Consider the following discrete probability distribution. x 15 22 34 40 p(x = x) 0.08 0.52 0.25 0.15

a. is this a valid probability distribution?
c. what is the probability that the random variable x is less than 39? (round your answer to 2 decimal places.) probability
d. what is the probability that the random variable x is strictly between 9 and 24? (round your answer to 2 decimal places.) probability
e. what is the probability that the random variable x is greater than 17? (round your answer to 2 decimal places.) probability

     a. Yes.

c. 0.85

d. 0.60

e. 0.92

Step-by-step explanation:

Given discrete probability distribution :

    x       15      22     34    40

P(X = x) 0.08  0.52  0.25   0.15

a. The sum of the all probabilities in an distribution is 1.

here, Total sum = P(X= 15) + P(X=22) + P(X=34) + P(X=40)

= 0.08 + 0.52+  0.25   +0.15=1

∴ Yes , It is a valid probability distribution.

c. The probability that the random variable X is less than 39 : P(X<39)

= P(X= 15) + P(X=22) + P(X=34)

= 0.08 + 0.52+  0.25 = 0.85

∴ The probability that the random variable X is less than 39 = 0.85

d. The probability that the random variable X is strictly between 9 and 24

:P( 9<X<24)= P(X= 15) + P(X=22)

= 0.08 + 0.52 =0.60

∴ The probability that the random variable X is strictly between 9 and 24=0.60

e. The probability that the random variable X is greater than 17 : P(X>17)

= P(X=22) + P(X=34) + P(X=40)

=0.52+  0.25  +0.15=0.92

∴ The probability that the random variable X is greater than 17=0.92

step-by-step explanation:

first, eliminate the parenthesis:

3

x

+

2

−

4

x

second, group like terms:

3

x

−

4

x

+

2

third, combine like terms:

(

3

−

4

)

x

+

2

→

−

1

x

+

2

→

−

x

+

2