1. Warm-up Page 356 #27-30 Submit answers to Socrative
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R.E.D.

2. Transforming and Combining Random Variables
So far, we have looked at settings that involve a single random variable. Many interesting statistics problems require us to examine two or more random variables.
Let’s investigate the result of adding and subtracting random variables.
Transforming and Combining Random Variables

3. Transforming and Combining Random Variables
Let X = the number of passengers on a randomly selected trip with Pete’s Jeep Tours.
Y = the number of passengers on a randomly selected trip with Erin’s Adventures.
Define T = X + Y. What are the mean and variance of T?
Transforming and Combining Random Variables
Passengers xi 2 3 4 5 6
Probability pi
0.15
0.25
0.35
0.20
0.05
Mean µX = Standard Deviation σX = 1.090
Passengers yi 2 3 4 5
Probability pi
0.3
0.4
0.2
0.1
Mean µY = Standard Deviation σY = 0.943

4. Transforming and Combining Random Variables
How many total passengers can Pete and Erin expect on a randomly selected day?
Since Pete expects µX = 3.75 and Erin expects µY = 3.10 , they will average a total of = 6.85 passengers per trip. We can generalize this result as follows:
Transforming and Combining Random Variables
Mean of the Sum of Random Variables
For any two random variables X and Y, if T = X + Y, then the expected value of T is
E(T) = µT = µX + µY
In general, the mean of the sum of several random variables is the sum of their means.
How much variability is there in the total number of passengers who go on Pete’s and Erin’s tours on a randomly selected day? To determine this, we need to find the probability distribution of T.

5. Transforming and Combining Random Variables
The only way to determine the probability for any value of T is if X and Y are independent random variables.
Transforming and Combining Random Variables
Definition:
If knowing whether any event involving X alone has occurred tells us nothing about the occurrence of any event involving Y alone, and vice versa, then X and Y are independent random variables.
Probability models often assume independence when the random variables describe outcomes that appear unrelated to each other.
You should always ask whether the assumption of independence seems reasonable.
In our investigation, it is reasonable to assume X and Y are independent since the siblings operate their tours in different parts of the country.

6. Combining Random Variables page 366
Let T = X + Y. Consider all possible combinations of the values of X and Y.

7. Combining Random Variables page 366
Let T = X + Y. Consider all possible combinations of the values of X and Y.
Recall: µT = µX + µY = 6.85
= (4 – 6.85)2(0.045) + … +
(11 – 6.85)2(0.005) =
Note:
What do you notice about the
variance of T?

8. Transforming and Combining Random Variables
As the preceding example illustrates, when we add two independent random variables, their variances add. Standard deviations do not add.
Transforming and Combining Random Variables
Variance of the Sum of Random Variables
For any two independent random variables X and Y, if T = X + Y, then the variance of T is
In general, the variance of the sum of several independent random variables is the sum of their variances.
Remember that you can add variances only if the two random variables are independent, and that you can NEVER add standard deviations!

9. Transforming and Combining Random Variables
We can perform a similar investigation to determine what happens when we define a random variable as the difference of two random variables. In summary, we find the following:
Transforming and Combining Random Variables
Mean of the Difference of Random Variables
For any two random variables X and Y, if D = X - Y, then the expected value of D is
E(D) = µD = µX - µY
In general, the mean of the difference of several random variables is the difference of their means. The order of subtraction is important!
Variance of the Difference of Random Variables
For any two independent random variables X and Y, if D = X - Y, then the variance of D is
In general, the variance of the difference of two independent random variables is the sum of their variances.

10. Transforming and Combining Random Variables
Combining Normal Random Variables
So far, we have concentrated on finding rules for means and variances of random variables. If a random variable is Normally distributed, we can use its mean and standard deviation to compute probabilities.
An important fact about Normal random variables is that any sum or difference of independent Normal random variables is also Normally distributed.
Transforming and Combining Random Variables
Example
Page Mr. Starnes likes between 8.5 and 9 grams of sugar in his hot tea. Suppose
the amount of sugar in a randomly selected packet follows a Normal distribution with mean 2.17 g and standard deviation 0.08 g. If Mr. Starnes selects 4 packets at random, what is the probability his tea will taste right?
Let X = the amount of sugar in a randomly selected packet.
Then, T = X1 + X2 + X3 + X4. We want to find P(8.5 ≤ T ≤ 9).
P(-1.13 ≤ Z ≤ 2.00) = – =
There is about an 85% chance Mr. Starnes’s tea will taste right.
µT = µX1 + µX2 + µX3 + µX4 = = 8.68

11. µD = -3(.06) + -2(.15) + -1(.14) + 0(.35) + 1(.09) + 3(.21) = .1
µX = 2.7 X = µY = 2.6 Y = .917
Page 379 #48
µD = -3(.06) + -2(.15) + -1(.14) + 0(.35) + 1(.09) + 3(.21) = .1 X 1 2 5
P(X) .2 .5 .3 Y 2 4
P(Y) .7 .3
D = X - Y
µD = µX - µY = 2.7 – 2.6 = .1
D = X - Y -1 -3 -2 3 1
P(X-Y)
(.2)(.7)
(.2)(.3)
(.5)(.7)
(.5)(.3)
(.3)(.7)
(.3)(.3)
Both methods give the same answer.
D = X - Y -3 -2 -1 1 3
P(X-Y)
.06
.15
.14
.35
.09
.21

