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Chapter 6: Q5 (page 240)

Continuous Uniform Distribution. In Exercises 5–8, refer to the continuous uniform distribution depicted in Figure 6-2 and described in Example 1. Assume that a passenger is randomly selected, and find the probability that the waiting time is within the given range.

Greater than 3.00 minutes

Short Answer

Expert verified

The probability for waiting time larger than 3.00 minutes is 0.4.

Step by step solution

01

Given information

The graph shows the complete area under the uniform distribution curve, which is equal to 1.

02

State the relationship between area and probability

As the area under the density curve is 1, it can be inferred that there is a one-to-one correspondence between the area and the probabilities.

Thus, the probability that the waiting time is greater than 3 is equal to the area under the curve in the shaded region, as shown in the graph below.


03

Find the probability

The probability that the waiting time is greater than 3.00 minutes is the area of the shaded rectangle with length 0.2 and width is 5-3=2.

Thus,

PX>3.00=Length×widthoftheshadedarea=0.2×2=0.4

So, the probability of selecting a passenger randomly when the waiting time is greater than 3.00 minutes is 0.4.

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Most popular questions from this chapter

Standard normal distribution, assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. In each case, Draw a graph, then find the probability of the given bone density test score. If using technology instead of Table A-2, round answers to four decimal places.

Less than 0

Continuous Uniform Distribution. In Exercises 5–8, refer to the continuous uniform distribution depicted in Figure 6-2 and described in Example 1. Assume that a passenger is randomly selected, and find the probability that the waiting time is within the given range.

Between 2 minutes and 3 minutes

Standard Normal DistributionIn Exercises 17–36, assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. In each case, draw a graph, then find the probability of the given bone density test scores. If using technology instead of Table A-2, round answers

to four decimal places.

Between -2.00 and 2.00.

Birth Weights Based on Data Set 4 “Births” in Appendix B, birth weights are normally distributed with a mean of 3152.0 g and a standard deviation of 693.4 g.

a. What are the values of the mean and standard deviation after converting all birth weights to z scores using z=x-μσ?

b. The original birth weights are in grams. What are the units of the corresponding z scores?

Standard Normal Distribution. Find the indicated z score. The graph depicts the standard normal distribution of bone density scores with mean 0 and standard deviation 1.
See all solutions

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