Question

Write a python function average_sample_median(n,P), that return the average of 1000 samples of size n sampled from the distribution P.

* Sample run :

print(average_sample_median(10,[0.2,0.1,0.15,0.15,0.2,0.2]))

print(average_sample_median(10,[0.3,0.4,0.3]))

print(average_sample_median(10,P=[0.99,0.01])

* Expected Output :
3.7855
2.004
1

Answer 1

Here Iam providing the answer.Hope this helps you.If you have any doubts comment me i will clarify you.Please Upvote if you like my work.Hope you will like this.Thank you :- )

We do this problem in 2 ways.Iam providing 2 ways use what you want :- )

1st Way Using Numpy module:

import numpy as np
def average_sample_median(n, P):  
    return np.median( np.random.choice ( np.arange (1, len(P) + 1 ), n, p = P ))
print(average_sample_median(10,[0.2,0.1,0.15,0.15,0.2,0.2]))
print(average_sample_median(10,[0.3,0.4,0.3]))
print(average_sample_median(10,[0.99,0.01]))

2nd Way using random module:-

import random
def average_sample_median(n,P):
    times = 1000
    median_sum = 0
    for i in range(times):
        samples = random.choices(P, k=n)
        sum_p = 0
        for j, p in enumerate(samples):
            sum_p += p
            if sum_p == 0.5:
                median_sum += j + 1.5
                break
            if sum_p > 0.5:
                median_sum += j + 1
                break     
    avg_median = median_sum / times
    return avg_median
print(average_sample_median(10,[0.2,0.1,0.15,0.15,0.2,0.2]))
print(average_sample_median(10,[0.3,0.4,0.3]))
print(average_sample_median(10,[0.99,0.01]))

Output:-