Suppose that the waiting time for a license plate renewal at a local office of a state motor vehicle department has been found to be normally distributed with a mean of 30 minutes and a standard deviation of 8 minutes. What is the probability that a randomly selected individual will have a waiting time between 15 and 45 minutes?
P(15 P[(15-30)/8] P(-1.88 "Check the handout Cumulative Areas under the Standard Normal Curve"
.9699-.0301=
.9398
. T/F

1. The probability that a randomly selected individual will have a waiting time between 15 and 45 minutes is:

2. The result 0.9699 - 0.301 = 0.9398 is true.

Explanation:

The relevant information to solve the problem is:

   1.  The waiting time has been found to be normally distributed

   2. The mean is 30 minutes

   3. The standard deviation is 8 minutes

   4. The question is What is the probability that a randomly selected individual will have a waiting time between 15 and 45 minutes?

You must calculate the Z-scores for both 15 min and 45 min, to use a standard distribution table to look up the Z-score, and the area between those values.

It is important to know that there are two types of standard distribution tables: tables that show values that represent the AREA to the LEFT of the Z-score, and tables that show values that represent the AREA to the RIGHT of the Z-score". You can use either of them.

1. Calculate the Z-score for X = 15 min:

You may round - 1.875 to - 1.88 to use the most common tables, which have only two decimals for the Z-scores.

Then, find the area to the left of Z-score = - 1.88. It is 0.0301. You can also use the function DISTR.NORM.ESTAND.N, which comes with Excel, the parameters would be (-1.88,1).

And the area to the right is 1 - 0.0301 = 0.969.

2. Calculate the Z-score for X = 45min

Given the symmetry of the standard normal distribution, the area to the right of Z-score = 1.875 is the same as the area to the left of Z-score = - 1.875. Rounding again 1.875 to 1.88, the area is 0.0301.

Hence, the area in between those Z-scores is 0.9699 - 0.0301 = 0.9398.

1. The probability that a randomly selected individual will have a waiting time between 15 and 45 minutes is:

2. The result 0.9699 - 0.301 = 0.9398 is true.

the answer is: c.that person might be arrested and deported

filling w-4 form incorrectly usually would rise suspicions from government's auditors. when they notice such mistakes, they would send their teams to investigate the situation further.

if those investigation teams believe that the incorrectness is done on purpose (with malicious intent), that person can be arrested or deported