Question

Flight 202's arrival time is normally distributed with a mean arrival time of 10:30 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 between 10:00 p.m. and 11:00 p.m.

The probability is__

Answer 1

Answer:

The probability is 0.953

Step-by-step explanation:

We know that the mean $\mu$ is:

$\mu=10:30\ p.m$

The standard deviation $\sigma$ is:

$\sigma=0:15\ minutes$

The Z-score is:

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

We seek to find

$P(10:00\ p.m.<x<11:00\ p.m.)$

The Z-score is:

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

$Z=\frac{10:00-10:30}{0:15}$

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

$Z=-2$

The score of Z =-2 means that 10:00 p.m. is -2 standard deviations from the mean. Then by the rule of the 8 parts of the normal curve, the area that satisfies the condition of 2 deviations from the mean has percentage of 2.35%

and

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

$Z=\frac{11:00-10:30}{0:15}$

$Z=\frac{0:30}{0:15}$

$Z=2$

The score of Z =2 means that 11:00 p.m. is 2 standard deviations from the mean. Then by the rule of the 8 parts of the normal curve, the area that satisfies the condition of 2 deviations from the mean has percentage of 2.35%

$P(10:00\ p.m.<x<11:00\ p.m.)=100\%-2.35\%-2.35\%$

$P(10:00\ p.m.<x<11:00\ p.m.)=95.3\%$

$P(10:00\ p.m.<x<11:00\ p.m.)=0.953$