A trucking firm delivers appliances for a large retail operation. The packages (or crates) have a mean weight of 306 lb. and a variance of 2209. (Give your answers correct to four decimal places.)

(a) If a truck can carry 3940 lb. and 25 appliances need to be picked up, what is the probability that the 25 appliances will have an aggregate weight greater than the truck's capacity? Assume that the 25 appliances represent a random sample.


(b) If the truck has a capacity of 7810 lb., what is the probability that it will be able to carry the entire lot of 25 appliances?


I will try and work if I just knew the correct formula.

To solve these problems, we can use the central limit theorem and the properties of the normal distribution.

(a) To find the probability that the 25 appliances will have an aggregate weight greater than the truck's capacity, we need to calculate the probability of the sample mean exceeding the truck's capacity.

First, we need to find the mean and standard deviation of the sample mean. The mean of the sample mean is the same as the mean of the individual appliances, which is given as 306 lb.

The variance of the sample mean can be calculated by dividing the variance of the individual appliances by the sample size. In this case, the variance is given as 2209 lb^2, and the sample size is 25. So, the variance of the sample mean is 2209/25 = 88.36 lb^2.

Now, we can calculate the standard deviation of the sample mean by taking the square root of the variance: √(88.36) ≈ 9.397 lb.

Next, we standardize the truck's capacity using the sample mean and standard deviation. The standardized value (Z-score) is calculated as (truck's capacity - sample mean) / standard deviation. In this case, the truck's capacity is 3940 lb. So, the Z-score is (3940 - 306) / 9.397 ≈ 369.53.

Finally, we use a standard normal distribution table or calculator to find the probability that the sample mean exceeds the truck's capacity. Since the Z-score is very large, the probability is essentially zero.

(b) To find the probability that the truck will be able to carry the entire lot of 25 appliances, we need to calculate the probability of the sample mean being less than or equal to the truck's capacity.

Using the same mean and standard deviation of the sample mean as in part (a), we standardize the truck's capacity: (7810 - 306) / 9.397 ≈ 759.62.

Again, we use a standard normal distribution table or calculator to find the probability that the sample mean is less than or equal to the truck's capacity. Since the Z-score is very large, the probability is essentially one.

So, the probability that the truck will be able to carry the entire lot of 25 appliances is very close to 1, or 100%.

Remember to use a standard normal distribution table or calculator to find the probabilities associated with the Z-scores.