The weight of a randomly selected shipping container follows an unknown distribution with mean 12 tons and standard deviation 2 tons. Your cargo ship always carries a load of 100 randomly selected shipping containers.
a. What is the probability the load weighs more than 1,231 tons?
b. What is the 90th percentile of load weights?
a. To find the probability that the load of 100 shipping containers weighs more than 1,231 tons, we can use the Central Limit Theorem. According to the Central Limit Theorem, the sum (or average) of a large number of independent and identically distributed random variables will be approximately normally distributed, regardless of the shape of the original distribution.
In this case, we have a large number of shipping containers (100) and assume that they are independent and identically distributed. The weight of each shipping container has a mean of 12 tons and a standard deviation of 2 tons. The sum of the weights of the 100 containers will also follow a normal distribution.
First, we need to calculate the mean and standard deviation of the sum of the weights of the 100 containers.
The mean of the sum of the weights (mu) is equal to the number of containers (n) multiplied by the individual mean (12 tons):
mu = n * mean = 100 * 12 = 1200 tons.
The standard deviation of the sum of the weights (sigma) is equal to the square root of the number of containers (n) multiplied by the individual standard deviation (2 tons):
sigma = sqrt(n) * stdev = sqrt(100) * 2 = 20 tons.
Now, we can standardize the value of 1,231 tons to find the z-score using the formula:
z = (x - mu) / sigma.
z = (1231 - 1200) / 20 = 31 / 20 = 1.55.
To find the probability that the load weighs more than 1,231 tons, we can use a standard normal distribution table (also known as the z-table) or a statistical software to find the area to the right of the z-score of 1.55.
By looking up the z-score in the z-table, we find that the area to the right of 1.55 is approximately 0.0606.
Therefore, the probability that the load weighs more than 1,231 tons is approximately 0.0606 or 6.06%.
b. The 90th percentile of load weights represents the weight below which 90% of the loads fall. To find this percentile, we need to find the z-score that corresponds to the 90th percentile of the standard normal distribution.
Using the standard normal distribution table (z-table) or statistical software, we can find the z-score that corresponds to the 90th percentile. The z-score corresponding to the 90th percentile is approximately 1.28.
Now, we can use the z-score formula to find the actual value at the 90th percentile:
x = z * sigma + mu.
x = 1.28 * 20 + 1200 = 25.6 + 1200 = 1225.6 tons.
Therefore, the 90th percentile of load weights is approximately 1,225.6 tons.
To solve this problem, we can use the properties of the normal distribution since the sample size (100 containers) is large enough.
a. To find the probability that the load weighs more than 1,231 tons, we first need to standardize the value using the z-score formula:
z = (x - μ) / σ
where x is the value we want to standardize, μ is the mean, and σ is the standard deviation. In this case, x = 1,231 tons, μ = 12 tons, and σ = 2 tons.
Now, let's calculate the z-score:
z = (1,231 - 12) / 2 = 609.5
Since we are interested in the probability that the load weighs more than 1,231 tons, we need to find the area under the normal curve to the right of this z-score. We can use a standard normal table or a calculator to find the corresponding probability.
P(X > 1,231) ≈ 1 - P(Z ≤ 609.5)
Note that the probability is very close to 1 since the value is extremely large. Anyway, the probability is practically 1.
b. The 90th percentile represents the value below which 90% of the load weights fall. To find the 90th percentile, we need to find the z-score that corresponds to the 90th percentile and then convert it back to the original scale.
We can use the inverse normal distribution function (also known as the probit function) to find the z-score corresponding to a given percentile. In this case, we are interested in the 90th percentile, which means 90% of the load weights fall below a certain value.
Z = invNorm(0.90) ≈ 1.282
Now, we can convert this z-score back to the original scale using the formula:
x = μ + Z * σ
x = 12 + 1.282 * 2 = 14.564
Therefore, the 90th percentile of the load weights is approximately 14.564 tons.