72.8k views
3 votes
W23 Normal Distriouriu Suppose the flight time between Montreal and Toronto follows a normal distribution with an average of 60 minutes and a standard deviation of 4 minutes. Mary is taking a flight tomorrow morning between these two cities. What is the probability that Mary's filght time will be between 50 minutes and 63 minutes? (Use only 3 decimal places)

1 Answer

4 votes

Final answer:

The probability that Mary's flight time will be between 50 minutes and 63 minutes is 0.767, calculated by finding the z-scores for each time, looking up the corresponding probabilities in the standard normal distribution table, and subtracting the smaller probability from the larger one.

Step-by-step explanation:

The question asks for the probability that Mary's flight time from Montreal to Toronto will be between 50 minutes and 63 minutes given that the flight time follows a normal distribution with an average (mean) of 60 minutes and a standard deviation of 4 minutes. We will use the properties of the normal distribution to calculate this probability.

First, convert the flight times to z-scores using the formula: z = (x - μ) / σ, where x is the flight time, μ is the mean, and σ is the standard deviation. The z-score for 50 minutes is (50 - 60) / 4 = -2.5, and for 63 minutes, it is (63 - 60) / 4 = 0.75.

Next, look up these z-scores in the standard normal distribution table, or use a calculator with normal distribution functions to find the probabilities. The probability of a z-score being less than -2.5 is approximately 0.0062, and the probability of a z-score less than 0.75 is approximately 0.7734.

To find the probability of the flight time being between 50 and 63 minutes, subtract the smaller probability from the larger one: 0.7734 - 0.0062 = 0.7672.

Therefore, the probability that Mary's flight time will be between 50 minutes and 63 minutes is 0.767 (rounded to three decimal places).

User Rogier Werschkull
by
8.5k points