The times that a cashier spends processing each person’s transaction are independent and identically distributed random variables with a mean of µ and a variance of σ2. Thus, if Xi is the processing time for each transaction, E(Xi)=μ and Var(Xi)=σ2. Let Y be the total processing time for 100 orders: Y=X1+X2+⋯+X100a) What is the approximate probability distribution of Y, the total processing time of 100 orders?b) Suppose for Z∼N(0, 1), a standard normal random variable: P(a&lt;Z&lt;b)=100(1−α)%. Using your distribution from part (a), show that an approximate 100(1−α)% confidence interval for the unknown population mean μ is: (Y−10bσ100)&lt;μ&lt;(Y−10aσ100)I have done part (a) and ended up with Y∼N(100μ, σ2)using the central limit theorem, but I am unsure of where to start in part (b).

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The times that a cashier spends processing each person’s transaction are independent and identically distributed random variables with a mean of µ and a variance of σ2. Thus, if Xi is the processing time for each transaction, E(Xi)=μ and Var(Xi)=σ2. Let Y be the total processing time for 100 orders: Y=X1+X2+⋯+X100a) What is the approximate probability distribution of Y, the total processing time of 100 orders?b) Suppose for Z∼N(0, 1), a standard normal random variable: P(a&amp;lt;Z&amp;lt;b)=100(1−α)%. Using your distribution from part (a), show that an approximate 100(1−α)% confidence interval for the unknown population mean μ is: (Y−10bσ100)&amp;lt;μ&amp;lt;(Y−10aσ100)I have done part (a) and ended up with Y∼N(100μ, σ2)using the central limit theorem, but I am unsure of where to start in part (b).

xxkrnjangxxed9q · Accepted Answer

Step 1The goal of the exercise to illustrate that you may derive a confidence interval that might not be the standard symmetric one (the usual conventional one that uses x― as a pivot).So starting with some a and b such that P(a&amp;lt;Z&amp;lt;b)=(1−α), we can write(1−α)=P(a&amp;lt;Z&amp;lt;b)=P(a&amp;lt;Y−100μ10σ&amp;lt;b)=P(10aσ&amp;lt;Y−100μ&amp;lt;10bσ)=P(−Y+10aσ&amp;lt;−100μ&amp;lt;−Y+10bσ)=P(Y−10aσ100&amp;gt;μ&amp;gt;Y−10bσ100)=P(Y−10bσ100&amp;lt;μ&amp;lt;Y−10aσ100)As stated at the beginning, this is a general confidence interval not necessarily symmetric. IF one takes a=Φ−1(α2) and b=Φ−1(1−α2), one gets the standard/conventional symmetric confidence interval. But the point is, if the only requirement is the interval having probability (1−α) without requiring symmetry, then there are infinitely many such intervals possible.

stecchiniror7 · Accepted Answer

Step 1If X1,X1,⋯,X100 are a random sample from a distribution with mean μ and variance σ2, then T=∑1=1100 as you say, and (by independence) V(T)=100σ2,. Hence, SD(X―)=σ100,, where X―=Tn=T100.By the Central Limit Theorem, Z=X―−μσ100 should have nearly a standard normal distribution, as you say. Thus, one has0.95≈P(−1.96&amp;lt;Z&amp;lt;1.96)=P(−1.96&amp;lt;X―−μσ100&amp;lt;1.96)=P(X―−1.96σ100,&amp;lt;μ&amp;lt;X―+1.96σ100),where the last inequality is found my manipulating the event in the penultimate term to isolate μ as in the last term. The last term is the basis for saying that X―±1.96σ100 is an approximate 95% confidence interval for μ. [I believe you may have mis-copied the last formula in part(b), and I hope the specific case for α=0.05 helps you put it right and to see how to derive the correct CI.]

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