Question

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.

Answer #1

X ~ Binomial (n,p)

Where n = 10 , p = 0.66

Binomial probability distribution is

P (X) = ^{n}C_{x} p^{x} (1 -
p)^{n-x}

a)

P (X = 5) = ^{10}C_{5} 0.66^{5}
0.34^{5}

= **0.1434**

b)

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

= ^{10}C_{6} 0.66^{6}
0.34^{4} +^{10}C_{7} 0.66^{7}
0.34^{3} +^{10}C_{8} 0.66^{8}
0.34^{2} +^{10}C_{9} 0.66^{9} 0.34
+^{10}C_{10} 0.66^{10} 0.34^{0}

= **0.7730**

c)

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

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

= ^{10}C_{0} 0.66^{0} 0.34^{10}
+^{10}C_{1} 0.66^{1} 0.34^{9}
+^{10}C_{2} 0.66^{2} 0.34^{8}
+^{10}C_{3} 0.66^{3} 0.34^{7}

= **0.0220**

59% 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.

47% 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.

61% 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.
(a) P(5) = ?
(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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less than eight
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