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3 years ago

HELP PLEASE HURRY WILL MARK BRAINLIEST!!!!!!

Mathematics

1 answer:

LiRa [457]3 years ago

8 0

Answer:

c . 63

Step-by-step explanation:

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

Licemer1 [7]

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:

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

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

$P(2$

7 0

3 years ago