Download 5. The sum of Normally distributed random variables.

Class 13 August 3, 2015
STT-315-204
Quiz 3 will be held on Wednesday, August 5.
The program of the Quiz
1. Uniform Distribution. Example 4.27, p.243,
Exercises 132-134, 137.
2. Exponential Distribution. Example 4.28, p.245,
Exercises 135.136,138.
3. Normal Distribution. Standard Normal: Exercises
4.84-4.89, p.233. General Normal: Example 4.24,
p.228.
4. Normal Approximation to Binomial: Example 4.25,
p.231.
5. The sum of Normally distributed random
variables.
SAMPLE QUIZ 3
08/03/15
STT-315-204
B -2015
SUMMER Name
Quiz 3
1. Using the table of Standard Normal in a.-c.
calculate the following areas under the
Standard Normal curve
a. (2 pts.) To the left of z=1.7
A . .9554
0.5179
B. 0.5808
E. 0.4824
C. 0.9821
b.(2 pts.) Between z=0 and z=1.51.
D.
A. .4192
B. 0.9192
0.4452
E. 0.9452
C. 0.4345
D.
c. (2 pts.) Between -1.51 and 1.51.
A. .0.8690
B. 0.8364
0.8904
E. 0.9452
C. 0.4345
D.
d.(2.5) Find a z0 such that the total area to the
left of z0 is 0.8340.
A.
0.97
.0.89
B. 0.83
C. 0.43
d) 0.71 e)
e. (2.5pts.) Find a z0 such that the total area to
the left of z0 is 0.3446
a) 0.90
b) 0.40 c) 0.43
d) -0.40 e) - 0.90
2. The lifetime of fans of a brand of Diesel engine is
subject to the exponential distribution with   20
(in thousands of hours).
a. (3 pts.) Find the probability that a randomly
chosen fan will last at least 30 .
P( X  30)  e

30
20
 0.223.
b.(4 pts.) What is the value of the lifetime that
places a fan into the 2% of best fans (i.e. find
an x such that P( X  x)  0.02) ?
We are to find x such that
e


x
20
 0.02. To solve this equation , apply ln to both sides :
x
 ln 0.02 . Therefore x  20  ln 0.02  78.2.
20
c. (4 pts.) What is the value of the lifetime that
places a fan into the 5% of worst fans ?
Now we are finding an x such that
e

x
20
 0.95. Still applying ln to both sides we get 
x  1.025.
.
x
 ln 0.95, and
20
3. Assume a random variable X has an
exponential distribution with parameter  .
a. (3 pts.) Find the probability P(  2  X    2 )
that X is within 2 standard deviations of  .
The desired probability is
P(  X  3 )  P(0  X  3 ) (because X is a positive r.v.)
1 e

3

 1  e 3  0.95.
b. (3 pts.) Now compute the same probability as
2
in a. for the normal case when X ~ N ( ,  ).
.
P(  2  X    2 )  P(2  Z  2 )  0.95 .
4. (3 pts.) Assume X has a Uniform distribution
on [100,200]. What are  and  ? Find the
same probability as in a. and b.:
X
X
P(  2  X    2 )
Solution. We have
 X 150 and  X  28.87 . Therefore,
P(   2  X    2 )  P(150  2  28.87, 150  2  28.87)
P(92.26  X  207.74)  P(100  X  200) 1
5. Let X be the casino win percentage in 100
roulette plays on black/red. It can be shown
that X~N(5.26, 102) .
a. (3 pts.) Find P(X>0) (This is the probability
that casino wins money.)
P( X  0)  P( Z 
0  5.26
)  P( Z  0.53)  0.7019.
10
b. (3 pts.) Find P(5<X<15).
P(5  X  15)  P(
5  5.26
15  5.26
Z
)  P(0.03  Z  0.97)  0.346.
10
10
c. (3 pts.) Find P(X<1).
P( X  1)  P(Z 
1  5.26
)  P( Z  0.426)  0.3336.
10
Sums of Normally distributed Independent
Random Variables
6. A restaurant has three sources of revenue:
eat in orders, takeout orders, and the bar. The
daily revenue from each source is normally
distributed with mean and standard deviation
shown in the table below.
Mean
Standard Dev.
Eat in
$5,780
$142
Takeout
641
78
Bar
712
72
a. Will the total revenue on a day be
normally distributed?
b. What are the mean and standard
deviation of the total revenue on a
particular day?
c. What is the probability that the revenue
will exceed $7,000 on a particular day?
Solution. a. According to a math theorem, a
sum of normally distributed independent
random variables is a Normally distributed
random variable. So, the answer is “Yes”.
b. E(X1+ X2 + X3)= 5780+641+712= 7133.
V(X1+ X2 + X3)= V(X1)+ V(X2)+ V(X3)= 31432.
Sigma (X1+ X2 + X3)=177.3
c. P(X1+ X2 + X3> 7000) = P(Z> 70007133/177.3)= P(Z> -0.75)=0.7734.
Normal Approximation to Binomial with
the continuity correction
7. An advertising research study indicates that
40% of the viewers exposed to an
advertisement try the product during the
following four months. If 100 people are
exposed to the ad, what is the probability that
at most 35 of them will try the product in the
following four months?
Denote X = (The number of people out of
100 who will try the product in the
following four months)
Obviously, X~B(100,0.4). We have to find
P( X  35) .
The Normal approximation to Binomial says that
if X ~ B(n, p ), n is big , p is small, then
k  np  0.5
P( X  k ) appr.  P( Z 
)
np(1  p)
In our case n  100, p  0.4. We have to calculate
35  40  0.5
P( X  35)  P( Z 
)  P( Z  0.918)  0.178
24
The straightforward Binomial table gives
0.179.
Quiz 3 Formulas
1. If X has an exponential distribution, then
P( X  a)  e

