Question

​% 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 ​ (a) exactly​ five, (b) at least​ six, and​ (c) less than four.

Answer 1

Complete Question:

41% 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 ​ (a) exactly​ five, (b) at least​ six, and​ (c) less than four.

Answer:

a) P(exactly 5) = 0.209

b) P(at least six) = 0.183

c) P(less than four) = 0.358

Step-by-step explanation:

Sample size, n = 10

Proportion of adults that have very little confidence in newspapers, p = 41% p = 0.41

q = 1 - 0.41 = 0.59

This is a binomial distribution question:

$P(X=r) = nCr p^{r} q^{n-r}$

a) P(exactly 5)

$P(X=5) = 10C5 * 0.41^{5} 0.59^{10-5}\\P(X=5) = 10C5 * 0.41^{5} 0.59^{10-5}\\P(X=5) = 252 * 0.01159 * 0.072\\P(X=5) = 0.209$

b) P(at least six)

$P(X \geq 6) = P(6) + P(7) + P(8) + P(9) + P(10)$

$P(X\geq6) = (10C6 * 0.41^6*0.59^4) + (10C7*0.41^7*0.59^3) + (10C8*0.41^8*0.59^2) + (10C9 *0.41^9*0.59^1) + (10C10 *0.41^{10})\\P(X\geq6) = 0.1209 + 0.0480 + 0.0125 + 0.0019 + 0.0001\\P(X\geq6) = 0.183$

c) P(less than four)

$P(X < 4) = 1 - [x \geq 4]$

$P(X<4)= 1 - [P(4) + P(5) + P(x \geq 6)]$

$P(X <4)= 1 - [(10C4*0.41^4*0.59^6) + 0.209 + 0.183]\\P(X <4)= 0.358$