Final answer:
To find P(0 < x < 4) given a continuous probability density function f(x)=4x^3 defined from 0 to 2, we actually look at the entire probability of the defined range, which is 1. Due to possible typos in the question, we've assumed the function's upper bound is 2 as beyond that the probability is 0, and the integral from 0 to 2 for a valid probability density function is 1.
Step-by-step explanation:
To answer the student's question: Given the probability density function f(x)=4x3 over the interval 0 ≤ x ≤ 2, what is P(0 < x < 4)? We need to first recognize that the function provided cannot be the correct probability density function over the interval 0 ≤ x ≤ 4 as it was initially stated for the interval 0 ≤ x ≤ 2. This appears to be a typographical error in the question. Assuming the student meant the interval from 0 to 2, we will approach the question accordingly.
To find the probability that x is between 0 and 4, given a continuous probability distribution, we need to calculate the area under the probability density function from 0 to 4. However, since our function is defined only up to 2, and the values for a probability density function must be between 0 and 1, we face two issues. First, the probabilities for values beyond the defined range (2 to 4 in this case) are 0. Second, the area under the curve of a probability density function from 0 to its upper bound (in this case, 2) is 1.
Therefore, the correct approach to find the probability for the range 0 < x < 4 when the function is only defined up to x = 2 is to realize that P(0 < x < 4) is equivalent to the total probability of the defined interval, which is P(0 < x ≤ 2), and hence it equals 1. So, if the provided interval was indeed from 0 to 4, the probability sought (P(0 < x < 4)) would be 1 as the entire range of the function is being considered.