Question

The scheduled arrival time for a daily flight from Boston to New York is 9:35 am. Historical data show that the arrival time follows the continuous uniform distribution with an early arrival time of 9:12 am and a late arrival time of 9:46 am. a. After converting the time data to a minute scale, calculate the mean and the standard deviation of the distribution. (Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) b. What is the probability that a flight arrives late (later than 9:35 am)? (Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.)

Answer 1

Answer: a) mean= 569 = 9:29am

SD= +/- 17

b) P= 11/34 = 0.32

Step-by-step explanation:

since it's a continuous uniform distribution all time between 9:12 and 9:46 have equal frequency of occurrence.

9:12= 9×60min + 12= 552mins

9:46= 9×60min + 46 = 586mins

9:35= 9×60min + 35 = 575mins

Therefore, the mean is given by,

Tm= (552+586)/2= 569mins = 9:29am

Standard deviation = √{(552-569)sq + (586-569)sq}/2

= √(17×17)+(17×17) /2

=17 =+/-17

Probability of arriving late = 586-575 / 586-552 = 11/34

= 0.32