probability distribution help?

Question

Let x be the number of minutes after 11:00 that a bus leaves the bus station. Assume that the distribution of times is approximately normal w/ a mean of fifteen and standard deviation of 4 minutes.

(1) if a person gets to the bus station at 11:10, what is the probability thtat the person has missed the bus?

(2) If a person is willing to risk a 20% chance of not making the bus, what is the maximum number ofminutes after 11:00 that the person can reach the station?

(3) What time should the person reach the station to have a 50/50 chance of catching the bus?

I AM SO CONFUSED! Your help is truly appreciated!

Answer 1

the mean is 15 ... sd = 4...

the standardized formula is z = (x-15)/4

now, x < 10, then
z = (10-15)/4 = -1.25
P(x<10) = P(z<-1.25) = 1 - 0.8944 = .1056

now, we find z1 first such that...

P(z < z1) = 0.2 .. z1 = -0.84
x1 = 4(-0.84) + 15 = 11.64
that means after 11 minutes and 38.4 seconds...


for the last one... clearly it must be at 11:15...


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Answer 2

From the given, (X-15)/4 is a standard normal variable.

(a) If X = 10, then t = (10 -15)/4 = -1.25. Find -1.2 along the side of the standard normal table, then move across that row to the entry in the column under "5" at the top. The z-value is .1056. This is the probability that the bus left before 11:10, and hence the probability that the person arriving at 11:10 has missed the bus.

(b) Find .2000 (or close to it, or do linear interpolation) in the body of the standard normal table and find the corresponding t-value. It should be about -.84 (or -.8418 if you do interpolation)

(x-15)/4 = -.84
x = 15 - (4)(.84) = 15 - 3.36 = 11.64

(c) This can be calculated as in (b), but because the normal curve is symmetric about its mean, we can obtain the answer immediately: 11:15.