Question

# 66​% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults....

66​% 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.

X ~ Binomial (n,p)

Where n = 10 , p = 0.66

Binomial probability distribution is

P (X) = nCx px (1 - p)n-x

a)

P (X = 5) = 10C5 0.665 0.345

= 0.1434

b)

P( X >= 6) = P (X = 6) + P (X = 7) + P (X = 8) + P (X = 9) + P (X = 10)

=  10C6 0.666 0.344 +10C7 0.667 0.343 +10C8 0.668 0.342 +10C9 0.669 0.34 +10C10 0.6610 0.340

= 0.7730

c)

P( X < 4) = P (X <= 3)

= P (X = 0) +P (X = 1) +P (X = 2) +P (X = 3)

= 10C0 0.660 0.3410 +10C1 0.661 0.349 +10C2 0.662 0.348 +10C3 0.663 0.347

= 0.0220

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