Question

53% of U.s adults have very little confidence in newspapers, you randomly select 10 U.s adults who have very little confidence in newspapers is (a) five, (b) at least six and (c) less than four
a) p(5)=

Answer 1

Solution:

Given,

p = 53% = 0.53

1 - p = 1 - 0.53= 0.47

n = 10

X follows the Binomial(10 , 0.53)

Using binomial probability formula ,

P(X = x) = (_n C _x) * p^x * (1 - p)^{n - x} ; x = 0 ,1 , 2 , ....., n

a)

P(X = 5) = (₁₀C ₅) * 0.53⁵ * (0.47)^{10 - 5}

= 0.24169584164

P(X = 5) = 0.24169584164

b)

P(At least 6)

= P(X 6)

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

= (₁₀C ₆) * 0.53⁶* (0.47)^{10 - 6} + (₁₀C ₇) * 0.53⁷ * (0.47)^{10 - 7}  + (₁₀C ₈) * 0.53⁸ * (0.47)^{10 - 8}  + (₁₀C ₉) * 0.53⁹ * (0.47)^{10 - 9} + (₁₀C ₁₀) * 0.53¹⁰ * (0.47)^{10 - 10}  

=  0.22712552494 + 0.14635444161 + 0.06188924525 + 0.01550888888 + 0.0017488747

= 0.45262697538

P(At least 6) = 0.45262697538

c)

P(Less than 4)

= P(X < 4)

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

=  (₁₀C ₀) * 0.536⁰* (0.47)^{10 - 0} + (₁₀C ₁) * 0.53¹ * (0.47)^{10 - 1} + (₁₀C ₂) * 0.53² * (0.47)^{10 - 2} + (₁₀C ₃) * 0.53³* (0.47)^{10 - 3}

= 0.00052599132 + 0.00593139151 + 0.0300986569 + 0.09050943637

= 0.1270654761

P(Less than 4) = 0.1270654761