribution better, with some deviation in the left tail. With different choices for sigma, you could match the upper or lower tail, but not both at the same time.

In summary I would say that neither model captures the whole distribution, so you might have to

1) look for another analytic model,

2) choose one that captures the part of the distribution that is most relevent, or

3) avoid using an analytic model altogether.

```
In [ ]:
```