Nathan VanHoudnos

10/17/2014

- Homework comments
- Checkpoint #12 results
- Lecture 11 (covers pp. 138-155)

- Homework comments
- Checkpoint #12 result
- Lecture 11 (covers pp. 128-155)

to fill in

The random variable X, representing the number of accidents in a certain intersection in a week, has the following probability distribution:

```
x | 0| 1| 2| 3| 4| 5|
-------------------------------------------
P(X=x)| 0.20| 0.30| 0.20| 0.15| 0.10| 0.05|
```

By the third day of a particular week, 2 accidents have already occurred in the intersection. What is the probability that there will be less than a total of 4 accidents during that week?

By the third day of a particular week, 2 accidents have already occurred in the intersection. What is the probability that there will be less than a total of 4 accidents during that week?

```
x | 0| 1| 2| 3| 4| 5|
-------------------------------------------
P(X=x)| 0.20| 0.30| 0.20| 0.15| 0.10| 0.05|
```

“Two accident have already occurred” implies that at least two accidents will occur this week. We condition on \( X \ge 2 \).

The “less than a total of 4” implies \( X < 4 \)

Therefore we want \( P(X < 4|X \ge 2) \).

By definition \( P(X < 4|X \ge 2) \) is

\[ P(X < 4|X \ge 2) = \frac{P( X < 4 \text{ and } X \ge 2 )}{P(X \ge 2)} \]

The only counts less than 4 and greater than or equal to 2 are 2 and 3:

\[ P( X < 4 \text{ and } X \ge 2 ) = P(X =2 ) + P(X = 3) \]

By the third day of a particular week, 2 accidents have already occurred in the intersection. What is the probability that there will be less than a total of 4 accidents during that week?

```
x | 0| 1| 2| 3| 4| 5|
-------------------------------------------
P(X=x)| 0.20| 0.30| 0.20| 0.15| 0.10| 0.05|
```

\[ \begin{aligned} P(X < 4|X \ge 2) & = \frac{P(X =2 ) + P(X = 3)}{P(X \ge 2)} \\ & = \frac{.2 + .15}{.20 + .15 +.10 +.05} \\ & = \frac{.35}{.5} = .70 \end{aligned} \]

- Homework comments
- Checkpoints #12 result
- Lecture 11 (covers pp. 128-155)

**Checkpoint 13: Expection and variance rules**(finish today, slides marked with R are review)- Checkpoint 15: Normal random variables

Let \( Y = X + 6 \) be a new random variable.

Is \( E[Y] = E[X] + 6 \)?

\[ \begin{aligned} E[Y] & = \sum y \cdot P(Y=y) \\ & = \sum (x + a) \cdot P(X + a = x + a) \\ & = \sum (x + a) \cdot P(X = x) \\ & = \left( \sum x \cdot P(X=x) \right ) + \left( \sum a \cdot P(X=x) \right) \\ & = E[X] + a \cdot \left(\sum P(X=x) \right) \\ & = E[X] + a \end{aligned} \]

Let \( Y = 4X \) be a new random variable.

Is \( E[Y] = 4 \cdot E[X] \)?

\[ \begin{aligned} E[Y] & = \sum y \cdot P(Y=y) \\ & = \sum (x b ) \cdot P(X b = x b) \\ & = \sum (x b) \cdot P(X = x) \\ & = b \cdot \left( \sum x \cdot P(X=x) \right ) \\ & = b \cdot E[X] \end{aligned} \]

We can combine these rules:

\[ \begin{aligned} E[Y] & = E[a + b X] \\ & = a + E[ b X] \\ & = a + b \cdot E[X] \end{aligned} \]

At StatsPie, the number of toppings on the typical pizza is a random variable X having a mean value of 3. The cost of a pizza is $10 plus $1.50 per topping. How much does the average customer spend on a pizza?

\[ \begin{aligned} E[X] & = 3 & Y & = 10 + 1.5 X \end{aligned} \]

\[ \begin{aligned} E[Y] & = E[10 + 1.5 X] \\ & = 10 + 1.5 \cdot E[X] \\ & = 10 + 1.5 \cdot 3 = 14.50 \end{aligned} \]

implying the average pizza costs $14.50.

Let the **variance** of the random variable \( X \) be defined as

\[ \text{Var}[X] = E\left[ \left(X - E[X]\right)^2 \right] \]

i.e. the **expected value** of the **squared distance from the mean**.

Using the rules for expected values we derive the following identity:

\[ \text{Var}[X] = E[X^2] ~ - ~ \left( E[X] \right)^2 \]

which is easier to calculate by hand.

