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Vehicle speed on a particular bridge in China can be modeled as normally distributed ("Fatigue Reliability Assessment for Long-Span Bridges under Combined Dynamic Loads from Winds and Vehicles" \(J\). of Bridge Engr., 2013: 735-747). a. If \(5 \%\) of all vehicles travel less than \(39.12 \mathrm{~m} / \mathrm{h}\) and \(10 \%\) travel more than \(73.24 \mathrm{~m} / \mathrm{h}\), what are the mean and standard deviation of vehicle speed? [Note: The resulting values should agree with those given in the cited article.] b. What is the probability that a randomly selected vehicle's speed is between 50 and \(65 \mathrm{~m} / \mathrm{h}\) ? c. What is the probability that a randomly selected vehicle's speed exceeds the speed limit of \(70 \mathrm{~m} / \mathrm{h}\) ?

Short Answer

Expert verified
Mean: 58.3 m/h, SD: 11.657 m/h. Probabilities are 0.480 (50-65 m/h) and 0.158 (>70 m/h).

Step by step solution

01

Identify Normal Distribution Parameters

The problem states that vehicle speeds follow a normal distribution. We have:- 5% of vehicles traveling less than 39.12 m/h.- 10% of vehicles traveling more than 73.24 m/h.Let \( \mu \) be the mean and \( \sigma \) be the standard deviation. We will use z-scores to solve for \( \mu \) and \( \sigma \).
02

Use the Z-Score Formula

For a normal distribution, the z-score corresponding to the 5% percentile is typically -1.645 and for the 90% percentile (10% above) is 1.2816.- For the lower 5%: \( z_1 = \frac{39.12 - \mu}{\sigma} = -1.645 \)- For the upper 10%: \( z_2 = \frac{73.24 - \mu}{\sigma} = 1.2816 \)
03

Solve the System of Equations

We have two equations:1. \(39.12 = \mu - 1.645 \sigma\)2. \(73.24 = \mu + 1.2816 \sigma\)Subtract the first equation from the second to eliminate \( \mu \) and solve for \( \sigma \):\[ 73.24 - 39.12 = \mu + 1.2816 \sigma - (\mu - 1.645 \sigma) \]\[ 34.12 = 2.9266 \sigma \]\[ \sigma = \frac{34.12}{2.9266} \approx 11.657 \]
04

Find the Mean

Substitute \( \sigma = 11.657 \) back into one of the original equations:Using \(39.12 = \mu - 1.645 \times 11.657\):\[ 39.12 = \mu - 19.1786 \]\[ \mu = 39.12 + 19.1786 \]\[ \mu = 58.2986 \]So, the mean \( \mu \) is approximately 58.3 m/h.
05

Calculate Probability for 50 to 65 m/h

To find the probability that a vehicle's speed is between 50 and 65 m/h, calculate the z-scores:- For 50 m/h: \( z_3 = \frac{50 - 58.3}{11.657} = -0.712 \)- For 65 m/h: \( z_4 = \frac{65 - 58.3}{11.657} = 0.575 \)Look up these z-scores in a standard normal distribution table to find the probabilities:- \( P(Z < -0.712) \approx 0.237 \)- \( P(Z < 0.575) \approx 0.717 \)The probability that a vehicle’s speed is between 50 and 65 m/h is:\[ P(-0.712 < Z < 0.575) = 0.717 - 0.237 = 0.480 \]
06

Calculate Probability for Exceeding 70 m/h

To find the probability that a vehicle exceeds 70 m/h, calculate the z-score:- For 70 m/h: \( z_5 = \frac{70 - 58.3}{11.657} \approx 1.001 \)Find \( P(Z > 1.001) \) using the standard normal distribution table:- \( P(Z > 1.001) = 1 - P(Z < 1.001) \approx 1 - 0.842 \)- \( P(Z > 1.001) \approx 0.158 \)The probability that a vehicle’s speed exceeds 70 m/h is approximately 0.158.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Z-Scores
Z-scores, often known as standard scores, are a way to describe a data point's distance from the mean of a distribution. The z-score is expressed in terms of standard deviations and is a handy tool for understanding where a particular value lies within a normal distribution. Whenever you calculate a z-score, you're essentially asking, "How many standard deviations away is this value from the mean?"

In a normal distribution:
  • A z-score of 0 means the data point is exactly at the mean.
  • A positive z-score indicates that the data point is higher than the mean.
  • A negative z-score shows that the data point is below the mean.
These scores are crucial when you need to solve for unknowns like the mean (\(\mu\)) or the standard deviation (\(\sigma\)), as seen in this exercise. By using z-scores, we solve the percentiles given by the problem (like 5% and 10%) to set up equations that allow us to derive these unknown parameters.
Mean and Standard Deviation
The mean and standard deviation are two fundamental concepts in statistics, especially when dealing with normal distributions. The mean (\(\mu\)) is the average value of all data points in a distribution and serves as a central point around which data is dispersed. Calculating the mean involves summing up all speeds and dividing by the number of data points.

