Math/Stat 304
Solutions: TEST III
1. The dartboard represented on the left is a square
whose sides are of length 6. The three circles are all
centered at the center of the board and are of radii 1, 2,
3, respectively. Darts landing within the circle of radius
1 score 30 points, those landing outside this circle, but
within the circle of radius 2, are worth 20 points, and
those landing outside the circle of radius 2 but within
the circle of radius 3, are worth 10 points. Darts that do not land within the
circle of radius 3 do not score any points. Assuming that each dart that
Lucky throws will, independently of what occurred on his previous throws,
land on a point uniformly distributed in the square, find the probability of
each of the following events:
Solution: Let pj = probability of obtaining j points on a single throw of the dart.
Then p30 = /36; p20 = 4/36 – p30 =  /12; p10 = 
p0 = 1 –  /4
(a) Lucky scores exactly 20 on a throw of the dart.
p20 =  /12
(b) Lucky scores at least 20 on a throw of the dart.
p20 + p30 =  /12 + /36 =  /9
(c) Lucky scores 0 on a throw of the dart.
p0 = 1 –  /4
(d) The expected value of Lucky’s score on a throw of the dart.
0 p0 + 10p10 + 20p20 + 30p30 = 10( /12) + 30(/36) ≈
2
(e) Each of Lucky’s first two throws score at least 10.
(1 – p0)2 = (1 – (1 – /4))2 = (/4)2
(f) Lucky’s total score after two throws is exactly 30.
p0p30 + p10p20 + p20p10 + p30p0 = 2p0p30 + 2p10p20 =
2(1 –  /4) ( /36) + 2(5/36)(/12) ≈ 0.27
2. (a) A normal random variable X with mean 5 satisfies P(X < 0.9) = 0.2. Find
the approximate value of the variance of X.
Solution: Let Y = (X – 5)/.
Then P(X < 0.9) = P( Y < (0.9 – 5)/) = P(Y < -4.1/) = 1 – P(Y < 4.1/) =
1 – (4.1/).
Now since P(X < 0.9) = 0.2, we have (4.1 / ) = 0.8
Now, using the table, (0.85) = 0.8; we have 4.1/ = 0.85.
So  = 4.1/0.85 = 4.82, and var(X) =  2 = 23.3
(b) A radar unit is used to measure the speed of cars on a highway. The speeds are
normally distributed with a mean of 90 km/hr and a standard deviation of 10 km/hr
What is the probability that a car chosen at random is traveling at more than 100
km/hr?
Solution: P(X >100) = 1 – P(X ≤ 100) = 1 – P( X – 90)/10 ≤ (100 – 90)/10 = 1) =
1 – P(Z ≤ 1) = 1 – 0.8413 = 0.1587
3
3. Suppose that the moment generating function of a random variable X is given
1
3
4
4
3
by M(t) = e2t( + 𝑒 𝑡 ) .
(a) Find E(X).
1
3
4
4
3
1
3
4
4
2
Solution: M' (t) = 2e2t( + 𝑒 𝑡 ) + 3𝑒 2𝑡 ( + 𝑒 𝑡 ) (3/4 𝑒 𝑡 )
Hence E(X) = M' (0) = 2 + 9/4 = 17/4 = 4.25
(b) Find P(X ≤ 3).
Solution: Since X is clearly discrete (by the inversion theorem), we need only find
the coefficients of ekt for k = 2 and 3. Using the binomial theorem,


M (t )  e 2t 1 / 4 3  (3)(1 / 4 2 )(3 / 4)e t  higher powers of e 
1 2 t 9 3t
e 
e  ...
64
43
Hence P(X = 2) = 3/64 and P(X = 3) = 9/64. Finally P(X ≤ 3) = P(X = 2) + P(X =
3) = 1/64 + 9/64 = 5/32
 x  y if 0  x  1, 0  y  1
0 otherwise
4. Let the joint pdf of X and Y be given by: f X ,Y ( x, y)  
(a) Compute the covariance of X and Y.
1
1
2
Solution: f X ( x)   f X ,Y ( x, y) dy  ( x  y) dy  xy  y / 2
0
0
1
1
0
0
1
y 1
y0
Thus E[ X ]   xf X ( x) dx  x( x  1 / 2) dx  x 3 / 3  x 2 / 4  7 / 12
0
Now, by symmetry, E[Y] = E[X] = 7/12, also.
1 1
1 1
0 0
0 0
Next, E[ XY ]    xy f X ,Y ( x, y) dx dy    xy ( x  y) dx dy 1 / 3
 x  1/ 2
4
Hence cov(X, Y) = E[XY] – E[X]E[Y] = 1/3 – (7/12) = - 1/144
2
(b) Are X and Y independent? Explain.
Solution: No; since cov(X, Y) ≠ 0, X and Y are not independent rvs.
 1  2x
5. Let the pdf of X be given by: f X ( x)   2 e if 0  x  
0 otherwise

Let Y = √𝑋
(a) Find the cdf of Y. Sketch the graph of FY(y).
Solution: The support of Y is the interval [0, ∞). Now, for y < 0, FY(y) = 0.
For y > 0,
y2
FY ( y )  P(Y  y )  P( X  y )  P( X  y ) 
2
f
y2
X
( x) dx 


0
x
x


y2
1 2
2
e dx  e
 1 e
2
0
Hence
0 for y  0

FY ( y )  
y2

1  e 2 for y  0
1.0
0.8
0.6
0.4
0.2
2
2
4
6
8
y2
2
5
(b) Find the pdf of Y. Sketch the graph of fY(y).
Solution:
f Y ( y) 
0 for y  0
d

FY ( y )    y 2
dy
 ye 2 for y  0
0.4
0.3
0.2
0.1
3
2
1
1
2
Among all the occurrences possible in the universe the a priori probability of any particular one
of them verges upon zero. Yet the universe exists; particular events must take place in it, the
probability of which (before the event) was infinitesimal. At the present time we have no
legitimate grounds for either asserting or denying that life got off to but a single start on earth,
and that, as a consequence, before it appeared its chances of occurring were next to nil. ...
Destiny is written concurrently with the event, not prior to it... The universe was not pregnant
with life nor the biosphere with man. Our number came up in the Monte Carlo game. Is it
surprising that, like the person who has just made a million at the casino, we should feel strange
and a little unreal?
― Jacques Monod, Chance and Necessity