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?

Hello. I managed to solve (a)

Expected aggregate weight = 300 x 25 = 7500 lbs

Thus,

Since 7500 lbs is over the maximum capacity of 4000lbs,

The probability is 1.00
(It will definitely be 100% over the capacity)

I don't know how to solve (b) though. If you do, do share with me,

To find the probability in both parts (a) and (b), we need to use the concept of the sampling distribution of the sample mean.

The mean of the sample mean, denoted by μx̄, is equal to the population mean, denoted by μ. In this case, the population mean is 306 lb.

The standard deviation of the sample mean, denoted by σx̄, is equal to the population standard deviation divided by the square root of the sample size. In this case, the population standard deviation is the square root of the variance, which is √2209 lb, and the sample size is 25.

(a) To find the probability that the 25 appliances will have an aggregate weight greater than the truck's capacity of 3940 lb, we need to calculate the probability of the sample mean exceeding 3940 lb.

First, we need to calculate the standard deviation of the sample mean:
σx̄ = √(variance / sample size) = √(2209 / 25) = 9.354 lb

Next, we can calculate the z-score, which is the number of standard deviations the sample mean is away from the population mean:
z = (3940 - μ) / σx̄ = (3940 - 306) / 9.354 = 404.548

Using a z-table, we can find the probability corresponding to this z-score:
P(x > 3940) = 1 - P(x ≤ 3940)
= 1 - P(z ≤ 404.548)
= 1 - 1
= 0

Therefore, the probability that the 25 appliances will have an aggregate weight greater than the truck's capacity is 0.

(b) To find the probability that the truck will be able to carry the entire lot of 25 appliances with a capacity of 7810 lb, we need to calculate the probability of the sample mean being less than or equal to 7810 lb.

First, we need to calculate the z-score:
z = (7810 - μ) / σx̄ = (7810 - 306) / 9.354 = 757.452

Using a z-table, we can find the probability corresponding to this z-score:
P(x ≤ 7810) = P(z ≤ 757.452) ≈ 1

Therefore, the probability that the truck will be able to carry the entire lot of 25 appliances is approximately 1.