Nathan VanHoudnos

10/17/2014

- Homework comments
- Checkpoint #13 results
- Lecture 12

Question 6:

- (A) What is the IQ of a women 4 standard deviations from the mean? (5 pts)
- (E) What is the IQ of a man 4 standard deviations from the mean? (5 pts)

Was meant to be:

- (A) What is the IQ of a women 4 standard deviations above the mean? (5 pts)
- (E) What is the IQ of a man 4 standard deviations above the mean? (5 pts)

- Homework comments
- Checkpoint #13 results
- Lecture 12

- 92% correct
- Good job!

- Homework comments
- Checkpoint #13 results
- Lecture 12
- Finishing up Lecture 11
- Review

At StatsPie, the amount of cheese used to make a typical pizza is normally distributed with a mean value of 6 oz with a variance of ¼ oz. What is the probability that a pizza will use at least 7.34 oz of cheese?

Step 1: How many standard deviations is 7.34 away from the mean? \[ \begin{aligned} P(X \ge 7.34) & = P(Z \ge z) \\ z & = \frac{7.34 - 6}{\sqrt{1/4}} = 2.68 \end{aligned} \] implying 7.34 is 2.68 standard deviations from the mean.

We know that \( P(Z \ge 2.68 ) \) will be somewhere between 2.5% and 0.15%.

A famous result in calculus is that

**there is no closed form formula**for the area under a normal curve.Use a table (or software)

Gives the area to the left of \( z \).

To use:

- Express your z-score as
**less than**. - Find the
**row**with the first two digits of \( z \). - Find the
**column**with the third digit of \( z \) - Read off the value.

Best explained by looking at one.

Step 1:

\( P(Z \ge 2.68 ) = 1 - P(Z \le 2.68 ) \)

Step 2: \( 2.68 = 2.6 + 0.08 \) find \( 2.6 \)

z | .00 | .01 | .02 | .03 | .04 | .05 | .06 | .07 | .08 | .09 |
---|---|---|---|---|---|---|---|---|---|---|

2.5 | .9938 | .9940 | .9941 | .9943 | .9945 | .9946 | .9948 | .9949 | .9951 | .9952 |

2.6 | .9953 | .9955 | .9956 | .9957 | .9959 | .9960 | .9961 | .9962 | .9963 | .9964 |

2.7 | .9965 | .9966 | .9967 | .9968 | .9969 | .9970 | .9971 | .9972 | .9973 | .9974 |

We want:

z | .00 | .01 | .02 | .03 | .04 | .05 | .06 | .07 | .08 | .09 |
---|---|---|---|---|---|---|---|---|---|---|

2.6 | .9953 | .9955 | .9956 | .9957 | .9959 | .9960 | .9961 | .9962 | .9963 | .9964 |

Step 1: \( P(Z \ge 2.68 ) = 1 - P(Z \le 2.68 ) \)

Step 2: \( 2.68 = 2.6 + 0.08 \) find \( 2.6 \)

z | .00 | .01 | .02 | .03 | .04 | .05 | .06 | .07 | .08 | .09 |
---|---|---|---|---|---|---|---|---|---|---|

2.6 | .9953 | .9955 | .9956 | .9957 | .9959 | .9960 | .9961 | .9962 | .9963 | .9964 |

Step 3: \( 2.68 = 2.6 + 0.08 \) find \( 0.08 \)

z | .08 |
---|---|

2.6 | .9963 |

Solve:

\( P(Z \ge 2.68 ) = 1 - P(Z \le 2.68 ) = 1 - .9963 = 0.0037 \)

At StatsPie, the amount of cheese used to make a typical pizza is normally distributed with a mean value of 6 oz with a variance of ¼ oz. What is the probability that a pizza will use at least 7.34 oz of cheese?

\[ \begin{aligned} P(X \ge 7.34) & = P(Z \ge 2.68 ) \\ & = 1 - P(Z \le 2.68 ) \\ & = 1 - .9963 = 0.0037 \end{aligned} \]

implying that only 0.37% of pizzas will use at least 7.34 oz of cheese.

At StatsPie, the amount of cheese used to make a typical pizza is normally distributed with a mean value of 6 oz with a variance of ¼ oz. What is the probability that a pizza will use at least 7.34 oz of cheese?

Find the z-score:

```
(7.34-6)/sqrt(1/4)
```

```
[1] 2.68
```

Lookup \( P(Z \le 2.68) \)

```
pnorm( 2.68 )
```

```
[1] 0.9963
```

Find \( P(Z \ge 2.68) = 1 - (Z \le 2.68) \)

```
1 - pnorm( 2.68 )
```

```
[1] 0.003681
```

implying that 0.37% of pizzas use at least 7.34 oz of cheese.

At StatsPie, the amount of cheese used to make a typical pizza is normally distributed with a mean value of 6 oz with a variance of ¼ oz. How much cheese would the top 10% of pizzas contain?

