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

Answer:

The probability is 0.0015

Step-by-step 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:

$Z=\frac{x-\mu}{\sigma}$

We seek to find

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

The Z-score is:

$Z=\frac{x-\mu}{\sigma}$

$Z=\frac{7:15-6:30}{0:15}$

$Z=\frac{0:45}{0:15}$

$Z=3$

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

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