A company delivers packages, and the delivery time (𝑇)(T) for a package follows an exponential distribution with a mean of 30 minutes. Suppose the company delivers 𝑛n packages in a day. What is the minimum number of packages 𝑛n required to ensure that the probability that the average delivery time (𝑇‾)( T ) for the day lies between 28 and 32 minutes is at least 0.95? Use CLT to approximate.
Question
A company delivers packages, and the delivery time (𝑇)(T) for a package follows an exponential distribution with a mean of 30 minutes. Suppose the company delivers 𝑛n packages in a day. What is the minimum number of packages 𝑛n required to ensure that the probability that the average delivery time (𝑇‾)( T ) for the day lies between 28 and 32 minutes is at least 0.95? Use CLT to approximate.
Solution
To solve this problem, we will use the Central Limit Theorem (CLT), which states that the sum of a large number of independent and identically distributed random variables will be approximately normally distributed.
Step 1: Identify the given information The mean (μ) of the exponential distribution is 30 minutes. The standard deviation (σ) of an exponential distribution is the same as the mean, so σ = 30 minutes. We want to find the minimum number of packages (n) such that the probability that the average delivery time lies between 28 and 32 minutes is at least 0.95.
Step 2: Standardize the bounds We standardize the bounds by subtracting the mean and dividing by the standard deviation. For the lower bound (28 minutes), the standardized value is (28-30)/sqrt(30/n) = -2/sqrt(30/n). For the upper bound (32 minutes), the standardized value is (32-30)/sqrt(30/n) = 2/sqrt(30/n).
Step 3: Use the standard normal distribution We want the probability that the standardized value lies between -2/sqrt(30/n) and 2/sqrt(30/n) to be at least 0.95. Looking up these values in the standard normal distribution table, we find that the z-score for a probability of 0.95 is approximately 1.96.
Step 4: Solve for n We set up the following inequality: -1.96 ≤ 2/sqrt(30/n) ≤ 1.96. Solving for n, we get n ≥ (2*1.96)^2 * 30 = 230.4.
Step 5: Round up to the nearest whole number Since we can't deliver a fraction of a package, we round up to the nearest whole number. Therefore, the minimum number of packages n required to ensure that the probability that the average delivery time for the day lies between 28 and 32 minutes is at least 0.95 is 231.
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