ECON 3818

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# ECON 3818
## Chapter 3
### Kyle Butts
### 27 September 2021

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## Chapter 3: The Normal Distribution

---
# Normal Distribution
    
Normal curve is symmetric about the mean and bell-shaped:

Lots of data naturally follow this distribution

- heights of people, blood pressure, grades on a test

---
# Galton Board

.center[
<iframe width="560" height="315" src="https://www.youtube.com/embed/EvHiee7gs9Y" title="YouTube video player" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
]

---
# Sample Probability

<img src="data:image/png;base64,#ch3_files/figure-html/unnamed-chunk-1-1.svg" width="80%" style="display: block; margin: auto;" />
    
- What's the probability that the x value is less than -2.5 .hi[in our sample]?

---
# Population Probability

<img src="data:image/png;base64,#ch3_files/figure-html/unnamed-chunk-2-1.svg" width="80%" style="display: block; margin: auto;" />
    
- What's the probability that the x value is less than -2.5 .hi[in our population distribution]?

---
# Parameters of Normal Distribution

Normal distribution is described .it.bf[completely] by two parameters-- its mean `$\mu$` and variance `$\sigma^2$`

- The mean is located at the center of the symmetric curve
      - It is the same as the median 
- Changing `$\mu$` (without changing `$\sigma^2$`), moves the curve along the horizontal axis
- The variance describes the variability of the curve
- Higher variance means a flatter and wider distribution

---
# Different Variances
    
        
<img src="data:image/png;base64,#ch3_files/figure-html/multiple-vars-1.svg" width="90%" style="display: block; margin: auto;" />

---
# 68-95-99 Rule

.small[
- 68.2% of data is within `$\pm 1$` standard deviation of the mean
- 95.4% of data is within `$\pm 2$` standard deviation of the mean
- 99.6% of data is within `$\pm 3$` standard deviation of the mean
]

---
# Clicker Question

Suppose that the mean birthweight in the sample is 113 oz. with a standard deviation of `$\sqrt{484} \approx 22$` oz. Assuming babies' birthweight is normally distributed, how heavy are the middle 95% of babies?

---
# Normal Distribution Notation

If X is distributed normally, we denote it the following way:

`$$X \sim N(\coral{\mu}, \kelly{\sigma^2})$$`
 
- This notation tells us everything we need to know about the normal distribution
      
- The distribution has mean `$\coral{\mu}$`
- The distribution has variance `$\kelly{\sigma^2}$`

---
# Standard Normal Distribution

Standard normal distribution is a specific type of normal distribution

If a variable X follows a normal distribution with `$\coral{\mu} = 0$` and  `$\kelly{\sigma^2} = 1$`, we say that X follows the .hi.purple[standard normal distribution]
  
- Since it is so common, it is denoted as `$Z \sim N(0,1)$`

- It is easier to find out probabilities about normal distributions if they are in the standard form
      
- Therefore we often will .hi.purple[standardize] any general normal distribution to be a standard normal

---
# Properties of Standard Normal

Graph of the standard normal distribution has two important properties

Symmetric

`$$P(Z < -1) = P(Z > 1)$$`

Area under the curve sums to one

`$$P(Z < 1) + P(Z > 1) = 1$$`

---
# Probabilities: Cumulative Proportions

Suppose we want to know the likelihood of a baby being born underweight (less than 88 oz).

The data suggests `$BW \sim N(113, 22^2)$`, that is a mean of 113 and a standard deviation of 22. The probability of a baby being underweight is equal to `$P(BW \leq 88)$`. Graphically:

---
# Left-tail probability

This probability is called the .hi.ruby[left-tail probability] as it's every value .ruby[to the left].

---
# Right-tail Probability

If you want the .hi.ruby[right-tail probability], `$P(BW > 88)$`, you can use

`$$P(BW > 88) = 1 - P(BW < 88)$$`

---
# Standardization
    
If a variable X has any normal distribution, `$X\sim N(\mu,\sigma^2)$`, then the standardized variable:
`$$Z= \frac{X-\coral{\mu}}{\kelly{\sigma}} \sim N(0,1)$$`
We call the standardized value the .hi.purple[Z-score].

The Z-score is equivalent to the number of .hi.kelly[standard deviations] that `$X$` is away from the .hi.coral[mean].

---
# Standardization

Since Z-scores are measured in number of standard deviations, we can compare across samples without having to worry about units.

