Question

Large Marge's Trucking Company claims that the average weight of its delivery trucks when they are fully loaded is 5,550 pounds and the standard deviation is 220 pounds. Assume that the population follows a normal distribution. Forty trucks are randomly selected and weighed. a) What is the probability that the sample average weight of 40 fully loaded delivery trucks is more than 5580 pounds? (Round answer to 4 decimal places.) b) Within what limits will 96% of the sample average weights occur? (Round answers to 3 decimal places.) to c) What is the probability that the weight of a randomly selected fully loaded delivery truck is between 5450 and 5600 pounds? (Round answer to 4 decimal places.) The bank manager wants to show that the new system reduces typical customer waiting times to less than 6 minutes. One way to do this is to demonstrate that the mean of the population of all customer waiting times is less than 6. Letting this mean be fi, in this exercise we wish to investigate whether the sample of 107 waiting times provides evidence to support the claim that u is less than 6. For the sake of argument, we will begin by assuming that u equals 6, and we will then attempt to use the sample to contradict this assumption in favor of the conclusion that u is less than 6. Recall that the mean of the sample of 107 waiting times is x = 5.58 and assume that o, the standard deviation of the population of all customer waiting times, is known to be 2.26. (a) Consider the population of all possible sample means obtained from random samples of 107 waiting times. What is the shape of this population of sample means? That is, what is the shape of the sampling distribution of x? Normal because the sample is (Click to select) (b) Find the mean and standard deviation of the population of all possible sample means when we assume that u equals 6. (Round your answer to 4 decimal places.) Ug = 6,0 = (c) The sample mean that we have actually observed is x = 5.58. Assuming that u equals 6, find the probability of observing a sample mean that is less than or equal to 5.58. (Round your answer to 4 decimal places.) O

Answer 1

a) The probability that the sample average weight of 40 fully loaded delivery trucks is more than 5580 pounds is 0.0843. b) The limits within which 96% of the sample average weights occur are 5481.4 pounds and 5628.6 pounds. c) The probability that the weight of a randomly selected fully loaded delivery truck is between 5450 and 5600 pounds is 0.0847.

Step-by-step explanation:

a) To find the probability that the sample average weight of 40 fully loaded delivery trucks is more than 5580 pounds, we need to calculate the z-score for this value.

The formula for the z-score is z = (x - μ) / (σ / sqrt(n)), where x is the sample average weight, μ is the population mean, σ is the population standard deviation, and n is the sample size. Plugging in the values, we have z = (5580 - 5550) / (220 / sqrt(40)) = 1.3726. Finding the area to the right of this z-score using a z-table (or a calculator), we get a probability of 0.0843.

b) To find the limits within which 96% of the sample average weights occur, we need to find the z-scores that correspond to the 2% and 98% percentiles of the standard normal distribution. The z-score for the 2% percentile is -2.05, and the z-score for the 98% percentile is 2.05.

Using the formula z = (x - μ) / (σ / sqrt(n)), we can calculate the corresponding sample average weights:

lower limit = 5550 + (-2.05) * (220 / sqrt(40)) = 5481.4 pounds and upper limit = 5550 + (2.05) * (220 / sqrt(40)) = 5628.6 pounds.

c) To find the probability that the weight of a randomly selected fully loaded delivery truck is between 5450 and 5600 pounds, we need to calculate the corresponding z-scores.

The z-score for 5450 pounds is (5450 - 5550) / 220 = -0.4545, and the z-score for 5600 pounds is (5600 - 5550) / 220 = 0.2273.

Using the z-table, we find the area to the right of -0.4545 and the area to the right of 0.2273, and then subtract the smaller area from the larger area to get the probability:

P(-0.4545 < z < 0.2273) = P(z > -0.4545) - P(z > 0.2273) = 0.3259 - 0.4090 = 0.0847.