Question

The following represents the probability distribution of the number of classes a student takes in a spring semester.

Answer 1

Answer:

X        1       2       3         4          5          6

P(X) 0.115   a     0.422   0.312   0.034   0.015

If we want a probability distribution we need to satisfy two important conditions:

1) $P(X_i) \geq 0, \forall i=1,2,3,4,5,6$

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

And using the second condition we have this:

$0.115+a+0.422+0.312 +0.034 +0.015=1$

With a the probability associated with the value of X=2 and solving for a we got:

$a = 1-0.115-0.422-0.312-0.034-0.015= 0.102$

So then for this case $P(X=2) = 0.102$

Step-by-step explanation:

Let X the random variable that represent the number of classes that a student takes

For this case we have the following probability distribution given:

X        1       2       3         4          5          6

P(X) 0.115   a     0.422   0.312   0.034   0.015

If we want a probability distribution we need to satisfy two important conditions:

1) $P(X_i) \geq 0, \forall i=1,2,3,4,5,6$

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

And using the second condition we have this:

$0.115+a+0.422+0.312 +0.034 +0.015=1$

With a the probability associated with the value of X=2 and solving for a we got:

$a = 1-0.115-0.422-0.312-0.034-0.015= 0.102$

So then for this case $P(X=2) = 0.102$