I used python with the code below. Not sure what is wrong

T4-HW4.7. Floating Point and Softmax The softmax function normalizes a vector so its entries are positive and sum to 1. This is useful for creating a probability distribution across the vector, so it is commonly used as the final activation layer for neural networks (especially since the gradient of the softmax function is relatively easy to compute). The softmax function is defined as follows, for a vector z ERD where D is the length of the vector: ez; 0(z); for j = 1, ...,D D i=1 ez In essence, the numerator scales each term of the vector to some non-negative value, and then the denominator normalizes the vector to sum up to 1. Let z= [20 85.19757 86.82316 86.15644] Assume z is represented in single-precision floating-point. When calculating the result of the softmax function on z, what is the maximum value that zo can take to avoid overflow in the calculations? 20 = 88.588 x 0% Correct answer 20 = 88.427 Submitted answer incorrect: 0% Submitted at 2021-02-15 10:28:52 (CST) 20 = 88.588 X 0% In [1]: import numpy as np In [2] : max_val np.finfo(np. float32).max In [15]: N 1 = np.array([85.19757, 84.82316, 86.15644], dtype=np.float32) j = max_val - np.sum (np.exp(1)) print("The maximum permissible value of zo is " + str(np.log(j))) The maximum permissible value of zo is 88.587585 In [ ]: N