The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.4 minutes and a standard deviation of 3 minutes. Assume that the distribution of taxi and takeoff times is approximately normal. You may assume that the jets are lined up on a runway so that one taxies and takes off immediately after the other, and that they take off one at a time on a given runway.
(a) What is the probability that for 35 jets on a given runway, total taxi and takeoff time will be less than 320 minutes? (Use 3 decimal places.)
Incorrect: Your answer is incorrect.
(b) What is the probability that for 35 jets on a given runway, total taxi and takeoff time will be more than 275 minutes? (Use 3 decimal places.)
(c) What is the probability that for 35 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes? (Use 3 decimal places.)
To solve these questions, we can use the properties of the normal distribution. We know that the total taxi and takeoff time is a random variable, denoted as x, with a mean (µ) of 8.4 minutes and a standard deviation (σ) of 3 minutes.
To find probabilities, we can convert the problem into a standard normal distribution problem by standardizing the variable x using the z-score formula:
z = (x - µ) / σ
Now, let's solve each question step-by-step:
(a) What is the probability that for 35 jets on a given runway, the total taxi and takeoff time will be less than 320 minutes?
We need to find P(x < 320). To calculate this probability, we need to find the z-score corresponding to 320 minutes.
z = (320 - 8.4) / 3
z = 103.6 / 3
z = 34.53
Now, we can use a standard normal distribution table or a calculator to find the probability corresponding to this z-score.
Using the standard normal distribution table, the probability is practically 1 (to three decimal places) since the z-score is very large.
(b) What is the probability that for 35 jets on a given runway, the total taxi and takeoff time will be more than 275 minutes?
We need to find P(x > 275). Again, we need to standardize the value, this time for 275 minutes.
z = (275 - 8.4) / 3
z = 266.6 / 3
z = 88.87
Using the standard normal distribution table, the probability corresponding to this z-score is approximately 1 (to three decimal places).
(c) What is the probability that for 35 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?
We need to find P(275 < x < 320). To calculate this probability, we need to find the z-scores for both values.
For 275 minutes:
z1 = (275 - 8.4) / 3
z1 = 266.6 / 3
z1 = 88.87
For 320 minutes:
z2 = (320 - 8.4) / 3
z2 = 311.6 / 3
z2 = 103.87
Now, we can use the standard normal distribution table to find the probability that falls between z1 and z2.
Using the table, we can find P(z1 < z < z2) or P(88.87 < z < 103.87).
Finding the area under the standard normal curve between these two z-scores gives us the required probability.
Please note that using a standard normal distribution table might not give you the exact decimal places mentioned in the question.
I hope this explanation helps you understand how to solve these probability problems using the standard normal distribution.