Question

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 m/h and 10% travel more than 73.24 m/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 m/h? c. What is the probability that a randomly selected vehicle's speed exceeds the speed limit of 70 m/h?

Answer 1

a. The mean and standard deviation of vehicle speed can be found using the properties of the normal distribution. We can set up equations using the given percentiles and solve for the mean and standard deviation. b. The probability that a randomly selected vehicle's speed is between two given values can be found using the cumulative probability values for the corresponding z-scores. c. The probability that a randomly selected vehicle's speed exceeds a given speed limit can be found using the cumulative probability for the corresponding z-score.

Step-by-step explanation:

a. To find the mean and standard deviation of the vehicle speed, we can use the properties of the normal distribution. Let's denote the mean as μ and the standard deviation as σ. We are given that 5% of vehicles travel less than 39.12 m/h, which corresponds to the z-score -1.645 (since the cumulative probability for z-score -1.645 is 0.05). Using this information, we can set up the equation:
z = (X - μ) / σ
-1.645 = (39.12 - μ) / σ
Similarly, we can set up another equation for 10% of vehicles traveling more than 73.24 m/h:
z = (X - μ) / σ
1.282 = (73.24 - μ) / σ
Solving these two equations simultaneously, we can find the values of μ and σ. The resulting values should agree with those given in the cited article.

b. To find the probability that a randomly selected vehicle's speed is between 50 and 65 m/h, we need to find the cumulative probability for these values in the normal distribution. We can calculate the z-scores for the given speeds:
z1 = (50 - μ) / σ
z2 = (65 - μ) / σ
Then, we can use the cumulative probability values corresponding to the z-scores to find the probability.
P(50 ≤ X ≤ 65) = P(z1 ≤ Z ≤ z2) = P(Z ≤ z2) - P(Z ≤ z1)

c. To find the probability that a randomly selected vehicle's speed exceeds the speed limit of 70 m/h, we can use the cumulative probability for the z-score corresponding to 70 m/h. We calculate the z-score:
z = (70 - μ) / σ
Then, we can find the probability using the cumulative probability for this z-score.
P(X > 70) = 1 - P(Z ≤ z)

Answer 2

The mean vehicle speed is about 58.89 m/h with a standard deviation of around 11.64 m/h, while the probability of a randomly selected vehicle's speed falling between 50 and 65 m/h can be determined using their respective z-scores. Additionally, the likelihood of a vehicle's speed exceeding the 70 m/h speed limit can be calculated using the z-score for 70 m/h.

Let's denote the mean speed of vehicles as μ and the standard deviation as σ.

Given:

1. 5% of vehicles travel less than 39.12 m/h.

2. 10% of vehicles travel more than 73.24 m/h.

To find μ and σ:

a. Using z-scores for the given percentiles in a standard normal distribution:

For the 5th percentile (z = -1.645):

`\[ (39.12 - \mu)/(\sigma) = -1.645 \]`

For the 90th percentile (z = 1.282):

`\[ (73.24 - \mu)/(\sigma) = 1.282 \]`

Solving these two equations will give us the mean (μ) and standard deviation (σ).

Let's solve these equations simultaneously:

From the first equation:

`\[ (39.12 - \mu)/(\sigma) = -1.645 \]`

`\[ 39.12 - \mu = -1.645 * \sigma \]`

From the second equation:

`\[ (73.24 - \mu)/(\sigma) = 1.282 \]`

`\[ 73.24 - \mu = 1.282 * \sigma \]`

Now, we have a system of equations:

`\[ 39.12 - \mu = -1.645 * \sigma \]`

`\[ 73.24 - \mu = 1.282 * \sigma \]`

By solving these equations simultaneously, we can find the values of μ and σ.

Let's solve for μ and σ:

Adding both equations:

`\[ 39.12 + 73.24 - 2\mu = -1.645\sigma + 1.282\sigma \]`

`\[ 112.36 - 2\mu = -0.363\sigma \]`

`\[ -2\mu = -0.363\sigma - 112.36 \]`

`\[ \mu = 0.1815\sigma + 56.18 \]`

Now, substitute this expression for μ into one of the earlier equations:

`\[ 39.12 - (0.1815\sigma + 56.18) = -1.645\sigma \]`

`\[ -17.06 - 0.1815\sigma = -1.645\sigma \]`

`\[ -0.1815\sigma + 1.645\sigma = 17.06 \]`

`\[ 1.4645\sigma = 17.06 \]`

`\[ \sigma \approx 11.64 \, \text{m/h} \]`

Now that we have found σ, let's substitute it back to find μ:

`\[ \mu = 0.1815 * 11.64 + 56.18 \]`

`\[ \mu \approx 58.89 \, \text{m/h} \]`

Therefore, the mean (μ) is approximately 58.89 m/h, and the standard deviation (σ) is approximately 11.64 m/h.

b. Probability that a randomly selected vehicle's speed is between 50 and 65 m/h:

We'll use the z-score formula to standardize and find the probabilities for both speeds:

For 50 m/h:

`\[ z = (50 - 58.89)/(11.64) \approx -0.76 \]`

For 65 m/h:

`\[ z = (65 - 58.89)/(11.64) \approx 0.53 \]`

Now, using a standard normal distribution table or calculator, find the probabilities associated with these z-scores. Then, find the probability between these two z-scores by subtracting the smaller probability from the larger one.

c. Probability that a randomly selected vehicle's speed exceeds the speed limit of 70 m/h:

`\[ z = (70 - 58.89)/(11.64) \approx 0.96 \]`

Using the standard normal distribution table or calculator, find the probability associated with this z-score, which represents the probability of a vehicle exceeding 70 m/h.