For example:

- SAT scores are `$X\sim N(1500, 250^2)$`
- ACT scores are `$Y \sim N(20.8, 2.8^2)$`

You scored an 1860 on the SAT, your neighbor scored a 29 on the ACT. Who did better? Just compare Z-scores!

---
# Calculating Probabilities

Coming back to the birth-weight example, let's do a little standardizing.

How do we actually calculate `$P(BW\leq 88)$`  (when `$BW\sim N(113,22^2)$`)

First, standardize the distribution

`$$P(BW\leq88)=P(\frac{BW-\mu}{\sigma} \leq \frac{88-113}{22}) = P(Z \leq -1.14)$$`

Then, we actually have a big table of left-tail probabilities for the .it[standard] normal distribution

- Table is either left-tail or right-tail (.hi.purple[Z table on exam is left-tailed])

---
# Standard Normal Tables

<img src="data:image/png;base64,#stdnormtableex.png" width="90%" style="display: block; margin: auto;" />
    
Standard normal tables show the cumulative probability of different z-scores

- A table like this shows the .hi.purple[left-tail] probabilities. (One is available on the course site.)

- .large.hi[Be Careful!] Some (but not many) tables display .hi.purple[right-tail] probabilities.

---
class: clear

.pull-left[.center[
<img src="data:image/png;base64,#neg_z.png" width="80%" style="display: block; margin: auto;" />
]]
.pull-right[.center[
<img src="data:image/png;base64,#pos_z.png" width="80%" style="display: block; margin: auto;" />
]]

---
# Using a Z-Table
    
Back to our birth-weight example. We want

`$$P(Z < -1.14)$$`
    
.hi[Method 1]: if you have negative values in Z-Table:

- Look up Z = -1.14 in z-table

.hi[Method 2]: If you have only positive values in Z-Table:

`\begin{aligned}
    P(Z \leq -1.14) &= P(Z \geq 1.14) \\
    &= 1 - P(Z \leq 1.14) \\
    &= 1 - .8729 \\ 
    &= 0.1271
\end{aligned}`

---
# Normal Example

A company chooses its new entry-level employees from a pool of recent college graduates. The cumulative GPA of the candidates is used as a tie-breaker. GPAs for the successful interviewees are normally distributed, with a mean of 3.3 and a standard deviation of 0.4. What proportion of candidates have a GPA under 3.0?

---
# Clicker Question

Consider the scenario on the previous slide, where `$GPA \sim N(3.3,0.4^2)$`. What percent of candidates have a `$GPA$` above 3.9?

---
# Area In Between Z-Scores
Suppose we want to calculate `$P(-1.13 \leq Z \leq 0.3)$`
Graphically we want to calculate the following shaded area:

---
# Area In Between Z-Scores
In order to calculate the area between two z-scores,

`$$P(-1.13 \leq Z \leq 0.3) = P(Z \leq 0.3) - P(Z \leq -1.13)$$`

So we calculate:

- `$P(Z \leq 0.3) = 0.6179$`
- `$P(Z \leq -1.13) = 0.1292$`

`\begin{aligned}
P(-1.13 \leq Z \leq 0.3) &= P(Z \leq 0.3) - P(Z \leq -1.13) \\
&= 0.6179 - 0.1292 \\
&= 0.4887
\end{aligned}`

---
# Clicker Question

A typical college freshman spends an average of `$\mu=150$` minutes per day with a standard deviation of `$\sigma=50$` minutes, on social media. The distribution of time on social media is known be Normal. What is the probability a college freshman spends between 2 and 3 hours on social media?

---
# Using probability to calculate Z-score

So far, we've used z-scores to calculate probabilities (values inside the table)

In some cases, we will use probabilities to calculate z-scores (values outside the table)

---
# Example

Scores on the SAT verbal test follow approximately the `$N(515,109^2)$` distribution. How high must a student score in order to place in top 5% of all students taking the SAT?

---
# Example

Back to the example discussing the distribution of GPAs, where GPA  `$\sim N(3.3, 0.4^2)$`. If the company is interviewing 163 people, but only 121 can be hired, then what cut-off GPA should the company use?

---
# Clicker Question
Suppose that `$P(Z\leq z^*) = 0.025$`. Using a standard normal table, find `$z^*$`

---
# Review of Normal Distribution

Consider men's height to be distributed normally with a mean of 5.9 feet and a standard deviation or 0.4 feet.
Calculate the following:

- `$P(X>6.5)$`
      
- `$P(X>5)$`
      
- What is the top 10% of men's height?
      
- What is the bottom 20% of men's height?