We know that:

\[ P(X \ge x) = .1 \]

what is \( x \)?

\[ \begin{aligned} P(X \ge x) & = .10 \\ & = P(\frac{X - 6}{\sqrt{1/4}} \ge \frac{x - 6}{\sqrt{1/4}}) \\ & = P(Z \ge z ) = .10 \end{aligned} \]

To find the correct \( z \) value we express it as less than:

\[ P(Z \ge z) = .1 = 1 - P(Z \le z) \]

Implying that we need to lookup

\[ P(Z \le z) = 1 - .1 = .90 \]

in our table.

\[ P(Z \le z) = 1 - .0001 = .90 \]

z | .00 | .01 | .02 | .03 | .04 | .05 | .06 | .07 | .08 | .09 |
---|---|---|---|---|---|---|---|---|---|---|

1.2 | .8849 | .8869 | .8888 | .8907 | .8925 | .8944 | .8962 | .8980 | .8997 | .9015 |

1.3 | .9032 | .9049 | .9066 | .9082 | .9099 | .9115 | .9131 | .9147 | .9162 | .9177 |

That is somewhat unsatisfying…

- \( P(Z \le 1.28) = .8997 \) and
- \( P(Z \le 1.29) = .9015 \)

Choose \( 1.28 \) as its probability is closer to .90.

We wanted: \( P(Z \le z) = 1 - .1 = .90 \)

We have: \( P(Z \le 1.28) = .8997 \).

Just go with it. Therefore: \[ \begin{aligned} P(X \ge x) & = .10 \\ & = P(\frac{X - 6}{\sqrt{1/4}} \ge \frac{x - 6}{\sqrt{1/4}}) \\ & = P(Z \ge 1.28 ) = .10 \end{aligned} \]

So solve: \( \frac{x - 6}{\sqrt{1/4}} = 1.28 \)

\[ \frac{x - 6}{\sqrt{1/4}} = 1.28 \]

Implies \[ x = 6 + 1.28 \sqrt{1/4} = 6.64 \]

At StatsPie, the amount of cheese used to make a typical pizza is normally distributed with a mean value of 6 oz with a variance of ¼ oz. How much cheese would the top 10% of pizzas contain?

\[ P(X \ge x) = .1 \]

\[ x = 6 + 1.28 \sqrt{1/4} = 6.64 \]

implies that at least 6.64 oz of cheese are used to make the top 10% of pizzas.

At StatsPie, the amount of cheese used to make a typical pizza is normally distributed with a mean value of 6 oz with a variance of ¼ oz. How much cheese would the top 10% of pizzas contain?

\[ P(X \ge x) = .1 \]

Express the z-score probability as a less than:

\[ P(Z \ge z) = 1 - P(Z \le z) = .1 \]

\( P(Z \le z) = .90 \)

Look \( P(Z \le z) = .90 \) up in R

```
qnorm(.90)
```

```
[1] 1.282
```

implying that the z score is \( 1.282 \).

Convert \( z = 1.282 \) to the units of X.

```
6 + sqrt(1/4) * 1.282
```

```
[1] 6.641
```

implying that the top 10% of pizzas use at least \( 6.64 \) of cheese.

At StatsPie, the amount of cheese used to make a typical pizza is normally distributed with a mean value of 6 oz with a variance of ¼ oz. What is the probability of getting a pizza that has between 5.4 and 6.7 ounces of cheese?

\[ P(5.4 \le X \le 6.7) = ? \]

Draw a picture to express it in terms of “less than”:

\[ P(5.4 \le X \le 6.7) = P(X \le 6.7) - P(X \le 5.4) \]

\[ \begin{aligned} P(& 5.4 \le X \le 6.7) = \\ & P(X \le 6.7) - P(X \le 5.4) \end{aligned} \]

Since you already know how to solve the “less than” problems, you can solve this problem too.

\( P(X \le 6.7) = P(\frac{X-6}{\sqrt{1/4}} \le \frac{6.7-6}{\sqrt{1/4}}) = P(Z \le 1.4) \)

```
pnorm(1.4)
```

```
[1] 0.9192
```

\( P(X \le 5.4) = P(\frac{X-6}{\sqrt{1/4}} \le \frac{5.4-6}{\sqrt{1/4}}) = P(Z \le -1.2) \)

```
pnorm(-1.2)
```

```
[1] 0.1151
```

At StatsPie, the amount of cheese used to make a typical pizza is normally distributed with a mean value of 6 oz with a variance of ¼ oz. What is the probability of getting a pizza that has between 5.4 and 6.7 ounces of cheese?

\[ \begin{aligned} P(5.4 \le X \le 6.7) & = P(X \le 6.7) - P(X \le 5.4) \\ & = 0.9192 - 0.1151 \\ & = 0.8041 \end{aligned} \]

implying that 80% of pizzas contain between 5.4 and 6.7 oz of cheese.

Normal problems look something like this:

- Convert to a standard normal
- Express the probabilities in terms of less than
- Lookup either the probability or z-score in the table
- Convert back to the original distribution (if needed)

Let's talk about it.

The high levels of performance on the checkpoints and homeowork assingments suggests that the class has already mastered certain skills.

The goal of the midterm is to assess your knowledge. The midterm will

**touch briefly upon**the mastered skills,**focus on**on the (presummed) unmastered skills, and**stretch you**to use what you have learned in new ways.

Be sure to know how to:

- Interpret boxplots, scatterplots, five number summaries, and R's regression output
- Check independence of events and discrete random variables
- Calculate and interpret conditional probabilities
- Use Bayes' Rule and interpret the results
- Use a standard normal table to solve normal random variable problems