Help po Grade 11 Prob & Stat

[kriegrust]

Patulong po dito sa probability mass function... Salamat po

 

[kriegrust]

parang iba po kasi yung ginawang example dito. nag multiply ng 1/6 kada {1, 2, 3, 4, 5, 6,}

P(X=x) = { 1/6, if x = (1,2,3,4,5,6) 0, otherwise​

First condition​

f(x) = P(X=x) > 0 (should be greater than zero)​

P(X=x) = 1/6​

P(X=1)= 1/6 (greater than 0)

P(X=2)=1/6 (greater than 0)

P(X=3)=1/6 (greater than 0)

P(X=4)=1/6 (greater than 0)

P(X=5)=1/6 (greater than 0)

P(X=6)=1/6 (greater than 0)

Second condition​

Sum of all probabilities should be equal to 1​

[ P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) ]= 1​

substitute and add
1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 1

1=1 ​

Ganito po yung ginawa sa example

 

[kriegrust]

fx=kx​

where kx=1/10x​

since problem required 1/6

without x

all 6 values (1,2,3,4,5,6)= 1/6

fx=1/6​

any value of x (1,2,3,4,5,6) in fx = 1/6

1/10 x po pala yun kaya nag multiply sa 1/10 lahat. paano po yung ganito? multiply ko muna sa sarili nila bago ko multiply sa 2/9? hindi po kasi = 1 ang sagot

 

[kriegrust]

same process apply​

1st condition checking​

sub values (1,2,3) to x (squared)

P(X=1)= 2/9 (greater than 0)

P(X=2)=8/9 (greater than 0)

P(X=3)=18/9 or 2 (greater than 0)

2nd condition checking​

2/9 + 8/9 + 18/9 = 1

28/9 is not equal to 1

NOT a PMF

kapag hindi po equal to 1 ilalagay po is not a PMF tama po ba? same po ba dito sa isang ito?

 

[kriegrust]

yes panote as not a PMF since hindi nacomply second condition

same process

sub 1,2,3,4 to x/3

1/3
2/3
3/3
4/3

salamat po. akala ko kasi may gagawin pang iba dahil with support S nakalagay. medyo na confuse kasi ako sa example na ito.

 

[kriegrust]

welcome, unless otherwise specified na compute for value, follow mo yang steps and formula nyan, pero sa problem #4 more on proving kung pmf or not pmf?

Opo, show that each of the following is a PMF lang ang nasa instruction. Salamat po ulit