Download AP Statistics Section 7.2 C Rules for Means & Variances

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
 x2  (1  2.7) 2 (.2)  (2  2.7) 2 (.5)  (5  2.7) 2 (.3)  2.41
Build a new random variable X + Y and calculate the probabilities
for the values of X + Y.
3
4
7
5
6
9
3
4
5
6
7
9
.14 .35 .06 .15 .21 .09
P(3)  P(1  2)  .2  .7  .14
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:
    2.41  .84  3.25
2
x
2
y
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.
1 - 3
0 -2
3
1
 3  2 1 0 1 3
.06 .15 .14 .35 .09 .21
Calculate the variance of the random variable X – Y.
 x2 y  3.25
2
Note that the variance of the difference  x
y equals
the sum of the variances  and 
2
x
2
y
Rules for Means
Rule 1: If X is a random variable and a and b are
a  b x .
constants, then  a bx  _______
If a is added to each value of x, then a is added to the mean as well.
If each value of x is multiplied by b, then the mean is
multiplied by b as well.
Rules for Means
Rule 2: If X and Y are random variables, then
x   y
x   y
 x  y  ______and
 x  y  _______
Rules for Variances
Rule 1: If X is a random variable and a and b are
2 2
2
b
x


constants, then a bx ______
Adding a to each value of x does not change the variance.
Multiplying each value of x by b, multiplies the variance by b 2
Rules for Variances
Rule 2: If X and Y are independent random
variables, then
 x2 y   x2   y2 and  x2- y   x2   y2
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  .001g
 y2 x   y2   x2  .002 2  .0012  .000005
 y  x  .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 2  x  1 2  y  .5(2)  .5(2.001)  2.0005
 
2
z
2
1 x 1 y
2
2
 
2
 
2
 1  x2  1  y2  1 (.002) 2  1 (.001) 2  .00000125
2
2
2
2
 z  .00000125  .001118034
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
0
  10
  12.806
 x  y  110  100  10
 x  y  10 2  82  12.806
Calculator :
normalcdf(-10000,0,10,12.806)
.2174