Transcript 6.9 – Discrete Random Variables: Mean and Standard Deviation

6.9 – Discrete Random
Variables: Mean and Standard
Deviation
IBHLY2 - Santowski
(A) Mean of a Probability Distribution
for a discrete random variable, X, which can assume the
values x1, x2, x3, ....xn, the mean is also called the
expected value of X
on a very simple level, if there are n members in a given
event, and each of the n members has a probability of
occurring of p, then the expectation of occurrence of
any specific event is n x p
example  if we roll a die, how many 6's are expected
if we roll 120 times  since the event of rolling 6's has
a probability of occurring 1 in 6 times (i.e. p = 1/6) and
we roll n = 120 times, it would be expected that we get
the six 20 times (n x p = 120 x 1/6)
(B) Class Work
CLASSWORK to reinforce the idea of
expected value:
SL Math, Chap 29C, p715, Q1-9
HL Math, Chap 30C, p733, Q1-6
(C) Expected Value (Mean) by Formula
the expected value formula comes from the general observation
of n x p as before
n
now we write it as
  E ( X )  xi P( X  xi )

i 1
and we use  again as we are looking at the idea of a mean of a
population
xi represents specific outcomes
P(X=xi) represents the probability of xi occurring (or pi)
X represents the random variable we are concerned about
the expected value represents the Along term average@ of the
variable X. The effect of multiplying each value of x by its
probability, p, gives it a weighting  then summing all the
weighted values gives us an overall expectation for X
(C) Expected Value (Mean) by Formula
ex 1  Find the mean of the probability distribution given in the table
below:
x
4
P(X = x) 0.002
6
8
10
0.040
0.299
0.659
What does the data mean?  the specific outcome of 4 occurs with a
probability of 0.002, the outcome 8 occurs with a probability of 0.299,
the outcome 10 occurs with the highest probability of 0.659
 = E(X) = xiP(X=xi) = [(4)(0.002) + (6)(0.040) + (8)(0.299) +
(10)(0.659)] = 9.23
So the mean (expected value) is 9.23  so given the possible
outcomes and their associated probabilities, you can expect a value of
9.23
(C) Expected Value (Mean) by Formula
ex 2  Find the expected value of X, E(X), for the probability
distribution given by the formula
x
4 x
 4  1 
P( X  x)    
 x  3 
 2
 
 3
, x  0,1,2,3,4
To interpret the formula  we have 4 events occurring and the
probability for “success” in any one event is 1/3
At times, the data is easier to work with, if we have a chart/table:
X
0
P(X = x) 0.198
1
2
3
4
0.395
0.296
0.099
0.012
 = E(X) = xiP(X=xi) = [(0)(0.197) + (1)(0.395) + (2)(0.296) +
(3)(0.099) + (4)(0.012)] = 1.332
(C) Expected Value (Mean) by Formula
Find the expected value of X, E(X), for the probability distribution given
 4  3 
 

x  3  x 

P( X  x) 
, x  0,1,2,3
 7
 
 3
ANS = 1.713
(C) Expected Value (Mean) by Formula
ex 4  A committee of 3 people is to be selected
from 4 men and 2 women. Let X represent the
number of women chosen. Find the expected value
of X
ANS = 1
(D) Homework
SL Math Text, 29C, p716, Q10-14
HL Math Text, 30C, p734, Q7-9
Peter Smythe book, Mathematics
SL&HL, Section 14.2, p391, Q1-8
(E) Variance and Standard Deviation
recall that the way we calculated the population variance was to find the
square of the deviation of each point from the population mean
so in a similar manner, when we calculate the variance of a random
variable, X, we wind up subtracting  from our possible values of X (xi )
n
Thus
  Var ( X )  E ( X   )   ( xi   ) 2 P( X  xi )
2
2
i 1
and
  SD( X )  Var ( X ) 
n
 (x  )
i 1
i
2
P( X  xi )
where  = E(X)
the standard deviation gives a measure of the way in which the values
are spread about the mean
(E) Variance and Standard Deviation
other textbooks will express these formulas slightly differently, as they let pi
represent P(X = xi) so the formulas are written as :
n
   xi pi
i 1
n
   ( xi   ) pi
2
2
i 1
(E) Variance and Standard Deviation
we can make one simplification to the variance formula 
 2  Var ( X )
2
 E ( x  )
 (x
  (x

i
  ) pi
2
i
pi )   2
2
 E( X )  
2
2
 E ( X 2 )  [ E ( X )]2
(F) Examples
ex 1  Find Var(X) and SD(X), given the probability distribution below:
x
2
3
4
P(X = x)
0.3
0.5
0.2
We will set up a table to work through our calculation:
x
P(X=x)
xP(X=x)
x-
(x- )2
(x- )2P(X=x)
2
0.3
0.6
-0.9
0.81
0.243
3
0.5
1.5
0.1
0.01
0.005
4
0.2
0.8
1.1
1.21
0.242
  E( X )   xi P( X  xi )  29.
 = 0.490
(F) Examples
therefore, in our example, Var(X) = E(x - u)2 = 0.490
and SD(X) = (0.490)0.5 = 0.7
now, using our alternative formula of Var(X) = E(X2) [E(X)]2,
we first work out E(X2) as (x2)(P(X=x)) which is
then 22(0.3) + 32(0.5) + 42(0.2) = 8.9
so we get Var(X) = 8.9 - 2.92 = 0.49, which is the
same answer as when we used the complete table
(G) Examples
ex 2  Find the expected value (mean) and standard deviation
for the random variable, X, which is the event of rolling a single
die.
so
then
1 1 1 1 1 1
  E ( X )   xi pi  1   2   3   4   5   6   3.5
6 6 6 6 6 6
2
2
2 1 
2 1 
2 1 
2 1 
2 1 
2 1 
E ( X )   xi pi  1    2    3    4    5    6    15 1
6
6 6 6 6 6 6
so then Var(X) = E(X2) - [E(X)]2 = 15.166666.... - (3.5)2 =
2.91666666...
(G) Examples
ex 3  Find the mean and standard deviation of the random
variable, X, with the probability distribution given in the table:
x
0
1
2
3
P(X=x)
0.264
0.494
0.220
0.022
(G) Examples
ex 4  A debating team of 4 is to be chosen from 6 girls and 3
boys. Let X be the number of boys chosen. Find:
(i) E(X)
(ii) Var(X)
(iii) SD(X)
(H) Homework
SL Math text, Chap 29D, p720, Q1-8
HL Math text, Chap 30D, p737, Q1-8