Question

Sixty percent of U.S. adults trust national newspapers to present the news fairly and accurately. You randomly select nine U.S. adults. Find the probability that the number of U.S. adults who trust national newspapers to present the news fairly and accurately is: a. exactly five. b. at least six c. less than four.

Answer 1

Answer:

a) 0.251

b) 0.483

c) 0.0994

Step-by-step explanation:

We are given the following information:

We treat adult adults trusting national newspapers to present the news fairly and accurately as a success.

P(Adult trust) = 60% = 0.6

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 = 9

a) exactly five

$P(x = 15) = \binom{9}{5}(0.6)^{5}(1-0.6)^{9-5} = 0.2508 \approx 0.251$

b) at least six

We have to evaluate

$P(x \geq 6) = P(x = 6) + P(x = 7) + P(x=8) + P(x=9) \\= \binom{9}{6}(0.6)^6(1-0.6)^3 + \binom{9}{7}(0.6)^7(1-0.6)^2 + \binom{9}{8}(0.6)^8(1-0.6)^1 +\binom{9}{9}(0.6)^9(1-0.6)^0\\= 0.482609 \approx 0.483$

c) less than four

We have to evaluate

$P(x < 4) = P(x = 0) + P(x = 1) + P(x=2) + P(x=3) \\= \binom{9}{0}(0.6)^0(1-0.6)^9 + \binom{9}{1}(0.6)^1(1-0.6)^8 + \binom{9}{2}(0.6)^2(1-0.6)^7 +\binom{9}{3}(0.6)^3(1-0.6)^6\\= 0.099352 \approx 0.0994$