Question

In a typical day, 61% of U.S. adults go online to get news. You randomly select five U.S. adults. Find the probability that the number of U.S. adults who say they go online to get news is

a) Exactly 2.

b)At least two.

c)More than two

Answer #1

Solution

Given that ,

p = 0.61

q = 1 - p = 1 - 0.61=0.39

n = 5

Using binomial probability formula ,

(A)P(X = x) = (n C x) * p x * (1 - p)n - x

P(X = 2) = (5 C 2) * 0.61 2 * (0.39)3

=0.2207

probability0.2207

(B)

P(X 2 ) = 1 - P( x <2)

= 1 - P(X = 0) - P(X = 1)

= 1 -(5 C 0) * 0.61 0 * (0.39)5-(5 C 1) * 0.61 1 * (0.39)4

=1-0.0796

probability=0.9204

(C)

P(X > 2) = 1 - P(X 2)

= 1 - P(X = 0) - P(X = 1) - P(X = 2)

= 1 - (5 C 0) * 0.61 0 * (0.39)5-(5 C 1) * 0.61 1 * (0.39)4 -(5 C 2) * 0.61 2 * (0.39)3

= 1 - 0.3003

= 0.6997

Probability = 0.6997

61% of U.S. adults have very little confidence in newspapers.
You randomly select 10 U.S. adults. Find the probability that the
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(Round to three decimal places as needed.)
(b) P(x≥6) = ?
(Round to three decimal places as needed.)
(c) P(x<4) = ?
(Round to three decimal places as needed.)

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a P(5) round to 3 decimal place

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