Question

64% of US adults have very little confidence in newspapers you randomly select 10 US adults find the probability that the number of US adults who have very little confidence in newspapers is a exactly 5 be at least six and C less than four

Answer 1

A) 0.1636

B) 0.7292

C) 0.0305

Explanation:

This is a binomial probability distribution question that follows;

P(X = x) = nCx × p^(x) × q^(n - x)

In this case, p = 64% = 0.64

q = 1 - p = 1 - 0.64 = 0.36

Thus;

A) P(exactly 5) = P(5) = 10C5 × 0.64^(5) × 0.36^(10 - 5) = 0.1636

B) P(at least 6) = P(x ≥ 6) = P(6) + P(7) + P(8) + P(9) + P(10)

P(6) = 10C6 × (0.64^(6)) × (0.36^(10 - 6)) = 0.2424

P(7) = 10C7 × (0.64^(7)) × (0.36^(10 - 7)) = 0.2462

P(8) = 10C8 × (0.64^(8)) × (0.36^(10 - 8)) = 0.1642

P(9) = 10C9 × (0.64^(9)) × (0.36^(10 - 9)) = 0.0649

P(10) = 10C10 × (0.64^(10)) × (0.36^(10 - 10)) = 0.0115

Thus;

P(x ≥ 6) = 0.2424 + 0.2462 + 0.1642 + 0.0649 + 0.0115

P(x ≥ 6) = 0.7292

C) P(x < 4) = P(0) + P(1) + P(2) + P(3)

From online binomial probability calculator, we have;

P(x < 4) = 0.0305