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Fill in the P(X=x) values to give a legitimate probability distribution for the discrete random variable , whose possible values

Naddika [18.5K]

For a probability distribution to be represented, it is needed that P(X = 0) + P(X = 1) = 0.44. Hence one possible example is:

P(X = 0) = 0.40.

P(X = 1) = 0.04.

<h3>What is needed for a discrete random variable to represent a probability distribution?</h3>

The sum of all the probabilities must be of 1, hence:

P(X = 0) + P(X = 1) + P(X = 3) + P(X = 4) + P(X = 5) = 1.

Then, considering the table:

P(X = 0) + P(X = 1) + 0.15 + 0.17 + 0.24 = 1

P(X = 0) + P(X = 1) + 0.56 = 1

P(X = 0) + P(X = 1) = 0.44.

Hence one possible example is:

P(X = 0) = 0.40.

P(X = 1) = 0.04.

More can be learned about probability distributions at brainly.com/question/24802582

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6 0

2 years ago

I need this asap

Margarita [4]

Answer:

can you show the graph

Step-by-step explanation:

8 0

3 years ago

FP!!!!!!!!!!!!!!!!!!! if ur up ( 10 x 8 ) + 150

taurus [48]

Answer:

220

Step-by-step explanation:

first we have by Bracket of BODMAS rule then we have to multiply 10 *8 = 80

first we have by Bracket of BODMAS rule then we have to multiply 10 *8 = 80then we have to open bracket 80 +150

first we have by Bracket of BODMAS rule then we have to multiply 10 *8 = 80then we have to open bracket 80 +150 = 220

(10*8) +15

Solution

Solution= (80) +150

Solution= (80) +150= 80+150

Solution= (80) +150= 80+150=. 220

Solution= (80) +150= 80+150=. 220

8 0

3 years ago

Read 2 more answers

okay, genuinely does anyone know how this work? if you do could you please explain how you get the answer-

fomenos

Answer:

The pdf is not showing anything

Step-by-step explanation:I would help you otherwise.I am sorry.Don't let this get you down.I hope you have a good day.

4 0

3 years ago

Consider the sequence defined recursively by do = 0, an = an-1 + 3n – 1. a) Write out the first 5 terms of this sequence,

creativ13 [48]

Answer:

The first 5 terms of the sequence is 2,7,15,26,40.

Step-by-step explanation:

Given : Consider the sequence defined recursively by $a_0=0$ $a_n=a_{n-1}+3n-1$

To find : Write out the first 5 terms of this sequence ?

Solution :

$a_n=a_{n-1}+3n-1$ and $a_0=0$

The first five terms in the sequence is at n=1,2,3,4,5

For n=1,

$a_1=a_{1-1}+3(1)-1$

$a_1=a_{0}+3-1$

$a_1=0+2$

$a_1=2$

For n=2,

$a_2=a_{2-1}+3(2)-1$

$a_2=a_{1}+6-1$

$a_2=2+5$

$a_2=7$

For n=3,

$a_3=a_{3-1}+3(3)-1$

$a_3=a_{2}+9-1$

$a_3=7+8$

$a_3=15$

For n=4,

$a_4=a_{4-1}+3(4)-1$

$a_4=a_{3}+12-1$

$a_4=15+11$

$a_4=26$

For n=5,

$a_5=a_{5-1}+3(5)-1$

$a_5=a_{4}+15-1$

$a_5=26+14$

$a_5=40$

The first 5 terms of the sequence is 2,7,15,26,40.

5 0

3 years ago