Question

Civil engineers believe that W, the amount of weight (in unitsof 1,000 pounds) that a certain span of bridge can withstandwithout structural damage resulting is normally distributed, with amean of 400 and a standard deviation of 40. Suppose that the weight(again, in units of 1,000 pounds) of a car is a random variablewith mean = 3 and a standard deviation = 0.3. How many cars wouldhave to be on the bridge span for the probability of structuraldamage to exceed 0.1 ?

Answer 1

To find the number of cars that would cause structural damage to exceed 0.1 probability, we calculate the z-score for this probability and use the formula z = (W - 400) / 40.

Using the z-score formula, we find that the weight of the cars would have to be equal to or greater than 348,800 pounds.

Step-by-step explanation:

To find the number of cars that would have to be on the bridge span for the probability of structural damage to exceed 0.1, we need to calculate the z-score for this probability and then use the z-score formula.

The formula to calculate the z-score is: z = (x - μ) / σ, where x is the value we are interested in, μ is the mean, and σ is the standard deviation.

In this case, the z-score formula becomes: z = (W - 400) / 40.

Next, we need to find the x-value (W) for a given z-score that corresponds to a probability of 0.1. Using a standard normal distribution table, we find that the z-score for a 0.1 probability is approximately -1.28.

Plugging this value into the z-score formula and solving for W, we get: -1.28 = (W - 400) / 40.

Solving for W, we find that W ≈ 348.8. This means that the weight of the cars would have to be equal to or greater than 348,800 pounds (348.8 * 1000) for the probability of structural damage to exceed 0.1.