Question

62% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the number of U.S. adults who have very little confidence in newspapers is %E2%80%8B (a) exactly%E2%80%8B five, (b) at least%E2%80%8B six, and%E2%80%8B (c) less than four.

Answer 1

a) 0.1829

b) 0.6823

c) 0.0413

Explanation:

We are given the following information:

We treat adult having little confidence in the newspaper as a success.

P(Adult have little confidence) = 62% = 0.62

Then the number of adults follows a binomial distribution, where

`P(X=x) = \binom{n}{x}.p^x.(1-p)^(n-x)`

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 10

a) exactly 5

`P(x =5)\\\\=\binom{10}{5}(0.62)^5(1-0.62)^5\\\\=0.1829`

0.1829 is the probability that exactly 5 out of 10 U.S.adults have very little confidence in newspapers.

b) atleast six

`P(x \geq 6)\\\\ = P(x = 6) + P(x = 7)+P(x = 8)+P(x = 9)+P(x=10)\\\\= \displaystyle\sum \binom{10}{n}(0.62)^n(1-0.62)^(10-n), n =6,7,8,9,10\\\\= 0.6823`

0.6823 is the probability that atleast 6 out of 10 U.S. adults have very little confidence in newspapers.

c) less than four

`P(x <4)\\\\ = P(x = 0) + P(x = 1)+P(x = 2)+P(x = 3)\\\\= \displaystyle\sum \binom{10}{n}(0.62)^n(1-0.62)^(10-n), n =0,1,2,3\\\\= 0.0413`

0.0413 is the probability that less than 4 out of 10 U.S. adults have very little confidence in newspapers.