Suppose you have to cross a train track on your commute. The probability that you will have to wait for a train is 1/5, or .20. If you don’t have to wait, the com mute takes 15 minutes, but if you have to wait, it takes 20 minutes.

What is the expected value of the time it takes you to commute?
Is the expected value ever the actual commute time? Explain.

So, the expected value of the time it takes to commute is 16 minutes.

Step-by-step explanation:

Let X is time it takes you to commute. We calculate E(X)?

We know  that the probability that you will have to wait for a train is 1/5, or .20. If you don’t have to wait, the com mute takes 15 minutes, but if you have to wait, it takes 20 minutes.

The probability that you will have to wait for a train is :

The probability of not waiting for a train is:

So we get the following probability distribution:

We calculate E(X):

So, the expected value of the time it takes to commute is 16 minutes.

Expected value is not always the actual travel time, because it can sometimes take anywhere from 15 to 20 minutes.

answer: c, the green line

step-by-step explanation:

there is no black line, eliminate b.

simple interest implies periodic payments, but the red line shows no payments, eliminate d.

simple interest has equal payments. if paid say every day, the graph is about a straight line (actually a stairstep).

the yellow line is not straight, eliminate a.

check: the green line is straight, answer is c.

simple interest with positive interest rate gives a straight line sloping up to the right. green line:

g(t) = $100 + ($200-$100) ÷ (20 years) × (t years)

= $100 + (4%/year) × ($100)(t years)

incidentally, the yellow line curving upward represents continuously compounded interest

y = $100 × 2^(t years / (14 assuming the yellow line hits $200 at 14 years. 2^0 is 1, so y(0) = $100; 2^(14/14) is 2, so y(14) = $200.

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step-by-step explanation: