Probabilities and Statistics Exercises 1. Consider throwing 2 usual dice. In each play, we observe the number of dots that stay on top, sum the returned result and to that sum, we apply the rule of...

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Probabilities and Statistics Exercises
1. Consider throwing 2 usual dice. In each play, we observe the number of dots that stay
on top, sum the returned result and to that sum, we apply the rule of casting out
nines. For example, if one dice has 6 dots and the other 4 dots, we get a sum of 10
dots, and, casting out nines, we achieve a result of 1.
1.1 Identify the space of results and the underlined X random variable;
1.2 How many ways we have to obtain a result of 2, according to the procedure
described above?
1.3 Calculate the value of P(X = 2) and P(X <= 5="" |="" x="">= 2) [P -> probability]
1.4 Calculate the probability function and the probability distribution of X.
2. Consider the random variable X.
2.1 Prove that
[ is the variance of ].
2.2 Calculate the expected value and the variance of random variable X from exercise
1.
3. The number of clients that go, every hour, to a bank agency, is processed according
the Poisson distribution. We know that in one hour the probability of entering 15
clients is 0,031 and in the same period the probability of entering 16 clients is 0,062.
Now, calculate the mean of clients entrance in the bank agency by hour.
4. The joint probability function of two random discrete variables X and Y, with natural
values (not negative) included zero, is equal to if
and if , with in every cases.
4.1 Calculate…
4.1.1 Calculate the value of .
4.1.2 Calculate the value of P(X = 2, Y = 1).
4.1.3 Calculate the value of P(X >= 1, Y

4.1.4 The marginal probability functions of X and Y.
4.2 Verify if the random variables X and Y are dependent or not. Justify your answer
with all necessary calculations.


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Probabilities and Statistics Exercises 1. Consider throwing 2 usual dice. In each play, we observe the number of dots that stay on top, sum the returned result and to that sum, we apply the rule of casting out nines. For example, if one dice has 6 dots and the other 4 dots, we get a sum of 10 dots, and, casting out nines, we achieve a result of 1. 1.1 Identify the space of results and the underlined X random variable; 1.2 How many ways we have to obtain a result of 2, according to the procedure described above? 1.3 Calculate the value of P(X = 2) and P(X <=>= 2) [P -> probability] 1.4 Calculate the probability function and the probability distribution of X. 2. Consider the random variable X. 2.1 Prove that [ is the variance of ]. 2.2 Calculate the expected value and the variance of random variable X from exercise 1. 3. The number of clients that go, every hour, to a bank agency, is processed according the Poisson distribution. We know that in one hour the probability of entering 15 clients is 0,031 and in the same period the probability of entering 16 clients is 0,062. Now, calculate the mean of clients entrance in the bank agency by hour. 4. The joint probability function of two random discrete variables X and Y, with natural values (not negative) included zero, is equal to if and if , with in every cases. 4.1 Calculate… 4.1.1 Calculate the value of . 4.1.2 Calculate the value of P(X = 2, Y = 1). 4.1.3 Calculate the value of P(X >= 1, Y <=>

Answered Same DayDec 21, 2021

Answer To: Probabilities and Statistics Exercises 1. Consider throwing 2 usual dice. In each play, we observe...

Robert answered on Dec 21 2021
121 Votes
1. Two dice are thrown. Number of dots on top=sum of dots on
two dice. We apply casting of 9 ,
lets say con . so we know that
Con(1)=1, con(2)=2, con(9)=0 ,con(10)=1
Sample space=36
Sum
on
dice(k)
Occurrences Con(k)
Casting
of 9 on
sum
Random
variable
Xi
occurrence Probability
2 1 2 0 4 4/36
3 2 3 1 3 3/36
4 3 4 2 3 3/36
5 4 5 3 3 3/36
6 5 6 4 3 3/36
7 6 7 5 4 4/36
8 5 8 6 5 5/36
9 4 0 7 6 6/36
10 3 1 8 5 5/36
11 2 2
12 1 3
1.1 Space of results [0,1,2,3,4,5,6,7,8]
Random variable X= value after applying casting cause.
1.2 . we can obtain a result 2 by 3 ways
(1,1),(6,5),(5,6) as 11 gives also 2
And 2 give the result 2.
1.3 P(X=2)=3/36
Let EVENT A = X<=5 P(A)=20/36
Event B= X>=2 P(B)=29/36
P(A/B)=P(A and B)/P(B)
P(A and B)=13/36
So P(A/B)=13/29
1.4 probability function, x is non negative integer
P(x)=4/36 for x=0
=3/36 for 1 <=X<=3
= x-1/36 for4<=x<=7
=13-x/36 x=8
x 0 1 2 3...
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