Math 105/206 - Quiz 5, Mar 27 2015
IMPORTANT: Write your name AND student number somewhere on this sheet.
No calculators, books or notes. Please show your work to get full marks. (10 marks total)
Problem 1
The percentage of fourth-year students that graduates at UBC each year is a random variable X modelled by
a probability distribution function f (x) = 3(1 − x)2 for 0 ≤ x ≤ 1, and 0 otherwise.
• compute the CDF of X. (2 marks)
• compute the expected value of X. (2 marks)
• compute the variance of X. (2 marks)
Solution
Call the CDF F (x) =
Z
Rx
f (t)dt. If x ≤ 0, then clearly F (x) =
−∞
x
0
Z
Z
f (t)dt = 0 +
f (t)dt +
f (t)dt =
F (x) =
x
Z
−∞
−∞
0
0
x
Rx
−∞
0dt = 0. If 0 ≤ x ≤ 1 we get
x
3(1 − t)2 dt = − (1 − t)3 0 = 1 − (1 − x)3 .
R1
Rx
and if x ≥ 1, then F (x) = −∞ f (t)dt + 1 0dt = 1 + 0 = 1.
The expected value of X is the integral
Z
∞
1
Z
E(X) =
t · 3(1 − t) dt = 3
tf (t)dt =
−∞
2
0
1
t2 t4
t3 1
+ −2
= .
2
4
3 0 4
The variance of X is
Var(X) = E(X 2 ) − E(X)2
where
2
Z
∞
2
Z
−∞
2
2
t · 3(1 − t) dt = 3
t f (t)dt =
E(X ) =
1
0
so
1
Var(X) =
−
10
1
t3 t5
t4 1
+ −2
=
3
5
4 0 10
2
1
3
= .
4
80
Problem 2
Decide if each of the following sequences converges or diverges, and if they converge compute the limit. (1 mark
each)
n
1
an = −
3
an =
n(3n2 + 2)
n2 + 1
3
an = (2) n
an = en − n2
Solution
The first one converges to 0, since it is geometric with ratio −1/3.
The second one diverges, because it is a quotient of two polynomials, and the numerator has degree higher
than the denominator.
The third converges to 1 because the exponent 3/n converges to zero, and 20 = 1.
The fourth one diverges because en grows faster than n2 . Formally
n2
n
2
n
e −n =e 1− n
e
and n2 /en goes to 0 (for example you can apply l’Hôpital’s rule), so overall the thing inside the parentheses
tends to 1, and we get ∞ · 1 = ∞.
Problem 3
Show that the decimal number 0.999999
P · · · 9 ·k· · = a0.9 is equal to 1 by writing it as a sum of a geometric series
and using the formula for the sum, ∞
k=0 a · r = 1−r where |r| < 1. (4 marks)
Solution
We have
∞
9
9
9
9
9 X
0.9 =
+
+
+ ··· + n + ··· =
·
10 100 1000
10
10 k=0
1
10
k
=
9
1
·
1 = 1.
10 1 − 10