Unidad_3_Personal_Practice_Random_Variables_Part_1 pdf

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MODULE VARIABLES ALEATORIAS
Alliyah wants to be an astronaut. She decides to apply to NASA until she is
accepted but if she is rejected 3 times she will apply for a PhD program instead. Let
A denote acceptance and R denote rejection.
We previously determined that the sample space for this experiment would be: S
={A, RA, RRA, RRR}
Let X = the number of rejections that Alliyah experiences. If we get the outcome
RRA, what is the value of X?
3
1
2
0
Complete the following sentence:
If the outcome we get is RRR, then
X = 0
X = 3
X = 1 
X = 2
Good Job. The outcome RRA has two rejections, so the random variable
X (defined to be the number of rejections) is equal to 2. Note that this is
the only outcome for which X=2.
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Alliyah wants to be an astronaut. She decides to apply to NASA until she is
accepted but if she is rejected 3 times she will apply for a PhD program instead. Let
A denote acceptance and R denote rejection.
We previously determined that the sample space for this experiment would be: S
={A, RA, RRA, RRR}
Let X equal the number of rejections that Alliyah experiences.
Complete the following sentence:
X = 1 when the outcome we get is:
RA
RRA
AR
A
RRR
Recall again the sample space of our random experiment: S ={A, RA, RRA, RRR}
Let's define a different random variable.
Let the random variable Y be the number of applications that Alliyah submits.
What are the possible values of Y?
1, 2, 3, 4
1, 2, 3
0, 1, 2, 3, 4
0, 1, 2, 3
Correct! This is the only scenario where Alliyah gets rejected once. Her
first attempt was a rejection and her second attempt was an
acceptance. Great job!
Good job!
If Alliyah is accepted on her first attempt this would require only one
application: Y = 1
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A good rule of thumb is that  Discrete  random variables are
things we  count , while   random variables are things
we  measure.
Correct! A discrete variable has a countable set of values.
Correct! Continuous random variables are variables that we measure.
Continuous
A doctor's office wants to collect data from their maternity ward. A pregnant
woman is chosen at random from their practice. Decide whether each of the
following is a discrete or continuous random variable:
1. The number of  birth classes taken in preparation for labor 
Discrete
2. The mother's weight  Continuous
3. The number of previous children the woman has birthed 
Discrete
4. The exact time per week that the mother spends exercising 
Continuous
5. The number of alcoholic beverages that the woman consumes per week 
Discrete
If Alliyah is accepted on her second attempt this would require two
applications: Y = 2
If Alliyah is accepted on her third attempt or rejected all three times,
both of these scenarios would require 3 applications: Y = 3
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Good job! Number of classes is indeed a discrete random variable, since
it can take any value from a list of distinct values: 1, 2, 3, 4, etc.
Good job! The mother's weight is indeed continuous, since it can take
any value in an interval. It's true that when rounded to the nearest pound it
looks like it is discrete, but the exact weight of a person is continuous. You
cannot list all possible weights.
Good job! The mother's number of previous children is indeed a discrete
random variable, since it can take any value from a list of distinct values: 0,
1, 2, 3, 4, etc.
Good job! The exact time a woman spends doing prenatal exercise
during a week is indeed a continuous random variable, since it can take any
value in an interval. It's true that when rounded to the nearest hour or
minute it looks like it is discrete, but the exact time is continuous. You
cannot list all possible exact times.
Good job! The number of alcoholic beverages a mother drinks in a
typical week is indeed a discrete random variable, since it can take any value
from a list of distinct values: 0, 1, 2, 3, 4, etc.
A couple decides to have children until they have one boy and one girl, but not
more than four children.
The sample space is therefore:
                                          S = {BG, BBG, BBBG, BBBB, GB, GGB, GGGB, GGGG}
We are assuming that having a boy or a girl is equally likely [P(B)=P(G)=1/2], and
the that the child's gender in each birth is independent of the gender in the other
births.
Let the random variable X be the total number of children that the couple will
have.
Which of the following is the correct probability distribution of X?
