Stat 202: Lecture 12 (End of Lecture 11 + Review)

Nathan VanHoudnos
10/17/2014

Agenda

  1. Homework comments
  2. Checkpoint #13 results
  3. Lecture 12

Homework comments

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)

Agenda

  1. Homework comments
  2. Checkpoint #13 results
  3. Lecture 12

Checkpoint #13 results

  • 92% correct
  • Good job!

Agenda

  1. Homework comments
  2. Checkpoint #13 results
  3. Lecture 12
    • Finishing up Lecture 11
    • Review

A further example

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.

Extending the 68–95–99.7 rule

plot of chunk unnamed-chunk-1

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)

A standard normal table

plot of chunk unnamed-chunk-2

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

To use:

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

Best explained by looking at one.

Example

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

Example

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:

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

Pulling it together

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.

Using R

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

Using R

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?

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

pnorm( 2.68 )
[1] 0.9963

Using R

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

1 - pnorm( 2.68 )
[1] 0.003681

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

Going the other way

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 \( x \)?

Going the other way

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

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.

Going the other way

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

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.

Going the other way

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

Going the other way

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

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

Pulling it together

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.

Using R

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:

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

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

Using R

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?

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

qnorm(.90)
[1] 1.282

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

Using R

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?

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.

A further example

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”:

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

Picture

plot of chunk unnamed-chunk-8

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

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

Solution

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

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

Pulling it together

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 Summary

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)

Midterm

Let's talk about it.

Midterm

  • 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.

Some hints

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