\[ \begin{aligned} \text{Var}[X] & = E[X^2] ~ - ~ \left( E[X] \right)^2 \\ % E[X] & = \sum_{x \in S} x \cdot P(X = x) \\ % E[X^2] & = \sum_{x \in S} x^2 \cdot P(X = x) \end{aligned} \]

```
x | 0| 1| 2| 3| 4| 5|
--------+----+----+----+----+----+----|
P(X = x)| .28| .37| .23| .09| .02| .01|
```

\[ \begin{aligned} E[X] & = \sum_{x \in S} x \cdot P(X = x) \\ & = ~ 0 \times .28 + 1 \times .37 + 2 \times .23 \\ & ~ + \! 3 \times .09 + 4 \times .02 + 5 \times .01 \\ & = 1.23 \end{aligned} \]

\[ \begin{aligned} \text{Var}[X] & = E[X^2] ~ - ~ \left( E[X] \right)^2 \\ % E[X] & = \sum_{x \in S} x \cdot P(X = x) \\ % E[X^2] & = \sum_{x \in S} x^2 \cdot P(X = x) \end{aligned} \]

```
x | 0| 1| 2| 3| 4| 5|
--------+----+----+----+----+----+----|
P(X = x)| .28| .37| .23| .09| .02| .01|
```

\[ \begin{aligned} E[X^2] & = \sum_{x \in S} x^2 \cdot P(X = x) \\ & = ~ 0^2 \times .28 + 1^2 \times .37 + 2^2 \times .23 \\ & ~ + \! 3^2 \times .09 + 4^2 \times .02 + 5^2 \times .01 \\ & = 2.67 \end{aligned} \]

\[ \begin{aligned} E[X] & = 1.23 & E[X^2] & = 2.67 \end{aligned} \]

```
x | 0| 1| 2| 3| 4| 5|
--------+----+----+----+----+----+----|
P(X = x)| .28| .37| .23| .09| .02| .01|
```

\[ \begin{aligned} \text{Var}[X] & = E[X^2] - \left( E[X] \right)^2 \\ & = 2.67 - (1.23)^2 \\ & = 1.16 \end{aligned} \]

implying that college students change their majors on average 1.23 times with a **variance** of 1.16 “times-squared”.

Let the **standard deviation** be the square root of the variance.

\[ \text{sd}[X] = \sqrt{\text{Var}[X]} \]

Example:

\[ \begin{aligned} \text{sd}[X] & = \sqrt{\text{Var}[X]} \\ & = \sqrt{1.16} & = 1.078 \end{aligned} \]

implying that college students change their majors on average 1.23 times with a **standard deviation** of 1.08 times.

Using the rules for expectations, we can show that:

\[ \text{Var}[a + bX] = b^2 \cdot \text{Var}[X] \]

Therefore

\[ \text{sd}[a + bX] = b \cdot \text{sd}[X] \]

At StatsPie, the number of toppings on the typical pizza has a mean value of 3 with a standard deviation of ½. The cost of a pizza is $10 plus $1.50 per topping.

What are the mean and standard deviation of the distribution of pizza cost?

\[ \begin{aligned} E[X] & = 3 & \text{sd}[X] & = 0.5 \\ Y & = 10 + 1.5 X \\ E[Y] & = ? & \text{sd}[Y] & = ? \end{aligned} \]

\[ \begin{aligned} E[X] & = 3 & \text{sd}[X] & = 0.5 \\ Y & = 10 + 1.5 X \\ \end{aligned} \]

Recall that \( E[Y] = 14.50 \).

Find the standard deviation:

\[ \begin{aligned} \text{sd}[Y] & = \text{sd}[10 + 1.5 X] \\ & = 1.5 \cdot \text{sd}[X] \\ & = 1.5 \cdot 0.5 = .75 \end{aligned} \]

At StatsPie, the number of toppings on the typical pizza has a mean value of 3 with a standard deviation of ½. The cost of a pizza is $10 plus $1.50 per topping.

What are the mean and standard deviation of the distribution of pizza cost?

\[ \begin{aligned} E[Y] & = 14.50 & \text{sd}[Y] & = .75 \end{aligned} \]

implying that the pizza typically costs an average of $14.50 with a standard deviation of $0.75.

Let \( X \) and \( Y \) be random variables and define \( Z = X + Y \) and \( Q = X - Y \).

It can be shown that, in all cases:

\[ \begin{aligned} E[Z] & = E[X] + E[Y] \\ E[Q] & = E[X] - E[Y] \end{aligned} \]

And, **if and only if X and Y are independent**, that

\[ \begin{aligned} \text{Var}[Z] & = \text{Var}[X] + \text{Var}[Y]\\ \text{Var}[Q] & = \text{Var}[X] + \text{Var}[Y] \end{aligned} \]

**If and only if X and Y are independent**, then

\[ \begin{aligned} \text{Var}[Z] & = \text{Var}[X] + \text{Var}[Y]\\ \text{Var}[Q] & = \text{Var}[X] + \text{Var}[Y] \end{aligned} \]

Further note that, **iff X and Y are independent**
\[ \begin{aligned}
\text{sd}[Z] & = \sqrt{\text{Var}[X] + \text{Var}[Y]}\\
\text{sd}[Q] & = \sqrt{\text{Var}[X] + \text{Var}[Y]}
\end{aligned} \]

Standard deviations

DO NOTadd.

**Not covered in Stat 202.**

Answer:

Let \( X \) and \( Y \) be random variables and define \( Z = aX + bY + c \).

\[ \text{Var}[Z] = a^2 \cdot \text{Var}[X] + b^2 \cdot \text{Var}[Y] - 2ab \cdot \text{Cov}[X,Y] \]

where

\[ \text{Cov}[X,Y] = E\big[ \left( X - E[X] \right) \cdot \left( Y - E[Y] \right) \big] \]

- Homework comments
- Checkpoints #12 result
- Lecture 11 (covers pp. 128-155)

- Checkpoint 13: Exception and variance rules
**Checkpoint 15: Normal random variables**

IQ: 2 categories

IQ: 3 categories

and lots more…

and infinitely more…