A student has a class that is supposed to end at 9:00AM and another that is supp

Question

A student has a class that is supposed to end at 9:00AM and another that is supposed to begin at 9:15AM. Suppose the actual ending time of the 9AM class is normally distributed random variable (X1) with a mean of 9:02 and a standard deviation of 2.5 minutes and that the starting time of the next class is also a normally distributed random variable (X2) with a mean of 9:15 and a standard deviation of 3 minutes. Suppose also that the time necessary to get from one class to another is also a normally distributed random variable (X3) with a mean of 10 minutes and a standard deviation of 2.5 minutes. What is the probability that the student makes it to the second class before the second lecture starts? (Hint: Assume X1, X2 and X3 are independent also think Linear combinations)

Explanation / Answer

X1 ~ N(9:02 , 2.5^2)

X2 - N( 0:10 , 2.5^2)

X3 ~ N(9:15 , 3^2)

P(X3 > X1 +X2)

P(X3 -X1 -X2 >0)

E(X3-X1-X2) = 5

sd (X3-X1-X2) = sqrt(2.5^2 + 2.5^2 + 3^2) = 4.63680924^2

Z = ((X3-X1-X2) - 5)/4.63680924

P(X3 -X1 -X2 >0)

= P(Z > -5/4.63680924)

=P(Z > -1.0783277)

= 0.8596

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