The image presents a problem involving a probability density function (PDF) for arrival times at a terminal after 8:00 A.M. The function provided is:

$f(x) = \frac{e^{-x/14}}{14} \quad \text{for} \quad 0 < x$

We are asked to determine the probability of arriving before 8:10 A.M. on two or more of five days, assuming independence between the days.

Step-by-step solution:

1. Find the Probability of Arriving Before 8:10 A.M. on One Day:

To solve this, we need to find the cumulative probability of arriving within 10 minutes (since 8:10 A.M. is 10 minutes after 8:00 A.M.). The PDF given is for $x$, the time after 8:00 A.M. in minutes, so we integrate the PDF from 0 to 10:

$P(\text{before 8:10 A.M.}) = P(X \leq 10) = \int_0^{10} \frac{e^{-x/14}}{14} \, dx$

This is the cumulative distribution function (CDF) for an exponential distribution with a mean of 14. The CDF of an exponential distribution is:

$F(x) = 1 - e^{-x/\lambda}$

Here, $\lambda = 14$, so the probability of arriving before 8:10 A.M. (within 10 minutes) is:

$P(\text{before 8:10 A.M.}) = 1 - e^{-10/14}$

2. Compute the Probability:

Let me calculate the above expression for $P(\text{before 8:10 A.M.})$.The probability of arriving before 8:10 A.M. on a single day is approximately 0.510 (rounded to three decimal places).

3. Find the Probability of Arriving Before 8:10 A.M. on Two or More Days:

Now, we want to find the probability of arriving before 8:10 A.M. on two or more of five days, assuming independence.

This is a binomial probability problem, where:

• $n = 5$ (the number of days),
• $p = 0.510$ (the probability of arriving before 8:10 A.M. on a single day),
• $k \geq 2$ (we need to arrive early on at least two days).

The total probability can be found using the complement rule:

$P(\text{arriving before 8:10 A.M. on two or more days}) = 1 - P(0 \text{ or } 1 \text{ days})$

We will calculate $P(0 \text{ days})$ and $P(1 \text{ day})$ using the binomial formula:

$P(k \text{ days}) = \binom{n}{k} p^k (1-p)^{n-k}$

Let's compute $P(0 \text{ days})$ and $P(1 \text{ day})$.The probability of arriving before 8:10 A.M. on two or more of five days is approximately 0.825 (rounded to three decimal places).

Summary:

• The probability of arriving before 8:10 A.M. on a single day is $P = 0.510$.
• The probability of arriving before 8:10 A.M. on two or more of five days is $P = 0.825$.

Would you like further details or have any questions?

Here are five related questions you might explore:

1. How would the probabilities change if the time threshold was 8:15 A.M. instead of 8:10 A.M.?
2. What is the expected number of days you will arrive before 8:10 A.M.?
3. How would the probability change if you were looking at 10 days instead of 5?
4. How does the variance in arrival times affect the probability?
5. What is the probability of arriving before 8:10 A.M. on exactly 3 of the 5 days?

Tip: The binomial distribution is great for calculating probabilities of independent events over multiple trials.