Question

Airbus designs a jet that requires an average takeoff distance of 3,250 feet. Testing of this jet reveals a standard deviation of 300 feet. What is the probability that a jet of this model will

Answer 1

To find the probability that a jet of this model requires a takeoff distance of less than or equal to a certain value x, you can use the standard normal distribution. First, convert the takeoff distance to a z-score using the formula z = (x - mean) / standard deviation. Then, use a z-table to find the corresponding probability.

Step-by-step explanation:

To find the probability that a jet of this model requires a takeoff distance of less than or equal to a certain value x, we can use the standard normal distribution. First, we need to convert the given takeoff distance to a z-score, which measures the number of standard deviations an observation is from the mean.

The formula for calculating the z-score is z = (x - mean) / standard deviation. In this case, the mean is 3250 feet and the standard deviation is 300 feet.

By substituting the values into the formula, we get z = (x - 3250) / 300. Now, we can use a standard normal distribution table (also known as a z-table) to find the probability corresponding to the z-score. The z-table provides the cumulative probability up to a given z-score. In this case, we are interested in finding the probability up to the z-score calculated by the previous formula.

Let's say we want to find the probability that the jet requires a takeoff distance less than or equal to 3500 feet. We would calculate the z-score as z = (3500 - 3250) / 300 = 0.833. Using the z-table, we can find the corresponding probability as 0.7967.

This means that there is a 79.67% probability that a jet of this model will require a takeoff distance less than or equal to 3500 feet.

Answer 2

To calculate the probability that a jet will need 3,750 feet for takeoff, we need to find the z-score and use the z-table. The probability is 0.9525 (95.25%) that this model of jet will require that distance.

Step-by-step explanation:

To calculate the probability that a jet of this model will need 3,750 feet for a safe takeoff, we need to find the z-score associated with this value and then use the z-table to find the corresponding probability.

The z-score is calculated using the formula: z = (x - μ) / σ, where x is the value we want to find the probability for, μ is the mean, and σ is the standard deviation.

In this case, the mean is 3,250 feet and the standard deviation is 300 feet. Plugging these values into the formula, we get: z = (3,750 - 3,250) / 300 = 1.67 (rounded to two decimal places).

Now, we refer to the z-table to find the probability associated with a z-score of 1.67. From the table, we find that the probability is approximately 0.9525.

Therefore, the correct answer is: The probability is 0.9525 (95.25%) that this model of jet will require 3,750 feet for a takeoff.

Complete question:

Airbus designs a jet that requires an average takeoff distance of 3,250 feet. Testing of this jet reveals a standard deviation of 300 feet. What is the probability that a jet of this model will need 3,750 feet for a safe takeoff? Assume that the data are normally distributed. Use the z-source formula and the z-table below to calculate this probability. Round the z-score to two decimal places. NOTE: If the runway is not long enough for safe takeoff a serious incident can occur resulting in loss of life, injury to passengers, and damage to an aircraft cost millions of dollars.