Question

A student has a class that is supposed to end at 9:00am and another that is supposed to begin at 9:10am. Suppose the actual ending time of the 9am class is a normally distributed random variable (rv) X1 with mean 9:02am and standard deviation 1.5 minutes. Suppose that the starting time of the 9:10am class is a normally distributed rv X2 with mean 9:10am and standard deviation 1 minute. Suppose further that the time necessary to

Answer 1

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 $\sim$ N (2, 1.5)

B $\sim$ N (10, 1)

C $\sim$ 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 = $\sqrt{4.25}$

The standard deviation = 2.06155

$P(A+C-B) = p \Big (Z < \dfrac{0-(-2)}{2.06155} \Big )$

$P(A+C-B) = p \Big (Z \le 0.97014 \Big )$

$\mathbf{P(A+C-B) = 0.8340}$