Question

Suppose the approval rating for a local politician is about 42%, and so we choose a...

Suppose the approval rating for a local politician is about 42%, and so we choose a small sample of six (6) citizens randomly to learn more about why they are dissatisfied with the public servant. Assume the local population is large enough that choosing an individual to complete the survey does not affect the probability of selecting a new citizen that supports the politician.

Question 20 What is the probability that 1 or fewer citizens will say that they support the politician?

  1. 28.14%
  2. 20.35%
  3. 19.67%
  4. 91.86%
0 0
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Answer #1

Solution:

Given:

p = probability of the approval rating for a local politician = 0.42

n = sample size = 6

Since the local population is large enough that choosing an individual to complete the survey does not affect the probability of selecting a new citizen that supports the politician , selected sample citizens are independent.

Thus X = Number of citizens will say that they support the politician follows a Binomial distribution with parameters n = 6 and p = 0.42

Binomial probability formula :

P(X = x) =n Cx pul xqN-1

Where

n! nort! x (n-1)!

q = 1 - p = 1 - 0.42 = 0.58

We have to find:the probability that 1 or fewer citizens will say that they support the politician

P(X <1) = ........

P(X <1) = P(X = 0) + P(X = 1)

Thus find: P( X=0) and P( X =1)

P(X=0) =6 Co x 0.420 x 0.586–0

6! P(X = 0) = 0! (6-0)! O! (6_ X1X 0.586

6! P(X=0) = 1 x 6! X1 X 0.0380687

P(X=0) = 1x1 x 0.0380687

P(X =0) = 0.0380687

and

P(X = 1) =6 C1 x 0.421 x 0.586–1

61 P(X = 1) = TM 1! * (6 X 0.42 x 0.585

PY_16! P(X = 1) = 1 x 5! - x 0.42 x 0.0656357

6 x 5! P(X = 1) = 1 5 ! x 0.42 x 0.0656357

P(X = 1) = 6 x 0.42 x 0.0656357

P(X = 1) = 0.1654019

Thus

P(X <1) = P(X = 0) + P(X = 1)

P(X < 1) = 0.0380687 +0.1654019

P(X < 1) = 0.2034706

P(X < 1) = 0.2035

PX<1) = 20.35%

Thus correct answer is: B. 20.35%

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