import numpy as np

i = input()

n,m = list(map(int,i.split(' ')))

arr = np.array([list(map(int,input().split(' '))) for _ in range(n)])

print(np.mean(arr,axis=1)) print(np.var(arr,axis=0)) print(np.round(np.std(arr,axis=None),11))

For Python3 Platform

It isn't mentioned to round off the standard deviation to 11 decimals. I did that just to match with the given answer.

from numpy import array, mean, var, std

arr = []
n, m = map(int, input().split())

for _ in range(n):
    arr.append([*map(int, input().split())])
arr = array(arr)

print(mean(arr, axis=1))
print(var(arr, axis=0))
print(round(std(arr), 11))

I did two things to make it work for all the test cases:

1) commented out printoptions that were recommended in the previous examples, but were not needed in this question

2) added rounding to 11th decimal place to calculate std: print(numpy.round(numpy.std(arr, axis = None),11))

Hope thois helps

import numpy as np

i, j = map(int, input().split())

arr = np.array([list(map(int, input().split())) for _ in range(i)])

print(np.mean(arr, axis=1)) print(np.var(arr, axis=0))

std_value = np.std(arr, axis=None) if std_value == 0: print("0.0") else: print(f"{std_value:.11f}")