When n=16, the probability P(t>0.25)=[?]

To determine the probability \( P(t > 0.25) \) when \( n = 16 \), we'll assume that \( t \) follows a **t-distribution** with degrees of freedom \( \text{df} = n - 1 = 15 \). Here's a step-by-step explanation: ### Understanding the t-Distribution The t-distribution is symmetric and bell-shaped, similar to the standard normal distribution but with heavier tails. It's commonly used in scenarios where the sample size is small, and the population standard deviation is unknown. ### Calculating \( P(t > 0.25) \) for \( \text{df} = 15 \) 1. **Degrees of Freedom (df):** \[ \text{df} = n - 1 = 16 - 1 = 15 \] 2. **Using the t-Distribution Table:** Most t-distribution tables provide critical values for specific probabilities (e.g., 0.05, 0.025) but not for arbitrary points like 0.25. However, for values of \( t \) close to 0, the t-distribution approaches the standard normal distribution. 3. **Approximating Using the Standard Normal Distribution:** For \( t = 0.25 \) and \( \text{df} = 15 \), the t-distribution is very close to the standard normal distribution. The cumulative distribution function (CDF) for the standard normal at \( z = 0.25 \) is approximately 0.5987. 4. **Calculating the Tail Probability:** \[ P(t > 0.25) = 1 - P(t \leq 0.25) \approx 1 - 0.5987 = 0.4013 \] Given the slight differences between the t-distribution and the standard normal distribution for small \( t \) values and moderate degrees of freedom, this approximation is quite accurate. ### Conclusion \[ P(t > 0.25) \approx 0.401 \] **Therefore, the probability that \( t \) exceeds 0.25 when \( n = 16 \) is approximately 0.401.**