Answer:
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!
