Notes Chapter 6 - Sum/Difference of Random Variables
w/o November 14 2016
Rules for sum and difference of random variables
Rules for mean and standard deviation of a random variable: E(x) =  x =
( x   )
2
x
 x  p(x)
2
Var(x) =  x =
 p ( x)
 a  bx = a + b  x
Rules for linear transformation on a random variable:
Rules for sum and difference of random variables X and Y:
 2 a  bx  b 2 2 x
and  x  y   x   y
 x y = x   y
If X and Y are independent random variables, then:  2 X  Y =  2 x +  2 Y
 X Y   X2 Y
 a bx  b  x
and
 2 X  Y   2 x   2Y
and
 X Y   X2 Y
Notes and Examples
1. Let X be the random variable with the following probability distribution.
X
P(X)
5
0.05
10
0.30
15
0.25
20
0.25
25
0.15
(a) Enter the chart into the calculator, using one –variable stats .
Find the mean, standard deviation, and the variance:
mean __15.75_____ St. dev. ___5.761____ variance ____33.1875______
Remember: to get the variance square the standard deviation.
(b) Using the rules above, find the mean, standard deviation, and variation for the random variable “y”
where y = – 3 + 4X.
 abx  a  b x
 y  3  4 x
 abx  b  x
 y  4 x
 2  b 2 x2
a b x
 y2  4 2  x2
 3  4(15.75)
 4(5.761)
 16(33.1875)
 60
 23.044
 531
*Always write the formulas and show all substitutions.
Let’s Check are answers.
(c) In list 3, use the linear transformations to change random variable X to random variable Y.
That is let L3 = -3 + 4L1
(d) Then using one-variable stats, find the mean, standard deviation, and variance. (Notice: Do you get the same results?)
Use 1-Var Stats L3, L2
Example 2 Enter the following values in your calculator:
(assume each value is equally likely)
Enter X into L1 and Y into
Random variable X = 2 9 11 22
Random variable Y = 5 7
15
(a) Find the following using the calculator (one-variable statistics)
 x __11___
 x ___7.176_____  2 x ___51.5___
 y ___4.32____  2 y ____18.667___
 y __9___
(b) Add X and Y and put the values in list 3.
(You are going to add each x value to each y value – so you will enter 12 numbers: 2+5, 2+7, 2+15, 9+5, 9+7, ...)
Then using the calculator (One-variable)
Find:  X Y __20___  X Y _ 8.377____  2 X  Y __70.167 _
Now check with the rules above.
(c) Subtract X and Y (X – Y) and put values in list 3.
(You are going to add each x value to each y value – so you will enter 12 numbers: 2-5, 2-7, 2-15, 9-5, 9-7, ...)
Then using the calculator (one-variable)
Find:  X Y = __2___
 X Y = __8.377___
Example 3
Given:  x  10  x  2
and
 10  20
 10
2
X Y
= __70.167 _
Now check with the rules above.
 y  20  y  3 assume that it is known that X and Y are independent,
(b) Find  x 3 y
(a) Find  x y and  x y
 x y   x   y

 x y   x y
  x2   y2
and  x 3 y
 x3 y   x  3 y
 2 x 3 y   x23 y
 10  3(20)
 70
  x2   32y
 2 2  32
  x2  32  y2
 13
 2 2  9(3) 2
 85
Try the following problems on you own
– you may work together or in your groups.
- you will get to check your answers today
1) Given:  x  4 and  x = 1.2 Find the mean, standard deviation, and variance of Y where Y = 4 – 6x
2) Using the given from example 3 above find  4 x2 y and  4 x 2 y
3) You have two scales for measuring weights in a chemistry lab. Both scales give answers that vary a bit in repeated
weighing of the same item. If the true weight of a compound is 2 grams, the first scale produces readings X that have a
mean 2.000 grams and the standard deviation is 0.002 grams. The second scale’s readings Y have mean 2.001 grams and
standard deviation 0.001 grams. What are the mean and standard deviation of the difference Y – X between the
readings? (Assume the readings X and Y are independent)