A lorry carries heavy cartons of goods. The weights of theses cartons are distributed about a mean of 100 kg and with a standard deviation of 7 kg. Find how many cartons the lorry can carry so that the probability of the total load exceeding 4500 kg is less than 0.5.
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Solution
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To find out how many cartons the lorry can carry so that the probability of the total load exceeding 4500 kg is less than 0.5, we can use the concept of the Central Limit Theorem and the Normal Distribution.
Let's break the problem down step by step:
Step 1: Determine the mean and standard deviation of a single carton.
Given that the mean weight of the cartons is 100 kg and the standard deviation is 7 kg, the mean weight of a single carton is μ = 100 kg and the standard deviation is σ = 7 kg.
Step 2: Find the mean and standard deviation of the total load.
Since the lorry will carry multiple cartons, we need to find the mean and standard deviation of the total load. When random variables are summed, the means add up, but the variances (and therefore the standard deviations) add up in absolute terms.
The mean of the total load will be the mean weight of a single carton multiplied by the number of cartons, which we'll represent by N.
Therefore, the mean of the total load is μ_total = μ * N = 100N
The standard deviation of the total load is the square root of the variance of the sum of the individual cartons. Since the standard deviation of a single carton is σ = 7 kg, the variance of a single carton is σ^2 = 7^2 = 49 kg^2.
And since variances add up, the variance of the total load will be the variance of a single carton multiplied by the number of cartons, which we'll again represent by N.
Therefore, the variance of the total load is σ_total^2 = σ^2 * N = 49N.
Taking the square root of the variance gives us the standard deviation of the total load:
σ_total = √(σ_total^2) = √(49N) = 7√N
Step 3: Convert the problem to the standard normal distribution.
To find the probability of the total load exceeding 4500 kg, we need to convert this value to a z-score using the standard normal distribution. The formula for the z-score is:
z = (x - μ) / σ
In this case, x (the value we want to convert) is 4500 kg, μ_total (the mean of the total load) is 100N, and σ_total (the standard deviation of the total load) is 7√N.
Therefore, the z-score is:
z = (4500 - 100N) / (7√N)
Step 4: Find the probability.
Since we want the probability of the total load exceeding 4500 kg, we need to find the area under the curve to the right of the z-score.
Using a standard normal distribution table or a calculator, find the corresponding probability value for the z-score calculated in step 3. Let's call this probability P.
Step 5: Solve for the number of cartons.
Since we want the probability of the total load exceeding 4500 kg to be less than 0.5, we have the inequality:
P < 0.5
Substitute the value of P calculated in step 4 into the inequality and solve for N:
(4500 - 100N) / (7√N) < 0.5
This is a quadratic inequality. Solve it algebraically or graphically to find the range of values for N.
This will give you the number of cartons the lorry can carry so that the probability of the total load exceeding 4500 kg is less than 0.5.