Download Answers to Quiz #2 - Bryn Mawr College

Math 295
November 22, 2002
Exam # 2 Answers
1. ( 10 points) A discrete random variable X has the following probability function:
k
– 100
0
+ 100
+ 200
pX(k)
0.10
dev. from mean:
-200
squared deviation: 40000
0.10
0.50
0.30
-100
10000
0
0
+100
10000
a. What is its mean, E(X) ?
(-100)(.10) + (0)(.10) + (100)(.50) + (200)(.30) = 100
b. What is the standard deviation of X ?
E(X2)= (10000)(.10)+(0)(.10)+(10000)(.50)+(40000)(.30) = 18000
Var(X) = E(X2) – E(X)2 = 18000 – 10000 = 8000
So StDev(X) =
8000
Or: Extend table as above and note
StDev(X) = Root mean squared deviation =
(40000)(.10)+(10000)(.10)+0+(10000)(.30)  8000
2. ( 15 points ) Roll 12 standard six-sided dice. Define the random variable K to be the
number of sixes showing. (Note that K is always an integer from 0 to 12.)
a. What is E(K) ?
K has a binomial distribution with n=12, p=1/6. So,
E(K) = np = 2
b. What is Var(K) ?
Var(K) = np(1-p) = (12) (1/6) (5/6) = 5/3
c. Write an expression for P ( K = 3 ) . (You do not need to multiply it out.)
12 
n k
n k
3
9
  p (1  p)    (1/ 6) (5/ 6)
k 
3
1
3. ( 10 points ) The random variable Y has the density function
12 x(1  x)2 if 0  x  1,
f ( x)  
otherwise.
0
Write an expression for the variance of Y.
(Your expression should probably include some integrals. You do not need to
evaluate them for this problem. Otherwise be as explicit as you can.)

Var (Y )  E (Y 2 )  E (Y ) 2   y 2 fY ( y )dy 



12 y (1  y) dy  



3
2




yfY ( y )dy
12 y (1  y) dy 



2

2
2
2
4. ( 15 points ) The random variable W is normally distributed with mean  = 100 and
standard deviation σ = 10.
a. What is P ( 105 ≤ W ≤ 110 ) ?
If Z = (W-100)/10 then Z has a standard normal distribution.
So P( 105 ≤ W ≤ 110 ) = P( 0.5 ≤ Z ≤ 1 )
= P( Z ≤ 1 ) – P( Z ≤ 0.5 )
= (from table) 0.84135 – 0.69146 = 0.14989
b. If P ( W ≤ k1 ) = 0.05, what is k1 ?
From the table, P( Z ≤ -1.6448 ) = 0.05.
Since Z ≤ -1.6448 means W ≤ 100 – 10 * 1.6448 = 83.552,
we have P ( W ≤ 83.552 ) = 0.05 and k1 = 83.552.
c. If P ( W ≥ k2 ) = 0.05, what is k2 ?
From the table, P( Z ≥ +1.6448 ) = 1 – P( Z ≤ +1.6448 ) = 0.95,
and Z ≥ 1.6448 means W ≥ 116.448, so k2 = 116.448.
2
5. Consider a distribution with the density function
f ( x)  c e
1
2
x  
2
where  is unknown. …
( We are using c for the known constant that makes the density integrate to 1.
1
Actually, this is a normal distribution and c 
 .399 .)
2
We take four independent random draws from this distribution. They are:
x1 = 10.6, x2 = 10.0, x3 = 11.5, x4 = 11.9. (Note that x =11.0.)
a. ( 10 points ) Write an expression for the likelihood, L(), as a function of
x1, x2, x3, and x4.
2 
2 
2 
2

1
1
1
1
  2 x1     2 x2     2 x3     2 x4   
L(  )   c e
 c e
 c e
 c e











2

1
4
  2 xi   
or

c e

i 1 



4
2 
1
    xi    
2
i 1
 or fill in values for the x 's.
or
c4 e 
i










b. ( 5 points for answer ) What is the maximum likelihood estimate for  ?
( 10 points extra credit for a full derivation )
Answer: Same as x = 11.0.
Derivation: From the last form of the likelihood above, L() is
4
clearly maximized if S (  )    xi    is minimized. (This is
2
i 1
the same as the log likelihood with some constants dropped.)
Set the derivative to zero:
4
S '(  )   (2)  xi     0
i 1
4
  xi  4     x .
i 1
3
6. ( 15 points ) In a test of Acme All-Purpose Elixir on 100 randomly-selected subjects,
64 of the subjects reported that it cured whatever ailed them. Let’s count these
reports as successes.
trials: 100
successes: 64
Construct a 95% confidence interval for p, the true success rate.
 1   64 
64
64 
 1.96  
 
1 

100
 100   100  100 
or .64  .094
or
.546, .734  .
(Some idiot typed “1.9500” in the table where “1.9600”
belonged, so you’re free to use either value.)
4
7. ( 20 points ) Next, the same 100 subjects were tested for cholesterol levels.
The statistics of the general (untreated) population are known to be
untreated population:
0 = 190.0,
 = 50.0.
We want to test the hypothesis that the Elixir reduces cholesterol. Define:
 = true mean cholesterol of Elixir users
H0 : Cholesterol scores of test group are distributed in the same way as
those of the control group. In particular,  = 0.
H1 :  < 0.
Rejection region: X < x*.
Level of significance:  = 0.05
(“five percent”)
a. What should the x* be?
If H0 is true, then the sample variance X is roughly normal
(large sample) with mean 190 and standard deviation
1/100  50  =5.0. If H0 is true we want to reject with
probability 0.05, so we want
P( X < x* ) = 0.05.
So we should choose
x* = 190 – (1.6448)(5.0) = 181.776.
b. If it turns out that x = 179.0, what is the outcome of the test?
Since this value is in the rejection region, we reject H0. Put
another way: we claim statistically significant evidence (at the
5% level) that the treated group had a lower mean cholesterol.
(end of exam)
5
Cumulative Distribution Function (cdf)
for a Standard Normal Random Variable
(z)
z
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
0.02275
0.06681
0.15865
0.30854
0.5
0.69146
0.84135
0.93319
0.97725
-1.9500
-1.6448
-1.2815
-0.8416
0.8416
1.2815
1.6448
1.9500
0.025
0.05
0.1
0.2
0.8
0.9
0.95
0.975
(z)
z
6