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X 2 3 4
P(X=x) 1/4 1/4 1/2
X 2 3 4
P(X=x) 1/2 1/4 1/4
X 0 1 2 3 4
P(X=x) 1/16 11/16 1/8 1/16 1/16
Good Job. This is the correct probability distribution, which can be
obtained after the following summary of outcomes, probabilities, and
possible values of X:
Outcome Probability Value of X (number of
children)
BG 1/2 * 1/2 = 1/4 2
BBG 1/2 * 1/2 * 1/2 = 1/8 3
BBBG 1/2 * 1/2 * 1/2 * 1/2=
1/16
4
BBBB 1/16 4
GB 1/4 2
GGB 1/8 3
GGGB 1/16 4
GGGG 1/16 4
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Data were collected from a survey given to women on the number of children they
have. From that data, a probability distribution was constructed. The random
variable X is defined as the number of children born per family. It is shown below:
X 0 1 2 3 4 5
P(X=x) .28 .37 .23 .09 .02 .01
What is the probability that a randomly selected woman gave birth at least once?
.12
.35
.37
.65
.72
Given that a randomly selected person did give birth, what is the probability that
she gave birth more than three times?
.03
.04
.35
.72
Good job! At least once means one or more times. So P(gave birth at
least once) = P(X ≥ 1) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5).
This is .37 + .23 + .09 + .02 + .01 = .72. It is a little easier using
complements. P(gave birth at least once) = 1 - P(X = 0). This is 1 - .28,
which also is .72.
Good job! We want P(X > 3 | X ≥ 1) = .03 / .72 = .04.
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What is the probability that a randomly selected woman gave birth at most twice?
.12
.23
.35
.88
Recall the number of births per woman surveyed example:
Data were collected from a survey given to women on the number of children they
have. From those data, a probability distribution was constructed. The random
variable X is defined as the number of children born per family. It is shown below:
X 0 1 2 3 4 5
P(X=x) .28 .37 .23 .09 .02 .01
What is the mean of X?
2.5
0.1967
1/6 
1.23
Good job! We want P(gave birth at most twice). At most means that
number or less. P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2) = .28 + .37 + .23 =
.88. We also could have used complements. The complement of at most
2 is more than 2. So, P(X ≤ 2) = 1 - P(X > 2). P(X ≤ 2) = 1 - P(X > 2) = 1 - [ P(X
= 3) + P(X = 4) + P(X = 5) ]. The only advantage to using this method is
that the numbers (.09 + .02 + .01) are easier to add.
Good job! 
Indeed, the mean of X = (0)*(.28) + (1)*(.37) + (2)*(.23) + (3)*(.09) +
(4)*(.02) + (5)*(.01) 
= 0 + .37 + .46 + .27 + .08 + .05 = 1.23
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Data were collected froma survey given to women on the number of children they
have. From that data, a probability distribution was constructed. The random
variable X is defined as the number of children born per family. It is shown below:
X 0 1 2 3 4 5
P(X=x) .28 .37 .23 .09 .02 .01
Which of the following expressions represents the standard deviation of X?
[( − . ) (. )+( − . ) ( . )+( − . ) ( . )+( − . ) (. )+( − . ) ( . )+( − . ) ( .
( − )
(0 − 1 . 23)( . 28) + (1 − 1 . 23)( . 37) + (2 − 1 . 23)( . 23) + (3 − 1 . 23)( . 09) + (4 −
( . 28 − 1 . 23) (0) + ( . 37 − 1 . 23) (1) + ( . 23 − 1 . 23) (2) + ( . 09 − 1 . 23) (3)
(0 − 1 . 23) ( . 28) + (1 − 1 . 23) ( . 37) + (2 − 1 . 23) ( . 23) + (3 − 1 . 23) ( . 09) +
The following three histograms represent the probability distributions of the three
random variables X, Y, and Z.
Good job! Given that we solved for the mean showing that on average a
woman has 1.23 children. We can use this value to calculate the
Standard Deviation.
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Which of the three random variables has the largest standard deviation?
   X
   Y
   Z
   All three random variables have the same standard deviation.
   It is impossible to tell from the histograms.
A high-profile lawyer charges clients a $5000 retainer fee if she agrees to take on
their case and she bills an additional $500 per hour.
The average case takes 63 hours with a variance of 144.
If X is the number of hours worked on a case, an expression for the amount of
money that is collected from a case is:
Good job! All three random variables take the values 1, 2, 3, 4, 5,
and it is pretty easy to see (by symmetry) that the mean of all
three random variables is 3. The random variables are different,
though, with respect to how likely they are to have values that
are “far” from the mean. We see that out of the three random
variables, random variable Z is the most likely to have values 1
and 5 (which are the furthest from the mean), and therefore Z
has the largest standard deviation.
