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 Winter 20 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

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!

(a) the volume of the cone is 20/3 in3

hence, option a is correct.