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 4070 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 8050 lb., what is the probability that it will be able to carry the entire lot of 25 appliances?
To solve this question, we can use the concept of the sampling distribution of the sample mean.
(a) To find the probability that the aggregate weight of the 25 appliances will be greater than the truck's capacity of 4070 lb, we need to find the probability that the sample mean weight will be greater than 4070/25 = 162.8 lb.
The mean of the sample mean weight can be calculated as the mean weight of the population, which is given as 306 lb.
The standard deviation of the sample mean weight can be calculated as the standard deviation of the population divided by the square root of the sample size. The standard deviation of the population is the square root of the variance, which is given as sqrt(2209) = 47.0 lb.
Therefore, the standard deviation of the sample mean weight is 47.0 / sqrt(25) = 9.4 lb.
Now, we can use the z-score formula to find the probability:
z = (x - mean) / standard deviation,
where x is the value we want to find the probability for.
In this case, x = 162.8, mean = 306, and standard deviation = 9.4.
z = (162.8 - 306) / 9.4 = -14.4681 (approx)
Using a standard normal distribution table or a calculator, we can find the probability associated with the z-score of -14.4681. The probability can be found as 1 minus the cumulative probability of the z-score.
P(Z > -14.4681) = 1 - P(Z < -14.4681) ≈ 1 (since the z-score is extremely large, the probability is nearly 1)
Therefore, the probability that the 25 appliances will have an aggregate weight greater than the truck's capacity of 4070 lb is approximately 1.
(b) Similarly, to find the probability that the truck will be able to carry the entire lot of 25 appliances with a capacity of 8050 lb, we need to find the probability that the sample mean weight will be less than or equal to 8050/25 = 322 lb.
Using the same mean and standard deviation as in part (a), we can calculate the z-score as follows:
z = (322 - 306) / 9.4 ≈ 1.7021 (approx)
Using the standard normal distribution table or a calculator, we can find the cumulative probability associated with the z-score of 1.7021.
P(Z < 1.7021) ≈ 0.9564
Therefore, the probability that the truck will be able to carry the entire lot of 25 appliances with a capacity of 8050 lb is approximately 0.9564.