Final answer:
To find the smallest value of n such that P(Y(n) ≥ 0.99) ≥ 0.95, calculate the cumulative distribution function of Y(n) for different values of n and find the smallest value of n that satisfies the condition.
Step-by-step explanation:
To find the smallest value of n such that P(Y(n) ≥ 0.99) ≥ 0.95, we need to determine the distribution of Y(n) and calculate the probability.
Y(n) represents the maximum value of n random samples from the uniform distribution on the interval [0, 1]. Since the uniform distribution is continuous, the probability that Y(n) is greater than or equal to a certain value can be calculated using the cumulative distribution function (CDF).
In this case, we want to find n such that P(Y(n) ≥ 0.99) ≥ 0.95. This means we need to find the smallest value of n for which the cumulative distribution function of Y(n) at 0.99 is greater than or equal to 0.95.
By using the probability distribution of the maximum of n random samples from a uniform distribution, we can calculate the cumulative distribution function and find the smallest value of n that satisfies the given condition. The specific calculation involves finding the value of n for which the cumulative distribution function at 0.99 is equal to or greater than 0.95.
It requires mathematical calculations and statistical knowledge to solve this problem. I would recommend consulting a textbook or seeking help from a math tutor for a more detailed explanation and step-by-step solution.