AP Statistics Section 7.2 C
Rules for Means & Variances
Consider the independent random variables X and Y
and their probability distributions below:
2 .7
2 . 41
2 .6
. 84
 x  1(. 2 )  2 (. 5 )  5 (. 3 )  2 . 7
 x  (1  2 . 7 ) (. 2 )  ( 2  2 . 7 ) (. 5 )  ( 5  2 . 7 ) (. 3 )  2 . 41
2
2
2
2
Build a new random variable X + Y and calculate the probabilities
for the values of X + Y.
3
5
4
6
7
9
3
.14
4
.35
5
.06
6
.15
7
.21
P(3)  P(1  2)  .2  .7  .14
9
.09
Use your calculator to calculate the mean of the
random variable X + Y.
 x  y  5 .3
Note that the mean of the sum  x  y = ____
5 . 3 equals the
sum of the means  x   y =______________
2 .7  2 .6  5 .3 :
Use your calculator to calculate the variance of the
random variable X + Y.

2
x y
 3 . 25
Note that the variance of the sum equals the sum of
the variances:
 
2
x
2
y
 2 . 41  . 84  3 . 25
Repeat the steps above for the random variable X – Y.
1 -3
0
-2
3
1
 3  2 1 0 1
3
.06 .15 .14 .35 .09 .21
Verify  x  y   x   y .
. 1  2.7  2.6
.1  .1
Repeat the steps above for the random variable X – Y.
 3  2 1 0 1
3
.06 .15 .14 .35 .09 .21
1
-3
0
-2
3
1
Calculate the variance of the random variable X – Y.
 x  y  3 . 25
2
Note that the variance of the difference
the sum of the variances 
2
x
 x y
2
and 
equals
2
y
Rules for Means
Rule 1: If X is a random variable and a and b are
a  b .
constants, then   _______
a  bx
If a is added to each value
If each value
x
of x, then a is added to the mean as well.
of x is multiplied
multiplied
by b , then the mean is
by b as well.
Rules for Means
Rule 2: If X and Y are random variables, then
 
 

 ______and

 _______
x y
x
y
x y
x
y
Rules for Variances
Rule 1: If X is a random variable and a and b are
2
2
2
b



constants, then a  bx ______
x
Adding a to each value
Multiplyin g each value
of x does not change the variance.
of x by b , multiplies the variance by b
2
Rules for Variances
Rule 2: If X and Y are independent random
variables, then
 x y   x  
2
2
2
y
and  x - y  
2
2
x

2
y
Example: Consider two scales in a chemistry lab. Both scales give
answers that vary a little in repeated weighings of the same
item. For a 2 gram item, the first scale gives readings X with a
mean of 2g and a standard deviation of .002g. The second
scale’s readings Y have a mean of 2.001g and a standard
deviation of .001g.
If X and Y are independent, find the mean and standard
deviation of Y – X.
 y  x   y   x  2 . 001  2  . 001 g

2
yx


yx
2
y


2
x
 . 002
2
 . 001  . 000005
2
. 000005  . 002236
Example: Consider two scales in a chemistry lab. Both scales give
answers that vary a little in repeated weighings of the same item. For
a 2 gram item, the first scale gives readings X with a mean of 2g and a
standard deviation of .002g. The second scale’s readings Y have a
mean of 2.001g and a standard deviation of .001g.
You measure once with each scale and average the readings.
Your result is Z = (X+Y)/2. Find .
Note : z  1 x  1 y
2
2
 z  1  x  1  y  . 5 ( 2 )  . 5 ( 2 . 001 )  2 . 0005
2

2
z

z 
2
2
1 x 1 y
2
2
 1 
2
. 00000125
2
x
 1 
2
2
y
 2
 1
2
 2  (. 001 )
(. 002 )  1
 . 001118034
2
2
2
 . 00000125
Any linear combination of
independent Normal random
variables is also Normally
distributed.
Example: Tom and George are playing in the club golf
tournament. Their scores vary as they play the course
repeatedly. Tom’s score X has the N(110, 10) distribution and
George’s score Y has the N(100, 8) distribution. If they play
independently, what is the probability that Tom will score lower
than George?
Table :
P ( X  Y )  P ( X  Y  0)
0 - 10
z
  . 78
12.806
. 2177
  10
0
  12 . 806
Calculator
:
 x  y  110  100  10
normalcdf( -10000,0,1 0,12.806)
 x y 
.2174
10  8  12 . 806
2
2