Answer:
Yes the table shows a probability distribution
The mean is [tex]E(X) = 1.636 [/tex]
The standard deviation is [tex]\sigma = 0.845 [/tex]
Step-by-step explanation:
From the question we are told that
The data given is
x 0 1 2 3
P(x) 0.087 0.344 0.415 0.154
Generally we can evaluate [tex]P(x) * x[/tex] as
[tex]P(x) * x[/tex] 0 (1 * 0.344 ) (2 * 0.415) (3 * 0.154 )
=> [tex]P(x) * x[/tex] 0 0.344 0.830 0.462
Generally we can evaluate [tex]P(x) * x^2 [/tex] as
[tex]P(x) * x^2 [/tex] 0 [tex ](1^2 * 0.344 )[/tex] [tex ](2^2 * 0.415)[/tex] [tex ](3^2 * 0.154 )[/tex]
=> [tex]P(x) * x^2 [/tex] 0 [tex] 0.344 [/tex] [tex ]1.66 [/tex] [tex]1.386[/tex]
Generally the mean is mathematically represented as
[tex]E(X) = \sum [x* P(x)][/tex]
[tex]E(X) = 0 + 0.344+ 0.830+0.462 [/tex]
[tex]E(X) = 1.636 [/tex]
Generally the standard deviation is mathematically represented as
[tex]\sigma = \sqrt{V(X)}[/tex]
Here [tex]V(X)[/tex] is the variance which is mathematically represented as
[tex]V(X) = [E(X^2) ]- [E(X)]^2[/tex]
Here
[tex]E(X^2) = \sum [x^2 * P(x)][/tex]
[tex]E(X) =0 +0.344+1.66+1.386 [/tex]
[tex]E(X) = 3.3900 [/tex]
So
[tex]V(X) = 3.3900 - [1.636]^2[/tex]
[tex]V(X) = 3.3900 - 2.6764[/tex]
[tex]V(X) = 0.7136[/tex]
So
[tex]\sigma = \sqrt{0.7136}[/tex]
[tex]\sigma = 0.845 [/tex]
