The probability distribution of random variable, X, is defined as follows:
X 0 1 2 3 4
Probability 0 0.3 0.1 0.3 0.3
A) Explain if the above is a valid probability model.
B) Is the described random variable discrete or continuous?
C) What is the expected value (mean) of the probability distribution?
D) Calculate P(X = 3).
E) Calculate P(X < 4).
F) Calculate P(X > 0).
G) Calculate P(X = 5).
H) Calculate P(X = 0).
I) What is the total area under any density curve?

Respuesta :

Answer and Step-by-step explanation:

A) For the model to be a probability distribution, it has to follow two conditions:

1) The probability of each value of the discrete random variable is between, and included, 0 and 1:

2) The sum of all probabilities is 1;

In the table shown, the probabilities are from 0 to 0.3 - between 0 and 1;

Adding the probabilities: 0 + 0.3 + 0.1 + 0.3 + 0.3 = 1

Therefore, this model is a valid probability distribution model.

B) They are discrete because each value correspond to a finite number of possible values.

C) Expected value is calculated by

[tex]E(X) = \Sigma xP(x)[/tex]

[tex]E(X) = 0.0 + 1*0.3 + 2*0.1 + 3*0.3 + 4*0.3[/tex]

E(X) = 2.6

D) P(X=3) = P(3), which means probability of 3:

P(X=3) = 0.3

E) P(X<4): probabilities of values of X that are less than 4:

P(X<4) = P(0) + P(1) + P(2) + P(3)

P(X<4) = 0 + 0.3 + 0.1 + 0.3

P(X<4) = 0.7

F) P(X>0) = P(1) + P(2) + P(3) + P(4)

P(X>0) = 0.3 + 0.1 + 0.3 + 0.3

P(X>0) = 1

G) P(X=5) = P(5)

There is no probability of P(5) because the model doesn't "cover" that number.

H) P(X=0) = P(0)

P(X=0) = 0

I) The total area of any density curve is 1 because it represents all the possible values a variable can assume.

The answers are given as- A) Model is a valid probability distribution model.

B ) The described random variable is discrete.

C ) Mean E(X) = 2.6

D)  P(X=3) = 0.3

E )  P(X < 4) = 2.6

F )  P(X > 0) = 1

G ) There is no probability of P(5)

H ) P(X = 0) = 0

I ) The total area of any density curve is 1

What is probability?

Probability is defined as the ratio of the number of favorable outcomes to the total number of outcomes in other words the probability is the number that shows the happening of the event.

A) For the model to be a probability distribution, it has to follow two conditions:

In the table shown, the probabilities are from 0 to 0.3 - between 0 and 1;

Adding the probabilities: 0 + 0.3 + 0.1 + 0.3 + 0.3 = 1

Therefore, this model is a valid probability distribution model.

B) They are discrete because each value correspond to a finite number of possible values.

C) Expected value is calculated by

E(x) = ∑ P(x)

E ( x) = 0 + ( 1 x 0.3 ) + ( 2 x 0.1 ) + 3 ( 0.3 ) + 4 ( 0.3 )

E(X) = 2.6

D) P(X=3) = P(3), which means a probability of 3:

P(X=3) = 0.3

E) P(X<4): probabilities of values of X that are less than 4:

P(X<4) = P(0) + P(1) + P(2) + P(3)

P(X<4) = 0 + 0.3 + 0.1 + 0.3

P(X<4) = 0.7

F) P(X>0) = P(1) + P(2) + P(3) + P(4)

P(X>0) = 0.3 + 0.1 + 0.3 + 0.3

P(X>0) = 1

G) P(X=5) = P(5)

There is no probability of P(5) because the model doesn't "cover" that number.

H) P(X=0) = P(0)

P(X=0) = 0

I) The total area of any density curve is 1 because it represents all the possible values a variable can assume.

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