The standard deviation (\(\sigma\)) tells us how much individual data points vary or deviate from the mean. A smaller standard deviation indicates data points are close to the mean, while a larger one suggests a wider spread. In normal distributions, like our vehicle speeds, approximately 68% of the data falls within one standard deviation from the mean, 95% within two, and about 99.7% within three.

In the context of the problem, understanding these concepts helps us calculate the parameters using given percentile values and solve equations with z-scores. It allows us to model the vehicle speeds accurately based on the distribution's characteristics.
Probability Calculations in Normal Distribution
Probability calculations in a normal distribution allow us to find the likelihood of a random variable falling within a certain range. This is particularly useful when you have a bell-shaped curve where most outcomes are clustered around a mean, such as vehicle speeds on a bridge.

To determine probabilities:
  • First, identify the relevant z-scores for given values.
  • Use standard normal distribution tables or calculators to find probabilities associated with these z-scores.
  • Combine these probabilities to find the overall probability within or outside a range.
For example, if we want to know the probability that a vehicle's speed is between 50 and 65 m/h, we calculate the z-scores for these speeds and look up the probability of occurrences between these values. Also, to find the likelihood of speeds exceeding a specific limit, like 70 m/h, we need a similar approach by finding the complement of the cumulative probability at that z-score.

This process helps us understand the distribution's behavior, providing insights into vehicle flow and speed regulation on the bridge.

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Most popular questions from this chapter

Let the ordered sample observations be denoted by \(y_{1}, y_{2}, \ldots, y_{n}\left(y_{1}\right.\) being the smallest and \(y_{n}\) the largest). Our suggested check for normality is to plot the \(\left(\Phi^{-1}((i-.5) / n), y_{i}\right)\) pairs. Suppose we believe that the observations come from a distribution with mean 0 , and let \(w_{1}, \ldots, w_{n}\) be the ordered absolute values of the \(x_{i}^{\prime}\) s. A half-normal plot is a probability plot of the \(w_{i}^{\prime} s\). More specifically, since \(\quad P(|Z| \leq w)=P(-w \leq Z \leq w)=\) \(2 \Phi(w)-1\), a half-normal plot is a plot of the \(\left(\Phi^{-1} /\\{[(i-.5) / n+1] / 2\\}, w_{i}\right)\) pairs. The virtue of this plot is that small or large outliers in the original sample will now appear only at the upper end of the plot rather than at both ends. Construct a half-normal plot for the following sample of measurement errors, and comment: \(-3.78,-1.27,1.44\), \(-.39,12.38,-43.40,1.15,-3.96,-2.34,30.84\).

If \(X\) has an exponential distribution with parameter \(\lambda\), derive a general expression for the \((100 p)\) the percentile of the distribution. Then specialize to obtain the median.

Suppose that when a transistor of a certain type is subjected to an accelerated life test, the lifetime \(X\) (in weeks) has a gamma distribution with mean 24 weeks and standard deviation 12 weeks. a. What is the probability that a transistor will last between 12 and 24 weeks? b. What is the probability that a transistor will last at most 24 weeks? Is the median of the lifetime distribution less than 24 ? Why or why not? c. What is the 99th percentile of the lifetime distribution? d. Suppose the test will actually be terminated after \(t\) weeks. What value of \(t\) is such that only \(.5 \%\) of all transistors would still be operating at termination?

Suppose that blood chloride concentration (mmol/L) has a normal distribution with mean 104 and standard deviation 5 (information in the article "'Mathematical Model of Chloride Concentration in Human Blood," J. of Med. Engr: and Tech., 2006: 25-30, including a normal probability plot as described in Section 4.6, supports this assumption). a. What is the probability that chloride concentration equals 105 ? Is less than 105? Is at most 105 ? b. What is the probability that chloride concentration differs from the mean by more than 1 standard deviation? Does this probability depend on the values of \(\mu\) and \(\sigma\) ? c. How would you characterize the most extreme .1\% of chloride concentration values?

The lifetime \(X\) (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters \(\alpha=2\) and \(\beta=3\). Compute the following: a. \(E(X)\) and \(V(X)\) b. \(P(X \leq 6)\) c. \(P(1.5 \leq X \leq 6)\) (This Weibull distribution is suggested as a model for time in service in "On the Assessment of Equipment Reliability: Trading Data Collection Costs for Precision," J. of Engr. Manuf., 1991: 105-109.)

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