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5500
5000 + 500X
500 + 5000X
5500X
Given your answer to the previous question, and using the first set of rules for
means and variances, what is the mean and standard deviation of the amount of
money that is collected from a client?
The average case costs $36,500 and typically the cost is about $6000
away from that average.
The average case costs $346,500 and typically the cost is about $6000
away from that average.
The average case costs $36,500 and typically the cost is about $104.88
away from that average.
The average case costs $6,000 and typically the cost is about $6000
away from that average.
Good job! Indeed, there is a $5000 charge per case (regardless of the
number of hours spent), plus $500 for each hour spent working on the
case.
Great job! This is the correct answer. Indeed, An expression for the
amount of money that is collected from a case is: 5000 + 500x
The average case takes 63 hours with a variance of 144.
To find the mean charge per case we input 63 hours into the equation:
Mean = 5000 + 31,500 = $36,500
To solve for the standard deviation we use the rule for transforming the
variance and then convert that to the standard deviation by taking the
square root.
Variance = 500 (144) = (250,000)(144) = 36,000,000
SD = Square root of 36,000,000 = $6000
2
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There are corporate clients who use the law firm previously discussed. These
clients bill more hours but the law firm uses Junior Associates to work on the case
for these additional hours. These Junior Associates (represented by the random
variable J) tend to work for an average of 44 hours with a variance of 100 hours per
case  Recall that the Senior Associates (represented by the random variable S)
work for an average of 63 hours with a variance of 144 hours per case.
Let the random variable T be the total number of hours clocked per corporate
case. What is the mean of T?
107
75
53.5
54
Assuming that the Senior and Junior Associates work independently for corporate
cases, what is the standard deviation of T?
The square root of 44
15.62
244
22
A start up company hires two salespeople. 
Mercy is young but she is an amazing salesperson. She is successful at eventually closing the
deal with 75% of all of her potential clients.
Nick is somewhat successful. He was hired because his friend referred him to work at the
Good job!  Indeed, since 63 + 44 = 107 there is a mean total of 107 hours
per corporate case.
Correct!  Indeed, since S and J are independent we can add their
variances and then take the square root to solve for the standard
deviation: 100 + 144 = 244
Sq Rt of 244 is about equal to 15.62
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company. He is successful at eventually closing the deal with 45% of his potential clients.
The company wants to review sales performance metrics. They have more confidence in Mercy
so they decide to randomly follow 5 of her client outcomes to determine how she is
performing.
They decide that Nick may need more careful consideration so they decide to randomly select
10 of his client outcomes to determine how he is performing.
Let M equal the number of clients (out of 5) that Mercy closes successfully.
Let N equal the number of clients (out of 10) that Nick closes successfully.
Let S be the total number of clients that are successfully closed by these two sales associates.
Which of the following is true of M, N, and S?
M is binomial with n = 5 and p = .75
N is binomial with n = 10 and p = .45
S is not binomial.
All of the above.
Only the first two responses are true.
What is the probability that exactly 4 of Mercy's clients are closed successfully?
Answers may be rounded
75%
40%
38%
45%
Good job! M represents the number of sales that Mercy makes out of a
random sample of 5 → M is binomial with n = 5 and p = 0.75. N
represents the number of sales that Nick makes out of a random
sample of 10 → N is also binomial with n = 10 and p = 0.45. S represents
the total number of sales successfully closed by both associates. S is not
binomial, since the probability of “success” (making the sale) is not the
same for all the sales associates (.75 for Mercy and .45 for Nick).
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What is the mean and standard deviation of N (the number of sales out of 10 that
are successful)? Answers may be rounded
mean = 4.5, standard deviation = 1.57
mean = 10, standard deviation = .45
mean  = .45, standard deviation = 10
mean = 4.5, standard deviation = 2.48
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Good job! We need to find P(X = 4) where X is binomial with n = 5 and p
= 0.75. 
P(X = 4) =
( !)
( !( − )!)
( . 75 × . 25 ) = ( ) × ( . 79) = . 3955
The probability that Mercy makes exactly 4 sales is about equal to 40%.
Great job!
The mean can be computed as follows:  np = 10*.45 = 4.5
The standard deviation can also be computed:
𝑠𝑑 = 𝑛𝑝(1 − 𝑝) = (4 . 5 × . 55)
The standard deviation is about equal to 1.57
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