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Question 1a) Suppose you sample 100 times at random with replacement from a population in which 26% of the individuals are successes. Write a Python expression that evaluates to the chance that the sample has 20 successes.

Sagot :

fichoh

Answer:

from math import comb

n = 100

x = 20

p = 0.26

q = 0.76

print(comb(n, x)*(p**x)*(q**(n-x)))

Step-by-step explanation:

Given that :

Number of trials, n = 100

P(success), p = 26% = 0.26

P(success)' = 1 - p = 1 - 0.26 = 0.74

Chance that sample has 20 successes = x

This problem meets the condition for a binomial probability distribution :

p(x = 20)

Recall :

P(x = x) = nCx * p^x * q^(n-x)

Using python :

comb is an built in function which calculate the combination of two arguments it takes ; and returns the combination value.

** mean raised to the power and

* is used for multiplication

The Python code as per the given question is provided below.

Program explanation:

The number of trials,

  • 100

Probability of success,

  • 20% or 0.26

Size of array generated,

  • 2000

The output that shows chances of 20 success,

  • S

Program code:

import numpy as np

S=sum(np.random.binomial(100,0.26,2000)==20)/2000

S

Learn more about Python expression here:

https://brainly.com/question/21645017

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