California University encourages professors to consider using e-textbooks instead of the traditional paper textbooks. Many courses have adopted the new e-textbook option. Suppose that the random variable X represents the number of courses taken by a student during the Fall 19 semester at California University that provide an e-textbook option. The (partial) probability distribution for the random variable X is provided. X = # course with e-text option 0 1 2 3 4 Probability 0.30 0.25 0.15 0.10 California University encourages professors to consider using e-textbooks instead of the traditional paper textbooks. Many courses have adopted the new e-textbook option. Suppose that the random variable X represents the number of courses taken by a student during the Fall 19 semester at California University that provide an e-textbook option. The (partial) probability distribution for the random variable X is provided. X = # course with e-text option 0 1 2 3 4 5 Probability ___ ___ 0.30 0.25 0.15 0.10

Suppose the probability that only 1 course provides an e-textbook option is three times as likely as the probability of 0 courses providing an e-textbook option. Melissa is not your typical California University student, she is often looking for environmentally friendly options. She heard about the University's initiative and, when registering for courses, she decided to purposely look for classes that offered the e-textbook option.

Knowing that the number of courses she takes that offer an e-textbook option this semester is above the mean, what is the probability that all 5 of her classes have an e-textbook option?

Step-by-step explanation:

Hello!

The variable is:

X: number of coursers taken by students during Fall 19 semester at California University that provide an e-texbook option.

The following data represents the number of courses and their point probabilities:

X:     0; 1;    2;       3;      4;     5

P(X): ? ; ?; 0.30; 0.25; 0.15; 0.10

First step is to calculate the missing point probabilities corresponding to observations X=0 and X=1

Now remember that the total sum of probabilities of a variable is 1.

So P(0) + P(1) + P(2) + P(3) + P(4) + P(5) = 1

P(0) + P(1) + 0.30 + 0.25 + 0.15 + 0.10 = 1

P(0) + P(1) + 0.80= 1

P(0) + P(1) = 1 - 0.8

P(0) + P(1) = 0.2

Now acording to the text, the probability that 1 course offers an e-book option is three times as likely as the probability of 0 courses offerig it.

If P(0)= x then P(1)= 3x, then:

x + 3x= 0.2

4x= 0.2

x= 0.2/4

x= 0.05

Wich means that P(0)= 0.05 and P(1)= 0.15, and the probability distribution for the variable is:

X:       0;        1;      2;       3;      4;     5

P(X): 0.05 ;0.15 ; 0.30; 0.25; 0.15; 0.10

F(X): 0.05; 0.2   ; 0.5  ; 0.75; 0.90;   1

The average value for this variable is:

E(X)= ∑x*P(X)= (0*0.05)+(1*0.15)+(2*0.3)+(3*0.25)+(4*0.15)+(5*0.10)= 2.6

If all courses that the university offers are above the average, the probability that all courses offer e.book options is:

P(2.6≤X≤5)= P(X≤5) - P(X<2.6)= P(X≤5) - P(X≤2)= 1 - 0.5= 0.5

I hope it helps!

i think 1: 2

step-by-step explanation:

(2,20) means that the 2 is 2 hours and 20 is 20$. so there both 20$ for 2 hours only the same price. like it says. so, the dunk tank and the bounce house are the same price (20$) for the same amount of time (2 hours).

hope i