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Math 3C – Winter 2008
Continuous Distributions
Problem 1.
Determine a parameter c such that
f (x) =
0 for x < 2 c x−^4 for x ≥ 2
is a density function of some random variable X. Find the corresponding distribution
function F. Evaluate EX and Var(X).
Solution. A function f is a density function if and only if f is nonnegative and
∫ (^) +∞
−∞
f (x) dx = 1
Therefore we calculate
∫ (^) +∞
−∞
f (x) dx =
2
c x − 4 dx = c
[
x − 3
]x=+∞
x=
= c
c
24
This implies c 24 = 1, i.e.^ c^ = 24. The distribution function is defined by
F (x) =
∫ (^) x
−∞
f (t) dt
Therefore for x ≤ 2 we have F (x) = 0, while for x > 2 we calculate
F (x) =
∫ (^) x
2
24 t − 4 dt = 24
[
t − 3
]t=x
t=
x − 3
= 1 − 8 x − 3
Finally we can write
F (x) =
0 for x ≤ 2 1 − 8 x−^3 for x > 2
Using the formula
Eg(X) =
−∞
g(x)f (x) dx
we evaluate
EX =
−∞
xf (x) dx =
2
x · 24 x − 4 dx = 24
2
x − 3 dx =
[
x − 2
]x=+∞
x=
and
EX
2
−∞
x 2 f (x) dx =
2
x 2 · 24 x − 4 dx = 24
2
x − 2 dx =
[
−x − 1 ]x=+∞ x=
so that Var(X) = EX^2 − (EX)^2 = 12 − 9 = 3
Problem 2.
If a random variable X has the standard normal distribution, evaluate E|X| and Var(|X|)
and find P(− 2. 5 ≤ X < 2 .5).
Solution. A random variable is normally distributed with mean μ and standard deviation
σ if it has a density function given by the formula
f (x) =
σ
2 π
e
− (x−μ)
2 2 σ^2
In that case we also have fomulae
EX = μ and Var(X) = σ 2
The standard normal distribution has μ = 0 and σ = 1.
We calculate
E|X| =
−∞
|x| · f (x) dx =
−∞
|x| ·
2 π
e−^
x^2 (^2) dx =
−∞
(−x) ·
2 π
e − x 2 (^2) dx +
0
x ·
2 π
e − x 2 (^2) dx =
2 π
−∞
xe − x
2 (^2) dx +
2 π
0
xe − x
2 (^2) dx =
substitute t = −
x^2
2
, dt = −x dx
2 π
−∞
e t (−dt) +
2 π
0
e t (−dt) =
2 π
−∞
e t dt +
2 π
−∞
e t dt =
2 π
[
e t ]t= t=−∞
2 π
2 π
π
(a)
P(X ≤ 50) = P(
X − 75
) = P(Z ≤ − 2 .5) =
= F (− 2 .5) = 1 − F (2.5) = 1 − 0 .9938 = 0. 0062
Therefore, approximately 0.62% of students have score 50 or below.
(b)
P(X > 70) = P(
X − 75
) = P(Z > − 0 .5) = (by symmetry) =
= P(Z < 0 .5) = F (0.5) = 0. 6915
Therefore, approximately 69.15% of students have score greater than 70.
(c)
P(80 ≤ X ≤ 90) = P(
X − 75
) = P(0. 5 ≤ Z ≤ 1 .5) =
= F (1.5) − F (0.5) = 0. 9332 − 0 .6915 = 0. 2417
Therefore, approximately 24.17% of students have score between 80 and 90.
Problem 4.
A random number is chosen from the interval (− 2 , 3), with the uniform distribution.
What is the expected value of that number? Find the probability that its first digit
after the decimal point is 5.
Solution. The experiment is described by a random variable U that is uniformly dis-
tributed over the interval (− 2 , 3).
In general, the uniform distribution over the interval (a, b) has density
f (x) =
b−a for^ a < x < b 0 otherwise
and its expectation is EU = a+b
In our case a = −2, b = 3, so EU = −2+ 2 = 0.5 and the density function is
f (x) =
5 for^ −^2 < x <^3 0 otherwise
Denote the following subset of the interval (− 2 , 3):
A = {the first digit after the decimal point is 5}
It is easy to see that
A = (− 1. 6 , − 1 .5] ∪ (− 0. 6 , − 0 .5] ∪ [0. 5 , 0 .6) ∪ [1. 5 , 1 .6) ∪ [2. 5 , 2 .6)
Its “total length” is 5 · 0 .1 = 0.5. Therefore
P(U ∈ A) =
Problem 5.
Assume that a lifetime of a radioactive atom is exponentially distributed. If it is known
that the average lifetime of this atom is 5 days, find the probability that the atom will
still exist after 10 days.
Solution. The exponential distribution with parameter λ has density function
f (x) =
0 for x ≤ 0 λe−λx^ for x > 0
and its expectation is EX = 1 λ. In the problem we are given EX = 5 (in days), so 1 λ = 5, which gives^ λ^ =^
1
The probability that the atom “lives” more than 10 days is
P(X > 10) =
10
f (x) dx =
10
e − 15 x dx =
[
− 5 e − 15 x
]x=+∞
x=
(0 + 5e − 2 ) = e − 2 ≈ 0. 135335
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