10.1. We use the transformation r(x) = x/λ. Its inverse is s(y) = λy. If
f (x) is the density of X, then density of Y = r(X) is f (s(y))s0 (y). In our case
f (x) = e−x , hence density of Y is λe−λx , which is exponential(λ).
RR
R1R1
10.2. We have
f (x, y) dx dy = 1, hence c 0 0 x + y dx dy = 1. The
R1
integral is c 0 1/2 + y dy = c(1/2 + 1/2) = c, hence c = 1.
The probability is then equal to the integral
ZZ
(x + y) dx dy =
0≤x,y≤1,x+y<1/2
1/2
Z
!
1/2−y
Z
(x + y) dx
Z
1/2
2
(x /2 +
dy =
0
0
0
Z
0
1/2
1/2
(1/2 − y)2
+ (1/2 − y)y dy =
2
0
1/2
1
y y 3 1
1
1 y2
=
−
dy =
−
−
=
.
8
2
8
6 0
16 48
24
x=1/2−y
xy)|x=0
Z
dy =
10.3. They are not, because the shape of the set where the density is not
equal to zero is a triangle, hence the density can not be written as a product of
two densities f (x) and g(y).
10.4. The density is constant 1/π inside the unit circle and zero outside of
it. Marginal density fX (x) is equal on [−1, 1] to the length of the chord with
the first coordinate equal to x times 1/π, hence
p
fX (x) = 2 1 − x2 /π.
(x,y)
√1
The conditional density fY (y|X = x) = ffX
(x) is then equal to 2 1−x2 inside
the circle and zero outside.
10.5. The number of red jelly beans in a bag of 400 beans is binomial(400, 0.1).
Then the probability that there are at least 45 red beans is approximately equal
to
44.5 − 40
4.5
√
) = 1 − Φ(0.75)
P Z≥
= P (Z ≥
6
400 · 0.1 · 0.9
10.6. We need more than 66 rolls if the sum of the first 66 rolls is not larger
than 200. One number has expectation 7/2 and variance 35
12 . It follows that the
sum of the first 66 rolls is approximately normal with mean 33 · 7 = 231 and
35
variance 66 · 12
= 385
2 . The probability that the sum is less or equal to 200 is
approximately
!
200.5 − 231
P Z≤ p
= P (Z ≤ 2.1983) ≈ 1 − Φ(2.20) = 0.0139.
385/2
10.7. Observed
proportion is p̂ = 25/625 = 1/25. The 99% confidence
q
q
p̂(1−p̂)
p̂)
interval is then p̂ − 2.58
, p̂ + 2.58 p̂(1−
= [0.02, 0.06]
n
n
1
in
10.8. The 99% confidence interval for proportion p in this case is included
1.29
1.29
, p̂ + √
p̂ − √
= [p̂ − 0.0258, p̂ + 0.0258].
2, 500
2, 500
It follows that it is enough p̂ to be more than 0.528.
10.9. If probability of a rainy day is 1/3, then probability of having 12 or
less raining days out of 60 is approximately
!
12.5 − 20
P Z≤p
= P (Z ≤ −2.05) = 1 − Φ(2.05) = 0.0202.
60 · 1/3 · 2/3
Consequently, the evidence contradicts the assumption that the probability is
1/3.
2