Business, 17.02.2020 18:05 harveyangel123p2tjae

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

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Answer from: betsylancer8522

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

P [ X > (15 - 30)/8] - P [ X < (45 - 30)/8] = P( Z > -1.88) - P (Z < 1.88)

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?

Solution

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:

$Z-score=\frac{X-mean}{standard\text{ }deviation}$

$Z-score=\frac{15-30}{8}\approx -1.875$

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

Z-score=(45-30)/8=1.875

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.

Conclusion

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

P [ X > (15 - 30)/8] - P [ X < (45 - 30)/8] = P( Z > -1.88) - P (Z < 1.88)

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

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