MSIN0096 Exercise Problem Set 1
1. Suppose a company manufactures LED light. The average life of their LED lights is 50000
hours. A LED light is considered defective if its life is less than 35000 hours. The distribution
of LED light life hours is approximately normal with a standard deviation 12500 hours.
What’s the probability of a LED light being defective?
Solutions: Let X be the average life. According to the question, we know that X ⇠
N(50000, 12500). The defective probability is
P (X < 35000) = 0.1150697
> pnorm(35000, 50000, 12500)
[1] 0.1150697
2. A data scientist wants to study the relationship between Tube delay and weather. She collects
data on tube operation and London weather records in the past 365 days. Data are presented
in the following table.
Weather on-time minor delay sever delay
Cloudy 109 17 9
Rainy 89 14 11
Sunny 102 11 3
(a) What is the probability of minor delay?
(b) What is the probability of sunny?
(c) What is the probability of cloudy and on-time?
(d) What is the probability of severe delay on rainy day?
(e) What is the probability of severe delay on sunny day?
(f) what is the probability of sunny day given on on-time?
Solutions:
(a) the probability of minor delay is 17+14+11365
(b) the probability of sunny is 102+11+3365
(c) the probability of cloudy and on-time 109365
(d) the probability of severe delay on rainy day is 1189+14+11
(e) the probability of severe delay on sunny day is 3102+11+3
(f) the probability of sunny day given on on-time is 102109+89+102+3
3. A 10-second video advertisement is played before a Youtube clip. Users can skip the ads
anytime. The distribution of seconds watched is listed as below.
1
asoBoBMBkBBMBBBBSs⑤'
x p(X)
1 0.09
2 0.18
3 0.21
4 0.12
5 0.07
6 0.05
7 0.03
8 0.04
9 0.10
10 0.11
(a) Please compute CDF of X.
(b) Please compute the mean, variance and standard deviation.
(c) What could you conclude about user’s skipping patterns?
Solutions:
(a) CDF of X is listed below
x CDF
X < 1 0
1  X < 2 0.09
2  X < 3 0.27
3  X < 4 0.48
4  X < 5 0.60
5  X < 6 0.67
6  X < 7 0.72
7  X < 8 0.75
8  X < 9 0.79
9  X < 10 0.89
X 10 1
(b)
E(X) = 1⇤0.09+2⇤0.18+3⇤0.21+4⇤0.12+5⇤0.07+6⇤0.05+7⇤0.03+8⇤0.04+9⇤0.1+10⇤0.11 = 4.74
var(X) = E(X2) E(X)2
= 1 ⇤ 0.09 + 22 ⇤ 0.18 + 32 ⇤ 0.21 + 42 ⇤ 0.12 + 52 ⇤ 0.07 +
62 ⇤ 0.05 + 72 ⇤ 0.03 + 82 ⇤ 0.04 + 92 ⇤ 0.1 + 102 ⇤ 0.11 4.742
= 8.83
(c) People are more like to skip at the beginning or at the end.
4. Suppose a person is randomly drawn from a population and then tested for a disease. Let D
= 1 if the person has the disease and 0 otherwise. Let T = 1 if the person tests positive and
0 otherwise. Suppose
2
P (D = 0) = 0.99
P (T = 1|D = 0) = 0.01
P (T = 1|D = 1) = 0.97
(a) Calculate the joint distribution of T and D.
(b) What is P (D = 1|T = 0)? How about P (D = 0|T = 1)?
Solutions:
(a) The joint distribution is given by 4 following probabilities
P (D = 1, T = 0) = P (D = 1)P (T = 0|D = 1) = (1P (D = 0))(1P (T = 1|D = 1)) = 0.01⇤0.03
P (D = 1, T = 1) = P (D = 1)P (T = 1|D = 1) = (1P (D = 0))P (T = 1|D = 1) = 0.01⇤0.97
P (D = 0, T = 0) = P (D = 0)P (T = 0|D = 0) = P (D = 0)(1P (T = 1|D = 0)) = 0.99⇤0.99
P (D = 0, T = 1) = P (D = 0)P (T = 1|D = 0) = P (D = 0)P (T = 1|D = 0) = 0.99⇤0.01
(b) What is p(D = 1|T = 0)? How about p(D = 0|T = 1)?
P (D = 1|T = 0) = P (D = 1, T = 0)
P (T = 0)
=
P (D = 1, T = 0)
P (D = 1, T = 0) + P (D = 0, T = 0)
=
0.01 ⇤ 0.03
0.01 ⇤ 0.03 + 0.99 ⇤ 0.99
P (D = 0|T = 1) = P (D = 0, T = 1)
P (T = 1)
=
P (D = 0, T = 1)
P (D = 1, T = 1) + P (D = 0, T = 1)
=
0.99 ⇤ 0.01
0.99 ⇤ 0.01 + 0.01 ⇤ 0.97
3

学霸联盟