Transcript Lecture 9

CONTINUOUS RANDOM VARIABLES
Continuous random variables take all values in
an interval
Too many values! Impossible to list them all!
How to describe a continuous random variable?
Density curve
of random
variable X
Use density curve.
P(a ≤X < b)=area under the
density curve between a
and b.
Total area under density
curve =1.
P(a<X<b)
a
b
X
Values of X
Continuous density example -uniform distribution
X-time (min) taken by a fire truck to arrive at a fire after 911 call. Suppose
X is equally well anywhere between (2, 15) min.
X r.v. with uniform distribution on (2, 15).
Density curve for X has the same height over (2, 15): rectangle.
Area=1. Length (2, 15) is 13
height=1/13.
Density of X
P(5<X<8)
1/13
0
2
5
8
15
Probability that it takes between 5 and 8 min for the fire truck to arrive:
P( 5<X < 8) = Area between 5 and 8=3 x (1/13)=3/13.
MEAN AND VARIANCE OF A RANDOM VARIABLE
Measures of center and spread of a r.v.
X – discrete r.v.
X has probability
distribution:
x2 … xn
Values of the
r.v X
x1
probabilities
p1 p2 … pn
Center of the distribution: mean,
Expected value of X, EX or E(X):
  EX  x1 p1  x2 p2 
n
 xn pn   xi pi
i 1
NOTES: The mean represents the “long-run-average” value of X. If
we average many values of X, we expect to get a number close to
EX.
EX is a weighted average of the values of X, weights are the
probabilities of the values.
VARIANCE and STANDARD DEVIATION OF A RANDOM VARIABLE
Measures of spread around the mean μ of a random variable. X discrete rv
Variance of X
Take squared deviations from the mean and add them up with the same
weights as for the mean:
VarX=Var(X)= σ2 = (x1 – μ)2 x p1+ (x2 – μ)2 x p2 + … + (xn – μ)2 x pn=
n
  ( xi   ) 2  pi .
i 1
Standard deviation of X:
   2  VarX .
NOTES: Both variance and standard deviation of any random variable are
nonnegative.
If X is continuous use calculus to define and compute mean and variance.
NO WORRIES IN STAT152.
EXAMPLE
The number of defective chips produced in a month by a company
ABCHIP has the following probability distribution:
Number of defective chips
0
10
15
20
probabilities
0.1
0.2
0.5
0.2
a. What is the average number of defective chips this company produces in a
month?
b. What is the standard deviation of the number of defective chips this
company produces in a month?
Solution. a. X=# of defective chips produced by ABCHIP in a month.
EX= μ = 0 x 0.1 + 10 x 0.2 + 15 x 0.5 + 20 x 0.2 = 13.5
Note that the mean is not one of the possible values of X!
b. VarX= σ2 = (0-13.5)2 x 0.1 + (10 – 13.5)2 x 0.2 + (15 – 13.5)2x 0.5+(20-13.5)2x 0.2
= 30.25.
Standard deviation of X,   VarX   2  30.25  5.5.
EXAMPLE
Your winnings on a lottery are a random variable with the following
probability distribution:
Winnings in $
100 200 1,000
0
Probabilities
0.4
0.2
0.3
?
a. Find the probability that you win $1,000.
b. What is the probability that you will win at least $150?
c. What is your expected winning?
Solution. X= winnings in $.
a. Since sum of all probs=1, then P(X=1,000)=1-(0.4+0.3+0.2)=0.1.
b. P(X ≥ $150)=P(X=$200 or X=$1,000)=0.3 +0.1=0.4.
c. EX = 0 x 0.2 + 100 x 0.4 + 200 x 0.3 +1,000 x 0.1= 200.
You should expect to win about $200 if you play for a long time.
EXAMPLE
What if we consider that you must pay $10 to play the lottery. What are
your expected winning then? The probability distribution of the
winnings changes to accommodate the payment.
Winnings in $
90
190
990
-10
Probabilities
0.4
0.3
0.1
0.2
The negative $10 means you lose $10 with probability 0.2.
EX= -10 x 0.2 + 90 x 0.4 + 190 x 0.3 +990 x 0.1 = 190.
You should expect to win about $190 if you play for a long time.