(2) Compute \( \operatorname{Var}(Y) \).

4. Let \( X_1, \dots, X_{64} \) be i.i.d. random variables with \( E(X_i) = 5 \) and \( Var(X_i) = 4 \). Using the Central Limit Theorem, compute the approximation of \( P(\sum_{i=1}^{64} X_i > 346.32) \). (10%)

(1) At significance level \( \alpha = 0.05 \), test \( H_0: p = 0.1 \) vs. \( H_1: p \neq 0.1 \).

(2) Construct a \( 95\% \) confidence interval for \( p \).

II.1. The recruitment channels for a company are as follows: based on past experience, 50% of applicants learned of the openings through job banks, 10% through social media ads, 30% through employee referrals, and 10% through campus recruiting. The one-year turnover rates (the probability of an employee resigning within a year) are as follows: 10% for those from job banks, 20% for those from social media ads, 5% for those from employee referrals, and 30% for those from campus recruiting. If an employee is still with the company after one year, what is the probability that he (she) originally learned about the recruitment information through a job bank? (7%)

2. The annual returns of two ETFs, ETF A and ETF B, both follow a normal distribution, and are independent.ETF A $\sim N[\mu = 20\%, \sigma^2 = (24\%)^2]$; ETF B $\sim N[\mu = 15\%, \sigma^2 = (18\%)^2]$ Calculate the probability that the actual annual return of ETF B is higher than the actual annual return of ETF A. (6%)

(1) The probability of a Type I error. (6%)

(2) The power of the test when the true proportion is \( p = 0.5 \). (6%)

4. Regarding survey responses, people often hesitate to provide an affirmative answer. Suppose that in a population, 80% would truthfully answer "No" to a question, while 20% would truthfully answer "Yes." However, 70% of the individuals lie about their response. Let \( Y \) be the number of people who must be surveyed until the first affirmative ("Yes") response is obtained. Derive the probability distribution of \( Y \). (7%)

(1) The probability density function of \( W \). (12%)

(2) The expected value of \( W \), denoted as \( E(W) \). (6%)

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