Respuesta :
Considering the discrete distribution, it is found that the desired measures are given as follows:
a) 0.86.
b) 0.57.
c) 0.45.
d) 0.14.
e) 0.43.
f) 3.02.
g) 1.31.
What is the probability distribution?
According to the table, it is given by:
P(X = 1) = 0.14.
P(X = 2) = 0.29.
P(X = 3) = 0.12.
P(X = 4) = 0.31.
P(X = 5) = 0.14.
Item a:
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - P(X = 1) = 1 - 0.14 = 0.86[/tex]
Item b:
[tex]P(X \geq 3) = 1 - P(X < 3) = 1 - P(X = 1) - P(X = 2) = 1 - 0.14 - 0.29 = 0.57[/tex]
Item c:
P(X > 3) = P(X = 4) + P(X = 5) = 0.31 + 0.14 = 0.45.
Item d:
P(X < 2) = P(X = 1) = 0.14.
Item e:
[tex]P(X \leq 2) = P(X = 1) + P(X = 2) = 0.14 + 0.29 = 0.43[/tex].
Item f:
The mean is given by the sum of each outcome multiplied by it's respective probability, hence:
E(X) = 0.14(1) + 0.29(2) + 0.12(3) + 0.31(4) + 0.14(5) = 3.02.
Item g:
The standard deviation is given by the square root of the sum of the difference squared of each observation and the mean, multiplied by it's respective probabilities, hence:
[tex]\sqrt{V(X)} = \sqrt{0.14(1-3.02)^2 + 0.29(2-3.02)^2 + 0.12(3-3.02)^2 + ... + 0.14(5-3.02)^2} = 1.31[/tex]
More can be learned about discrete probability distributions at https://brainly.com/question/24855677
