Answer:
0.8340
Step-by-step explanation:
The aim of this question is to find the probability that the student makes it to the second class prior to when the lecture commences.
Provided that A, B, and C are independent of each other from the full complete question.
Then; we want P(A + C < B)
So A [tex]\sim[/tex] N (2, 1.5)
B [tex]\sim[/tex] N (10, 1)
C [tex]\sim[/tex] N (6, 1)
P(A + C - B < 0)
Since they are normally distributed( i.e. A + C - B)
Then;
E(A + C -B) = E(A) + E(C) - (B)
E(A + C -B) = 2 - 10 + 6
E(A + C -B) = -2
Var(A+C -B) = Var(A) + Var (B) + Var (C)
Var(A + C -B) = (1.5)² + (1)² + (1)²
Var(A + C -B) = 4.25
The standard deviation = [tex]\sqrt{4.25}[/tex]
The standard deviation = 2.06155
[tex]P(A+C-B) = p \Big (Z < \dfrac{0-(-2)}{2.06155} \Big )[/tex]
[tex]P(A+C-B) = p \Big (Z \le 0.97014 \Big )[/tex]
[tex]\mathbf{P(A+C-B) = 0.8340}[/tex]
