Question

Flight 202's arrival time is normally distributed with a mean arrival time of 6:30 p.m. and a standard deviation of 15 minutes. Use the eight-part symmetry of the area under a normal curve to find the probability that a randomly chosen arrival time is after 7:15 p.m.

The probability is__

Answer 1

The probability is 0.0015

Explanation:

We know that the mean
`\mu` is:

`\mu=6:30\ p.m`

The standard deviation
`\sigma` is:

`\sigma=0:15\ minutes`

The Z-score is:

`Z=(x-\mu)/(\sigma)`

We seek to find

`P(x>7:15\ p.m.)`

The Z-score is:

`Z=(x-\mu)/(\sigma)`

`Z=(7:15-6:30)/(0:15)`

`Z=(0:45)/(0:15)`

`Z=3`

The score of Z = 3 means that 7:15 p.m. is 3 standard deviations from the mean. Then by the rule of the 8 parts of the normal curve, the area that satisfies the conficion of 3 deviations from the mean has percentage of 0.15%

So

`P(x>7:15\ p.m.)=P(Z>3)=0.0015`