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Answer :
1) The probability distribution of X, the weight of a randomly selected shipment box, is normally distributed with parameters μ = 18 lbs and σ = 5.6 lbs. The probability that the weight of a randomly selected shipment box is less than 13 lbs is approximately 0.1515.
2) Using the Central Limit Theorem, the probability distribution of S, the total weight of a sample of 12 randomly selected shipment boxes, is approximately normally distributed with parameters μs = 12 * 18 = 216 lbs and σs = √[tex](12 * (5.6)^2) = 30.768 lbs[/tex]. The probability that the total weight of a sample of 12 randomly selected shipment boxes is less than 197 lbs is approximately 0.1217.
1) For the weight of a randomly selected shipment box X, the probability distribution is given as X ~ N(μ, σ), where μ = 18 lbs and σ = 5.6 lbs.
The probability that X is less than 13 lbs can be found using the standard normal distribution formula:
P(X < 13) = Φ((13 - μ) / σ)
Substituting the given values:
P(X < 13) = Φ((13 - 18) / 5.6)
P(X < 13) = Φ(-0.8929)
Using a standard normal distribution table or a calculator, we find that Φ(-0.8929) ≈ 0.1841.
So, the probability that the weight of a randomly selected shipment box is less than 13 lbs is approximately 0.1841.
2) According to the Central Limit Theorem, the distribution of the sample mean of a large sample size from any population will be approximately normally distributed.
For S, the total weight of a sample of 12 randomly selected shipment boxes, the parameters are:
μs = n * μ = 12 * 18 = 216 lbs
σs = √[tex](n * σ^2)[/tex] = √(12 * [tex](5.6)^2)[/tex] = 30.768 lbs
Now, to find the probability that the total weight of a sample of 12 randomly selected shipment boxes is less than 197 lbs, we standardize the variable S:
Z = (197 - μs) / σs
Z = (197 - 216) / 30.768
Z = -0.6164
Using a standard normal distribution table or a calculator, we find that Φ(-0.6164) ≈ 0.2708.
So, the probability that the total weight of a sample of 12 randomly selected shipment boxes is less than 197 lbs is approximately 0.2708.
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