Question

For questions 1-3, you must identify the probability distribution used. Additionally,

for every probability you calculate, you must show either the manual calculations used to
get it or the source from where you got the number (if an app was used, you must write
the commands used to get it).
1. The telephone lines serving an airline reservation office are all busy about 65% of the time.
a. If you are calling this office, what is the probability that you will complete your call on
the first try? The second try? The third try?
b. If you and a friend must both complete calls to this office, what is the probability that
a total of four tries will be necessary for both of you to get through?

Answer 1

1. a) The probability of completing the call on the first try is 35%. b) The probability that a total of four tries will be necessary for both you and your friend to get through is 12.25%. 2. a) The average time between two successive calls is 0.25 minutes. b) The probability that the next call occurs in less than 10 seconds is 21.59%. c) The probability that exactly five calls occur within a minute is 15.63%. d) The probability that fewer than five calls occur within a minute is 62.88%. e) The probability that more than 40 calls occur in an eight-minute period is 4.64%. 3. a) The random variable X represents the number of airplanes arriving at the airport per hour. b) The probability that there are exactly 100 arrivals and departures in one hour is 4.07%. c) The probability that there are at most 100 arrivals and departures in one hour is 36.31%.

Step-by-step explanation:

Question 1:

1. a. To find the probability that you will complete the call on the first try, we can subtract the probability of all lines being busy (65%) from 100%: 100% - 65% = 35%. So, the probability of completing the call on the first try is 35%.
   b. To find the probability that a total of four tries will be necessary for both you and your friend to get through, we can multiply the probability of completing the call on the first try for both of you: 0.35 * 0.35 = 0.1225 or 12.25%.

2. Question 2:

   1. a. The average time between two successive calls can be found by taking the reciprocal of the rate of calls per minute. In this case, the rate is 4 calls per minute, so the average time between two successive calls is 1/4 or 0.25 minutes.
   2. b. To find the probability that the next call occurs in less than 10 seconds, we convert 10 seconds to minutes (10/60) and use the exponential distribution: P(X < 10/60) = 1 - e^(-4 * 10/60) ≈ 0.2159 or 21.59%.
   3. c. The probability that exactly five calls occur within a minute can be found using the Poisson distribution with the mean rate of 4 calls per minute: P(X = 5) = (e^(-4) * 4^5) / 5! ≈ 0.1563 or 15.63%.
   4. d. The probability that fewer than five calls occur within a minute can be found by summing the probabilities of 0, 1, 2, 3, and 4 calls occurring: P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = e^(-4) + e^(-4) * 4 + e^(-4) * 4^2 / 2 + e^(-4) * 4^3 / 6 + e^(-4) * 4^4 / 24 ≈ 0.6288 or 62.88%.
   5. e. The probability that more than 40 calls occur in an eight-minute period can be found by summing the probabilities of 41, 42, 43, and so on up to infinity: P(X > 40) = 1 - P(X ≤ 40) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + ... + P(X = 40)) = 1 - (e^(-4) + e^(-4) * 4 + e^(-4) * 4^2 / 2 + ... + e^(-4) * 4^40 / 40!) ≈ 0.0464 or 4.64%.

3. Question 3:

   1. a. The random variable X represents the number of airplanes arriving at the airport per hour.
      b. To find the probability that there are exactly 100 arrivals and departures in one hour, we can use the Poisson distribution with the mean rate of arrivals and departures per hour. If the average is 120, then we can calculate P(X = 100) = (e^(-120) * 120^100) / 100! ≈ 0.0407 or 4.07%.
   2. c. To find the probability that there are at most 100 arrivals and departures in one hour, we can sum the probabilities of 0, 1, 2, ..., 100 arrivals and departures: P(X ≤ 100) = P(X = 0) + P(X = 1) + P(X = 2) + ... + P(X = 100) = (e^(-120) + e^(-120) * 120 + e^(-120) * 120^2 / 2 + ... + e^(-120) * 120^100 / 100!) ≈ 0.3631 or 36.31%.