# solutions: quiz 4

```MATH /stat 304
Solutions: QUIZ IV
1. [10 pts] Professor J. C. Oates has 394 students in her survey of gothic literature class at
Princeton. The scores on her midterm exam are normally distributed with mean of 62.8 and
variance of 72.1. How many students in the class can be expected to receive a score between 75
Solution: Let X = exam grade of a student chosen at random from Oates’ course.
We are given that X ~ N(2) = N(62.8, 72.1). Let Z = (X – m)/s = (X – 62.8)/√72.1
75−62.8
Thus P( 75 &lt; X &lt; 83) = P (
83−62.8
75−62.8
√72.1
√72.1
Φ(
) − Φ(
√72.1
&lt;
X−62.8
√72.1
&lt;
83−62.8
75−62.8
√72.1
√72.1
) = P(
&lt;Z&lt;
83−62.8
√72.1
)=
) = Φ(2.3789 ) − Φ(1.4368) = 0.0667 ≈ 26
Hence, the approximate number of students with grades between 75 and 83 is (0.0667)(394) =
26 students.
2. [15 pts] Suppose that X and Y are independent random variables, each uniformly distributed on
the unit interval.
(a) Find the joint pdf of X and Y.
Solution: The pdf of X is given by
1 if 0  x  1
f X ( x)  
0 otherwise
and the pdf of Y is given by
1 if 0  y  1
f Y ( y)  
0 otherwise
Since X and Y are independent, their joint pdf is given by:
1 if 0  x  1, 0  y  1
f X ,Y ( x, y )  f X ( x) f Y ( y )  
0 otherwise
(b) Find the cdf of Z = X + Y. Sketch.
Solution: Begin by observing that Z = X + Y achieves values between 0 and 2.
If 0 &lt; z ≤ 1, FZ(z) = P(Z ≤ z) = P(X + Y ≤ z) = P(Y ≤ -X + z) = area of triangle = &frac12; z2.
If 1 &lt; z ≤ 2, FZ(z) = P(Z ≤ z) = P(X + Y ≤ z) = P(Y ≤ -X + z) = area of square minus area of
triangle on upper right = 1 – &frac12; (2 – z)2
0 if z  0

2
(1 / 2) z if 0  z  1
 f X ,Y ( x, y)dA  1  (1 / 2)(2  z ) 2 if 1  z  2
x y z

1 if z  2

1.0
0.8
0.6
0.4
0.2
2
1
1
2
3
4
5
2
3
4
5
(c) Find the pdf of Z. Sketch.
Solution: fZ(z) = (d/dz) FZ(z) =
 z if 0  z  1

 z  2 if 1  z  2
0 otherwise

1.0
0.8
0.6
0.4
0.2
2
1
1
3. [20 pts] Assume that the joint pdf of random variables X and Y is given by:
c if 0  x  y  1

f X ,Y ( x, y )  
0 otherwise

(a) Find the value of c.
Solution:
1   f X ,Y ( x, y ) dA   c dA  c(area of triangle )  c (1 / 2)
R2
T
Thus c = 2.
(b)
Find the marginal pdf, fX (x), of X.
Solution:
2(1  x) if 0  x  1
f
(
x
,
y
)
dy

2
dy


 X ,Y
x
0 otherwise

f X ( x) 
1
(c) Find the conditional pdf of Y given that X = x .
Solution:
If fX(x) ≠ 0, then
f Y } X ( y | x) 
f X ,Y ( x, y )
f X ( x)
 1
if 0  x  y  1
f XmT ( x, y )
2



 1  x
2x
2(1  x) 
0 otherwise
(d) Find E(Y | X = &frac12; ).
Solution:

1
1
1
E (Y | X  1/ 2)   y fY } X ( y | 1/ 2) dy   y
dy   2 y dy 12  (1.2)2  3 / 4

1 / 2 1  1/ 2
1/ 2
Extra Credit:
Two points are selected randomly on a line of length L so as to be on opposite sides of the midpoint
of the line. That is, the two points X and Y are independent random variables such that X is
uniformly distributed on (0, L/2) and Y is uniformly distributed over (L/2, L).
Find the probability that the distance between the two points is greater than L/3.
Solution: The pdf of X is given by
2 / L if 0  x  L / 2
f X ( x)  
0 otherwise
and the pdf of Y is given by
2 / L if L / 2  y  L
f Y ( y)  
0 otherwise
Since X and Y are independent, their joint pdf is given by:
4 / L2 if 0  x  L / 2, L / 2  y  L
f X ,Y ( x, y)  f X ( x) f Y ( y)  
0 otherwise
We wish to find the probability that |X – Y| &gt; L/3, or equivalently, Y – X &gt; L/3, since Y ≥ X.
 f
X, Y
y  x  L/3
4
L2
( x, y ) d A 
4
L2
4
1 dA  L
2
(area of square  area of triangle ) 
y  x  L/3
 L2 1

4 7
  L / 32   1  
18 9
 4 2

The real voyage of discovery consists not in seeking new landscapes but in having new eyes.
- Marcel Proust
```