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ki77a [65]
2 years ago
7

Verify that the following function is a probability mass function, and determine the requested probabilities. f left-parenthesis

x right-parenthesis equals left-parenthesis 27divided by 13right-parenthesis left-parenthesis 1 divided by 3right-parenthesis Superscript x ,
Mathematics
1 answer:
Licemer1 [7]2 years ago
7 0

Answer:

f(x) = \frac{27}{13} (\frac{1}{3})^x , x=1,2,3

1) f(x_i) \geq 0, \forall x_i

2) sum_{i=1}^n P(X_i) =1

We can find the individual probabilities and we got:

f(1)= \frac{27}{13} (\frac{1}{3})^1 =0.6923

f(2)= \frac{27}{13} (\frac{1}{3})^2 =0.2307

f(3)= \frac{27}{13} (\frac{1}{3})^3 =0.0770

And the sum of the 3 values 0.6923+0.2307+0.0770= 1 so then we satisfy all the conditions and we can conclude that f(x) is a probability distribution.

P(X \leq 1) = P(X=1) =0.6923

P(X>1) = P(X=2) +P(X=3) = 0.2307+0.0770=0.3077

P(2

Step-by-step explanation:

For this case we have the following density function:

f(x) = \frac{27}{13} (\frac{1}{3})^x , x=1,2,3

In order to satisfty that this function is a probability mass function we need to check two conditions:

1) f(x_i) \geq 0, \forall x_i

2) sum_{i=1}^n P(X_i) =1

We can find the individual probabilities and we got:

f(1)= \frac{27}{13} (\frac{1}{3})^1 =0.6923

f(2)= \frac{27}{13} (\frac{1}{3})^2 =0.2307

f(3)= \frac{27}{13} (\frac{1}{3})^3 =0.0770

And the sum of the 3 values 0.6923+0.2307+0.0770= 1 so then we satisfy all the conditions and we can conclude that f(x) is a probability distribution.

And if we want to find the following probabilities:

P(X \leq 1) = P(X=1) =0.6923

P(X>1) = P(X=2) +P(X=3) = 0.2307+0.0770=0.3077

P(2

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