Nathan VanHoudnos

10/15/2014

- Homework comments
- Checkpoint #11 results
- Lecture 10 (covers pp. 124-137)

```
mean(hw3.grades/85)
```

```
[1] 0.9068
```

```
median(hw3.grades/85)
```

```
[1] 0.9412
```

- Homework comments
- Checkpoint #11 results
- Lecture 10 (covers pp. 124-137)

- 69/76 students
- on average 86% percent correct

(i) 31% of all Dalmatians have blue eyes.

(ii) 38% of all Dalmatians are deaf.

(iii) 42% of blue-eyed Dalmatians are deaf.

What is the probability that a randomly chosen Dalmatian is blue-eyed and deaf?

\( P(B \text{ and } D) = ? \)

Must use the **general multiplication rule**:

\( P(B \text{ and } D) = P(D|B)P(B) = .42 * .31 = .13 \)

Note that: \( P(D)P(B) = .38 * .31 = .118 \ne P(B \text{ and } D) \)

- Homework comments
- Checkpoint #11 results
- Lecture 10 (covers pp. 124-137)

**Checkpoint 12**: Random variables (RVs) and distributions**Checkpoint 13**: Expection and variance rules

random variable:A random variable assigns a unique numerical value to the outcome of a random experiment.

Consider the random experiment of flipping a coin twice. \( S = \{ HH, HT, TH, TT \} \)

Let \( X \) be the number of tails.

- HH implies \( X = 0 \)
- HT implies \( X = 1 \)
- TH implies \( X = 1 \)
- TT implies \( X = 2 \)

Consider getting data from a random sample on the number of ears in which a person wears one or more earrings.

Let \( X \) be the number of ears in which a randomly selected person wears an earring.

If the selected person

does not wear any earrings, then \( X = 0 \).

wears earrings in either the left or the right ear, then \( X = 1 \).

wears earrings in both ears, then \( X = 2 \).

Assume we choose a lightweight male boxer at random and record his exact weight. According to the boxing rules, a lightweight male boxer must weigh between 130 and 135 pounds, so the sample space here is \( S = [130, 135] \).

Let \( X \) be the weight of the boxer.

In this case, \( X \in [130,135] \).

random variable:A random variable assigns a unique numerical value to the outcome of a random experiment.

**Always** arise from a random experiment!

Two types of random variables:

**discrete**: possible values are from a list- “Things you count.”

**continuous**: possible values are from an interval- “Things you measure.” (w/ or w/o rounding)

Number of ears pierced?

**discrete**

Weight of a lightweight boxer?

**continuous**

Number of hours watching TV (rounded to the nearest hour)?

**continuous**(rounded is still continuous)

How many days per week do you drink soda?

**discrete**

How many ounces of soda per week do you drink?

**continuous**

The average weight of 18 year old males at basic training.

**continuous**

Consider the random experiment of flipping a coin twice. \( S = \{ HH, HT, TH, TT \} \)

Let \( X \) be the number of tails. Recall: \[ \begin{aligned} HH \rightarrow X & = 0 & HT \rightarrow X & = 1 \\ TH \rightarrow X & = 1 & TT \rightarrow X & = 2 \end{aligned} \]

What is the probability of \( X=0 \)?

\[ P( X = 0) = P( HH ) = \frac{1}{4} \]

Consider the random experiment of flipping a coin twice. \( S = \{ HH, HT, TH, TT \} \)

Let \( X \) be the number of tails. Recall: \[ \begin{aligned} HH \rightarrow X & = 0 & HT \rightarrow X & = 1 \\ TH \rightarrow X & = 1 & TT \rightarrow X & = 2 \end{aligned} \]

What is the probability of \( X=1 \)? \[ \begin{aligned} P( X = 1) & = P( HT \text{ or } TH ) \\ & = P( HT ) + P(TH ) \\ & = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \end{aligned} \]

Consider the random experiment of flipping a coin twice. \( S = \{ HH, HT, TH, TT \} \)

Let \( X \) be the number of tails. Recall: \[ \begin{aligned} HH \rightarrow X & = 0 & HT \rightarrow X & = 1 \\ TH \rightarrow X & = 1 & TT \rightarrow X & = 2 \end{aligned} \]

The distribution of \( X \):

```
x | 0 | 1 | 2 |
---------+-----+-----+-----|
P(X = x) | 1/4 | 1/2 | 1/4 |
```

The distribution of \( X \):

```
x | 0 | 1 | 2 |
---------+-----+-----+-----|
P(X = x) | 1/4 | 1/2 | 1/4 |
```

\( P(X = x) \) is read as

- the probability that
- the random variable \( X \) is
- equal to the value \( x \).

**Random variables** are **CAPITALIZED**.

**Values** (AKA realizations) are **lower case**.

Let \( X \) be a random variable and \( x \) a realization of that random variable.

- \( 0 \le P(X = x) \le 1 \) for all \( x \).

For example:

```
x | 0 | 1 | 2 |
---------+-----+-----+-----|
P(X = x) | 1/4 | 1/2 | 1/4 |
```

\[ \begin{aligned} 0 & \le P(X = 0) = \frac{1}{4} \le 1 \\ 0 & \le P(X = 1) = \frac{1}{2} \le 1\\ \end{aligned} \]

Let \( X \) be a random variable and \( x \) a realization of that random variable.

- \( 0 \le P(X = x) \le 1 \) for all \( x \).
- \( \sum_{x\in S} P(X = x) = 1 \) \( \quad \)

For example:

\[ \begin{aligned} \sum_{x \in S} P(X = x) & = P(X=0) + P(X=1) + P(X=2) \\ & = 1/4 + 1/2 + 1/4 = 1 \end{aligned} \]

**Note:** If summation notation is unfamiliar, check out Khan Academy here and here.

A coin is tossed three times. Let the random variable X be the number of tails. Find the probability distribution of X.

- Write out the sample space.
- Define each possible realization of \( X \) as an event.
- Find the probabilities of the events.

\[ \begin{aligned} S = \{ & HHH, HHT, HTH, HTT, \\ & THH, THT, TTH, TTT \} \end{aligned} \]

\[ \begin{aligned} X & = 0 \quad \{ HHH \} \\ X & = 1 \quad \{ HHT, HTH, THH \} \\ X & = 2 \quad \{ HTT, THT, TTH \} \\ X & = 3 \quad \{ TTT \} \end{aligned} \]

```
x | 0 | 1 | 2 | 3 |
---------+-----+-----+-----+-----|
P(X = x) | 1/8 | 3/8 | 3/8 | 1/8 |
```

An example: \[ P(X = x) = \frac{x+2}{25} \quad \quad x \in \{1,2,3,4,5\} \]

Check rule 1:

\[ 0 \le P(X=x) \le 1 \quad \text{ for all } x \]

Note that: \[ \begin{aligned} P(X = 1) & < P(X =5) \\ 0 \le \frac{1 + 2}{25} & < \frac{5+2}{25} \le 1 \end{aligned} \]

An example: \[ P(X = x) = \frac{x+2}{25} \quad \quad x \in \{1,2,3,4,5\} \]

Check rule 2:

\[ \begin{aligned} \sum_{x\in S} P(X = x) & = \sum_{x\in S} \frac{x+2}{25} \\ & = \frac{1}{25} \sum_{x\in \{1,2,3,4,5\}} \left( x +2 \right) \\ & = \frac{1}{25} 25 = 1\end{aligned} \]

\( P(X = x) = \frac{x+2}{25} \quad \quad x \in \{1,2,3,4,5\} \)

Distribution A

Distribution B

Distribution A

- values are farther from the mean
- larger standard deviation

Distribution B