Question

2. A survey found that 2 out of 5 Americans say he or she has visited a doctor in any given month. If 16 people are selected at random a) Find the probability that 5 or more have visited the doctor last month. b) What is the mean variance and standard deviation of this binomial distribution?

Answer 1

(a) Please note ⁿC_x = n! / [(n-x)!*x!]

Binomial Probability = ⁿC_x * (p)^x * (q)^n-x, where n = number of trials and x is the number of successes.

Here n = 16, p = 2/5 = 0.4, q = 1 – p = 0.6.

P(X 5) = 1 - P(0) + P(1) + P(2) + P(3) + P(4)

P(X = 0) = ¹⁶C₀ * (0.4)⁰ * (0.6)¹⁶ = 0.00028

P(X = 1) = ¹⁶C₁ * (0.4)¹ * (0.6)¹⁵ = 0.00301

P(X = 2) = ¹⁶C₂ * (0.4)² * (0.6)¹⁴ = 0.01505

P(X = 3) = ¹⁶C₃ * (0.4)³ * (0.6)¹³ = 0.04681

P(X = 4) = ¹⁶C₄ * (0.4)⁴ * (0.6)¹² = 0.10142

Therefore P(X 5) = 1 - (0.00028 + 0.00301 + 0.01505 + 0.04681 + 0.10142 ) = 1 - 0.1666 = 0.8334

____________________________________________

(b) Mean = n * p = 16 * 0.4 = 6.4

Standard deviation = Sqrt(n * p * q) = Sqrt(16 * 0.4 * 0.6) = 1.9596

____________________________________________