contestada

44​% 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.

Respuesta :

ogorwyne

The probability at the various points are

  • 0.02886
  • 0.2407
  • 0.2877

This is a binomial progression where

n = 10 persons

Probability = 44 percent

The probability of exactly five persons

P (X = 5) = 10C5x0.44⁵x 0.56⁵

= 252x0.01649x0.055

= 0.02886

2. At least six adults

10C6x0.44⁶x 0.56⁴ +10C7 x0.44⁷x 0.56³ +10C8 x0.44⁸ 0.56² +10C9 x0.44⁹ 0.56 +10C10x 0.44¹⁰x 0.56⁰

= 0.2407

3. Probability of less than 4

10C0 x 0.44⁰x 0.56¹⁰ + 10C1x 0.44¹x 0.56⁹ +10C2x 0.44²x 0.56⁸ +10C3 0.44³x 0.56⁷

= 0.2877

Read more on binomial progression here: https://brainly.com/question/13602562

Otras preguntas