Download 6.1-6.2 Review

6.1-6.2 Review
1.4.2016
6.1 Discrete and Continuous Random
Variables
• Let’s start with discrete random variables
• Remember, discrete means that there are a fixed number of possible
outcomes
•
•
•
•
•
•
Tossing a coin
Picking a card from a deck
Guessing a person’s eye color
Guessing what block/period someone has math class
Rolling a die
Etc.
Discrete Random Variables
• Let’s use the example of rolling a die
• So we say X is the number of dots that are showing on the die
• The possible outcomes are:
•1
2
3
4
5
6
• What is the probability of each of these outcomes?
Probability Distribution of Rolling a Die
Value
1
2
3
4
5
6
Probability
1/6
1/6
1/6
1/6
1/6
1/6
Probability Distribution of rolling two dice
• Possible outcomes:
•2 3
4
5
6
7
8
9
10
11
12
• But now, they don’t all have the same probabilities
• To find the probability distribution, we would find all possible
outcomes
• There will be 36 different combinations between the two dice (6 on each die,
so 6x6)
• Then we count how many of these outcomes have a sum of 2, a sum of 3, etc.
Probability Distribution of Rolling Two Dice
• Possible outcomes:
•2 3
4
5
6
7
8
9
10
11
12
• But now, they don’t all have the same probabilities
Value
2
3
4
5
6
7
8
9
10
11
12
Prob
1/36
1/18
1/12
1/9
5/36
1/6
5/36
1/9
1/12
1/18
1/36
Mean (Expected Value) of a Discrete Random
Variable
• Finding the mean of your discrete random variable is easy
• And, it is on your formula sheet for the AP exam
• You just have to know what you’re looking for
• To find the mean, you multiply each value by its corresponding
probability
• Then you add them all together
Mean of Discrete Random Variable
• So, in our example of rolling one die…
Value
1
2
3
4
5
6
Prob
1/6
1/6
1/6
1/6
1/6
1/6
• We would do (1)(1/6)+(2)(1/6)+(3)(1/6)+(4)(1/6)+(5)(1/6)+(6)(1/6)
• So your mean is 3.5
• Or, the expected value is 3.5
Using Your Calculator for Discrete Random
Variables
• Fortunately, we can make our calculator do some of the work for us
• If you enter the values into List1, and your probabilities into List2
• Then we will use the command 1-Var Stats L1,L2
Let’s Try an Example AP Problem
• This is #5 from 1999
• This one is hard—do your best
Continuous Random Variables
• Continuous random variables have an infinite number of possible
values
• Often occurs with measuring
• Although we usually round, most measurements are not exact
• If you weigh yourself every morning, the measurements are likely to vary from day to
day, even if only by .07 pounds
• Because there are an infinite number of possible values, the
probability of any particular outcome is infinitely small
• Or, in other words, the probability is 0
Continuous Random Variables
• Instead, we calculate the probability that the value is within a certain
range
• We typically do this using a normal distribution
• Example: in the United States, the mean height for men is 69.1 inches, with a
standard deviation of 2.9 inches
• We can then calculate the probability of being in a certain range
• Using either our standard normal table or our calculator
• I recommend using your calculator, but if you prefer the table, that is fine
Continuous Random Variables
• Example: in the United States, the mean height for men is 69.1 inches, with
a standard deviation of 2.9 inches
• We can then calculate the probability of being in a certain range
• Using either Table A or our calculator
• I recommend using your calculator, but if you prefer the table, that is fine
• So if we want to find the probability of being between 70 and 72 inches:
• Normalcdf(70,72,69.1,2.9)
• Convert 70 and 72 to z-scores: 0.31 and 1
• Then use the Table A to find area to the left of each
• Then do area to the left of 1 minus the area to the left of .31
6.2 Transforming Random Variables
• X is our random variable (discrete or continuous)
• If Y=a+bX, this means that we are adding a constant (a) to X and/or
multiplying X by a constant (b)
• The main thing that we are interested in here is how this changes the
mean and the standard deviation
6.2 Transforming Random Variables
• If Y=a+bX, this means that we are adding a constant (a) to X and/or
multiplying X by a constant (b)
• The main thing that we are interested in here is how this changes the
mean and the standard deviation
• New mean=a+b(old mean)
• Or, in other words, our new mean is our old mean times b, plus a
• New Stdev=|b|(Old stdev)
• Our new standard deviation is our old standard deviation times the absolute value of b
An Example
• The mean on the semester 1 final exam was a 58.7 (true story)
• The standard deviation was 12.8
• If I decided that my normal curve was too complicated, and I wanted to do
a curve just with adding and multiplying, then we could use this rule
• Let’s say that I multiplied all scores by 1.2 and then added 6
• So score of 60 would become a 78
• Our new mean would be 6+1.2(58.7)
• 76.44
• Our new standard deviation would be |1.2|*12.8
• 15.36
6.2 Combining Random Variables
• When we combine 2 random variables, we can either add them or
subtract them
• When we add X+Y, the new mean is just mean(X)+ mean(Y)
• When we subtract X-Y, the new mean is just mean(X)-mean(Y)
• When we add X+Y or subtract X-Y, the new VARIANCE is equal to
variance(X)+variance(Y)
• WE MUST USE VARIANCES, NOT STANDARD DEVIATIONS
• Remember, variance is equal to standard deviation squared
An Example
• X is a random variable that specifies the number of students with an
A in one class
• Mean 5.1, standard deviation of 4.2
• Y is a random variable that specifies the number of students with an A
in a different class
• Mean 10.3, standard deviation of 3.4
An Example
• X is a random variable that specifies the number of students with an
A in one class
• Mean 5.1, standard deviation of 4.2
• Y is a random variable that specifies the number of students with an A
in a different class
• Mean 10.3, standard deviation of 3.4
• If we want to know the expected number of A’s in the two classes
combined, we would add X+Y
• So the mean would be 5.1+10.3 = 15.4
• See next slide for standard deviation
An Example
• Finding the standard deviation
• X: mean=5.1 st.dev=4.2
• Y: mean=10.3 st.dev=3.4
• Variance of X is 4.2^2 = 17.64
• Variance of Y is 3.4^2 = 11.56
• So the new variance is 29.2
• To find the new standard deviation we take the square root of 29.2
• 5.4
• So our new variable has a mean of 15.4 and a standard deviation of 5.4
Another Example (An AP-type question)
• 3 runners are running a relay. Each runs 1 mile—the team’s overall time is
the sum of the 3 runners’ times
1.
Runner
Mean time
St. dev
Runner 1
4.5
.15
Runner 2
5
.2
Runner 3
4.2
.1
Runner 1 thinks that he can run faster than 4.4 minutes in the next race.
What is the probability of this?
2. What are the mean and standard deviation of the overall team time?
3. The team’s best time ever is 12.9. What is the probability that they beat
this time in their next race?
Answers
1. .252
2. mean=13.7, st.dev=.269258
3. .0015 (less than 1% chance)