The probability that X₁ will exceed 9 ounces is closest to A)0.3085 B)0.1587 C)0.1056 D)0.0256 E)0.0100

Question

Quizplus · Accepted Answer

We can use the Central Limit Theorem to approximate the distribution of the sum of the weights. Since each weight is uniformly distributed between 0 and 1 ounce, the expected value of one weight is E(X) = 0.5 and the variance is Var(X) = (1-0)^2/12 = 1/12. The sum of n weights has expectation E(S_n) = nE(X) = 0.5n and variance Var(S_n) = nVar(X) = n/12. Since we have n = 8 weights being added, the distribution of the sum of the weights can be approximated by a normal distribution with mean μ = 4 and variance σ^2 = n/12 = 2/3. To find the probability that X₁ exceeds 9 ounces, we can standardize the distribution: P(X₁ > 9) = P((X₁ - μ)/σ > (9 - μ)/σ) = P(Z > (9-4)/√(2/3)) where Z is a standard normal random variable = P(Z > 2.04) Using a standard normal table or calculator, we find that P(Z > 2.04) is approximately 0.0206, which is closest to choice (D) 0.0256.

The Probability That X1 Will Exceed 9 Ounces Is Closest

The Probability That X₁ Will Exceed 9 Ounces Is Closest