​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.

Question
Asked Nov 19, 2019
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​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.
 
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Expert Answer

Step 1

Given binomial parameters,

P = 41% =0.41 and n = 10.

Let ‘X’ be a binomial distribution random variable.

Formula for binomial probability is,

P" (1-P) ба)
Р(X-х)-
X
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P" (1-P) ба) Р(X-х)- X

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Step 2

Now,

(a)

P (Exactly five),

Ps (1-P)(x)
Р(X-х)%
х
10
PCX-5-(0.41) (1-0.41))
(0.41) (1-0.41)9)
0.28072760148
P(X 5) 0.2807
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Ps (1-P)(x) Р(X-х)% х 10 PCX-5-(0.41) (1-0.41)) (0.41) (1-0.41)9) 0.28072760148 P(X 5) 0.2807

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Step 3

(b)

P (at least ...

P(X 26){P(X-6)+P(X-7)+P(X-8)+P(X=9)+P(X=10)}
10
10
(0.41(1-0.41)0 (041) (1-0.41)041(1-041)
041(1-0.41) 10 (0.41)"(041)9
10-7
6
7
8
10
10-9
9
0.18344522029
P(X 2 6 0.1834 for 4 decimal places
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P(X 26){P(X-6)+P(X-7)+P(X-8)+P(X=9)+P(X=10)} 10 10 (0.41(1-0.41)0 (041) (1-0.41)041(1-041) 041(1-0.41) 10 (0.41)"(041)9 10-7 6 7 8 10 10-9 9 0.18344522029 P(X 2 6 0.1834 for 4 decimal places

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