12. (D) 2 = (X) 2 + (Y) 2 = (1.55) 2 + (.917) 2 = 3.24
µX = 2.7 X = µY = 2.6 Y = .917
Page 379 #48
(D) 2 = (-3-.1)2(.06) + … + (3-.1) = 3.25 X 1 2 5
P(X) .2 .5 .3 Y 2 4
P(Y) .7 .3
D = X - Y
(D) 2 = (X) 2 + (Y) 2 = (1.55) 2 + (.917) = 3.24
D = X - Y -1 -3 -2 3 1
P(X-Y)
(.2)(.7)
(.2)(.3)
(.5)(.7)
(.5)(.3)
(.3)(.7)
(.3)(.3)
These answers differ only due to rounding.
D = X - Y -3 -2 -1 1 3
P(X-Y)
.06
.15
.14
.35
.09
.21

14. a) Independent. Weather conditions a year apart should be independent.
b) Not Independent. Weather patterns tend to persist for several days; today’s weather tells us something about tomorrow’s.
Page 379 #50
c) Not independent. The two locations are very close together and would likely have similar weather conditions.

15. a) Randomly selected students would presumably be unrelated.
b) µF-M = 15 points F-M =
Page 380 #52
c) We cannot find the probability based on only the mean and standard deviation. Many different distributions have the same mean and standard deviation.

16. µT = -.5 + -.5 + … + -.5 (for 365 days) = 365(-.5) = -$182.50
a) Let T = the total winnings and xi = the winnings on the ith day.
µT = … (for 365 days) = 365(-.5) = -$182.50
Page 380 #54
b) Over the course of many years, you would expect to have total winnings (losses) of -$ per year, on average. But individual years will vary from this by $ on average.
Ç2T =(15.8)2 +(15.8) (15.8)2 = 365(15.8)2 =

17. (D)2 = (Y)2 + (X)2 = (1.136) 2 + (1.0) 2 = 2.290496
µX = 2.1 X = µY = 1.0 Y = 1.0
Page 380 #56
µD = µY - µX = 1.0 – 2.1 = -1.1
(D)2 = (Y)2 + (X)2 = (1.136) 2 + (1.0) = X 1 2 3 4
P(X) .1 .2 .3 Y 1 2 3
P(Y) .4 .3 .2 .1
On average, students make 1.1 more non-word errors than they do word errors. Individual differences will vary from this by 1.51 errors, on average.
D = Y - X
D  1.51

18. pg 381 #60
a) T = R1 + R2 is Normal with mean 350 ohms and standard deviation ohms.
b) P(345 < resistance < 355) =
normcdf(345, 355, µ = 350,  = ) = .8171

19. What is the probability that the scores differ by 5 or more points?
Let L and F denote the respective scores of Leona and Fred. The difference L – F has a Normal distribution with mean µL-F = 24 – 24 = 0 points and standard deviation L-F = points. Use this to find P(L-F<-5 or L-F>5)
Page 381 #62
State:
Plan:
Do:
Conclude:
P(L-F<-5 or L-F>5) = 1 - P(-5< L-F <5) = 1 - normcdf(-5, 5, µ = 0,  = ) = = .0771
There is about an 8% chance that one of the two friends will have to buy the other one a pizza.

20. What is the probability that Ken uses more than
What is the probability that Ken uses more than .85 ounces of toothpaste if he brushes his teeth six times?
Page 381 #64
Let T be the total amount of toothpaste and Xi be the amount used the ith time he brushes his teeth. T = X1 + X2 + X3 + X4 + X5 + X6. Each of the Xi has a N(.13, .02) distribution, and we will assume that the Xi are independent. So µT = 6(.13) = .78 ounces and (T)2 = 6(.02)2 = Since each of the Xi is normally distributed, then so is T. Putting all this together, T has N(.78, .049) distribution. Use this to find P(T>.85).
State:
Plan:

21. P(T>.85) = normcdf(.85, , µ = .78,  = .049 ) = .0764
Page 381 #64
There is about an 8% chance that Ken will use the entire tube of toothpaste on his trip
Do:
Conclude:

22. Read pg
Do pg 379 #49-65 odd
Video 6.3 (12 min)
Reading Guide 6.2 #6-16
Chapter 6 Vocabulary Flash Cards
Estimated Chapter 6 Test Date is Friday March 4.

23. Section 6.2 Transforming and Combining Random Variables
Summary
In this section, we learned that…
Adding a constant a (which could be negative) to a random variable increases (or decreases) the mean of the random variable by a but does not affect its standard deviation or the shape of its probability distribution.
Multiplying a random variable by a constant b (which could be negative) multiplies the mean of the random variable by b and the standard deviation by |b| but does not change the shape of its probability distribution.

24. Section 6.2 Transforming and Combining Random Variables
Summary
In this section, we learned that…
A linear transformation of a random variable involves adding a constant a, multiplying by a constant b, or both. If we write the linear transformation of X in the form Y = a + bX, the following about are true about Y:
Shape: same as the probability distribution of X.
Center: µY = a + bµX
Spread: σY = |b|σX

25. Section 6.2 Transforming and Combining Random Variables
Summary
In this section, we learned that…
If X and Y are any two random variables,
If X and Y are independent random variables
The sum or difference of independent Normal random variables follows a Normal distribution.