4. Discrete Random Variables
Author: Alberto Martı́n Zamora
Email: [email protected]
Website: www.icmat.es/miembros/amartin
4.1
A random variable is a variable that assumes numerical values associated
with the random outcomes of an experiment, where one and only one numerical
value is assigned to each sample point.
4.2
A discrete random variable can assume only a countable number of values, while
on the other hand, a continuous random variable takes values in an interval.
4.3
a Discrete
b Continuous
c Continuous
d Discrete
e Continuous
f Continuous
4.4
a Continuous
b Discrete
c Discrete (in minutes)
d Continuous
e Discrete
f Discrete
4.5
a Continuous
b Discrete
c Discrete
d Discrete
e Discrete
f Continuous
4.6
Discrete, because it can only take the values *, **, ***, **** and *****.
4.7
You are supposed to do this exercise.
4.8
It is a continuous variable.
4.9
You are supposed to do this exercise.
4.10
It is a discrete variable, a convicted drug dealer can only have an integer number
of prior felony arrests. Someone can not get ”half” arrested!
4.11
The number of patients he has to attend in a day.
4.12
The answers to the question: ”Who are you going to vote for POTUS?”
4.13
The number of babies born a given month.
4.14
The number of pictures drawn by each drawer.
4.15
We can write down a table which the values the variable can take and the associated probabilities, we can use a bar chart or a give a formula for the probability
of each outcome.
4.16
a) x can take the values −4, 0, 1, 3.
b) 1, with a probability of .4.
c) The probability that x is greater than 0 is .7.
d) 0.
4.17
You are supposed to do this exercise.
4.18
You are supposed to do this exercise.
4.19
a) .7
b) 1 - .7 = .3
c) 1
d) .2
e) P (x ≤ 11 or x > 12) = 1 − P (x = 12) = 1 − .2 = .8
4.20
a) P (x ≤ 0) = .65
b) P (x > −1) = .75
c) .85
d) .95
e) .7
f) .65
4.21
You are supposed to do this exercise.
4.22
a) x takes the values 2,3,4,5 with probabilities p(2) = .0408, p(3) = .1735, p(4) =
.6020, p(5) = .1837.
b) p(x = 5) = .1837
c) p(x ≤ 2) = p(x = 2) = .0408.
4.23
You are supposed to do this exercise.
4.24
a)
p(1) = .4
p(2) = .52
p(3) = .02
p(4) = .04
b) .06
In addition, check RStudio script Exercise 4 24.R.
4.25
a) GG, GB, BG, BB
b) Each sample point has probability .25
c)
p(0) = .25
p(1) = .5
p(2) = .25
d)
p(0) = .222
p(1) = .259 + .254 = .513
p(2) = .265
4.26
a) Because it can only take the values 0,1,2,3,4 and 5
b)
p(0) = .655 = 0.116029063
p(1) = 5 ∗ (.35) ∗ (.65)4 = 0.312385937
p(2) = 10 ∗ (.35)2 ∗ (.65)3 = 0.336415625
p(3) = 10 ∗ (.35)3 ∗ (.65)2 = 0.181146875
p(4) = 5 ∗ (.35)4 ∗ (.65) = 0.048770313
p(5) = .355 = 0.005252187
c) Clearly all the probabilities are positive, and we check that they add up to 1.
d) The probability is p(4) + p(5) = 0.0540225
4.27
a) p(1) = .23, because it means that in our first try we get a contaminated
cartridge.
b)p(5) = (.23) ∗ (.71)4 = 0.08085199, the exponent 4 means that we have first
to obtain 4 non contaminated cartridges before getting a contaminated one.
c) p(x ≥ 2) = 1 − p(1) = .77. The interpretation is simply, in order to have
x ≥ 2, the only requirement is that the first gun cartridge inspected is not contaminated, and this event has probability .77.
4.34
A measure of central tendency of the variables
4.35
No. The mean does not have to be an attainable value. For example, consider
a random variable that takes the values -1 and 1 with probability .5
4.36
In that case, we can use the Empirical Rule, so the probability is .95
4.37
Check RStudio Script Exercise 4 37.R
4.38
Check RStudio Script Exercise 4 38.R
4.39
Check RStudio Script Exercise 4 39.R
4.40
You are supposed to do this exercise.
4.41
Check RStudio Script Exercise 4 41.R
4.42
Check RStudio Script Exercise 4 42.R
4.43
You are supposed to do this exercise.
4.44
1 * .4 + 2*.54 + 3*.02 + 4 * .04 = 1.7
4.46
The net winnings is a random variable,x, that can take to values, -1 if you do
1
not win, and 6.999.999 if you win, with probabilities 22.999.999
23.000.000 and 23.000.000
respectively.
The mean is
6999999/2300000 − 1 ∗ 22999999/23000000 = 2.043478
this means that in average, computed over all participants, your investment gets
a return of roughly 2. However, since only 1 out of 23M get returns, it is not a
good way to get money.
4.50
1. The experiment consists of n identical trials.
2. There are only two possible outcomes on each trial. We will denote one
outcome by S (for Success) and the other by F (for Failure)
3. The probability of S remains the same from trial to trial. The probability is
denoted by p, and the probability of F is denoted by q = 1 − p.
4. The trials are independent.
5. The binomial random variable x is the number of S’s in n trials.
4.51
The formula for the probability distribution is
7
p(x) =
(.2)x (.8)7−x
x
4.52
a) 5
b) .7
4.53
a) 15
b) 10
c) 1
d) 1
e) 4
4.54
Check RStudio Script Exercise 4 54.R
4.55
You are supposed to do this exercise.
4.56
Check RStudio Script Exercise 4 56.R
4.57
Check RStudio Script Exercise 4 57.R
4.58
You are supposed to do this exercise.
4.59
Check RStudio Script Exercise 4 59.R
4.60
Check RStudio Script Exercise 4 60.R
4.61
Check RStudio Script Exercise 4 61.R
4.62
Check RStudio Script Exercise 4 62.R
4.63
Check RStudio Script Exercise 4 63.R
4.64
You are supposed to do this exercise.
4.65
Check RStudio Script Exercise 4 65.R
4.66
Check RStudio Script Exercise 4 66.R