Question

48% of US adults have very little confidence in newspapers you randomly selected 10 US adults. Find the probability that the number of US adults who have very little confidence in newspapers is exactly 5

Answer 1

Answer:

$P(x = 5) = 0.2441$

Step-by-step explanation:

Given

$p = 0.48$ --- proportion that have very little confidence

$n = 10$ --- sample

Required

Determine the probability that exactly 5 have very little confidence

This question will be solved using binomial probability:

$P(x) = ^nC_xp^x(1-p)^{n-x}$

In this case: x = 5.

So, we have:

$P(x = 5) = ^{10}C_5*0.48^5(1-0.48)^{10-5}$

$P(x = 5) = ^{10}C_5*0.48^5*0.52^5$

$P(x = 5) = 252*0.48^5*0.52^5$

$P(x = 5) = 0.2441$