Let a random sample of size 100 from a normal population with unknown mean \( \mu \) and variance 9 give the sample mean 5.608. Let \( \Phi(\cdot) \) denote the distribution function of the standard normal random variable. If \( \Phi(1.96) = 0.975 \), \( \Phi(1.64) = 0.95 \), and the uniformly most powerful unbiased test based on sample mean is used to test \[ H_0: \mu = 5.02 \quad \text{against} \quad H_1: \mu \neq 5.02, \] then the p-value equals
\[ f(x) = \begin{cases} e^{-x}, & x>0 \\ 0, & \text{otherwise} \end{cases} \]
Let \( X_{(n)} = \max\{ X_1, X_2, \dots, X_n \} \) for \( n \geq 1 \). If \( Z \) is the random variable to which\[ \{ X_{(n)} - \log n \}_{n \geq 1} \]
converges in distribution, as \( n \to \infty \), then the median of \( Z \) equals\[ \underline{\hspace{2cm}} \]
(round off to 2 decimal places).Let the joint distribution of random variables \( X_1, X_2, X_3 \) and \( X_4 \) be \( N_4(\mu, \Sigma) \), where
Then which one of the following statements is true?
\[ f(x) = \begin{cases} e^{-x}, & x>0 \\ 0, & \text{otherwise} \end{cases} \]
Let \( X_{(n)} = \max\{ X_1, X_2, \dots, X_n \} \) for \( n \geq 1 \). If \( Z \) is the random variable to which\[ \{ X_{(n)} - \log n \}_{n \geq 1} \]
converges in distribution, as \( n \to \infty \), then the median of \( Z \) equals\[ \underline{\hspace{2cm}} \]
(round off to 2 decimal places).