a

.
 X   ;  X  .
2. For the Uniform distribution on [c,d],
X 
cd
d c
; X 
.
2
12
3. Normal probabilities are reduced to Standard
Normal as follows:
P ( a  X  b)  P (
a  X
X
Z 
b  X
X
).
Material After Quiz 3
Plan for Class13,14:
1. Sampling distribution of the sample mean. The
Central Limit Theorem (CLT)
2. Confidence interval for unknown mean.
1. Sampling Distribution for the Sample mean.
Methods used are based on the CLT (the Central Limit
Theorem) and rules of the expectation and variance.
On class 6 we introduced a random variable X having the
meaning of a win in a card game. X was defined as
follows:
If a card drawn is either Jack or Queen, then X(s) = $5 ; if
s is either King or Ace, then X(s)=10$. Otherwise X(s)=-4.
Distribution of X:
X
-4
5
10
p
36/52
8/52
8/52
In the first row the possible values are given, in the second
row the corresponding probabilities.
The Expectation (Expected value, Mean) is defined as
follows:
E( X )   X 
n
px
i i
; Expectatio n also is denoted by E ( X ).
1
In our problem of drawing a card
n
8
8
36
E ( X )   X   pi xi   5  10  4 
  0.4615 .
52
52
52
1
The Variance is defined as
V (X ) 
n
 p (x  
i
1
i
X) 
2
n
px
2
i i
1
 X .
2
The last expression (right-hand-side) is the so called a
shortcut formula for variance.
Using it we find
n
V ( X )   pi xi2   X  16 
2
1
V ( X )  30.24,  ( X )  5.5.
36
8
8
 25
 100   0.46152  30.3.
52
52
52
Example 1. Assume we are playing now 2 games
independently,
X 1 , X 2 denote their wins respectively. Now consider the sum
S  X 1  X 2 of two games. Find P( S 1) .
Solution . S has the following distributi on :
S
-8
+1
p
0.479 0.213
6
10
15
20
0.213
0.024
0.048
0.024
The resulting table is called the Sampling Distribution of
the Sum.
The table gives us
Example 2. Find
P(S  1)  0.479.
P( X  1) .
More frequently the distribution of the following close
relative of S
is considered: (X1+X2) /2 which is called the Sampling
Distribution of the Sample Mean
(X1+X2)/2 -4
p
+1/2
0.479 0.213
3
5
7.5
10
0.213
0.024
0.048
0.024
So, the solution to Example 2 is:
P(
X1  X 2
 1)  0.692.
2
Example 3. Now we consider 100 games and put
X k  {The win in the k  th game} , k  1,...,100, and form
the sample mean :
X  ...  X n
X 1
(in our case n equals 100) . Find P( X  1).
n
Now the main question is: What is the distribution of
X
, or speaking in statistical language, what is the
sampling distribution of the sample mean ? CLT helps
solve out this problem.
A simplified version of the CLT which is frequently used in
applied statistics sounds as follows: The sum of many
independent identically distributed random variables
having a variance, has approximately a Normal
distribution.
So CLT tells us that the sampling distribution of X is
approximately Normal. It remains to find the mean and
variance of
X
.
The formulas for
 X and  X are :
X  X ;  X 
X
n
. (1)
We can easily prove the above formulas by use of the
following rules for expectations and variances: