Question

A nationwide survey of college seniors by the University of Michigan revealed that almost 70% disapprove of daily pot smoking. If 12 seniors are selected at random and asked their opinion, find the probability that the number who disapprove of smoking pot daily is:________. a. no less than 6. b. anywhere from 7 to 9. c. at most 5.

Answer 1

a) P(X≥6) = 0.9614

b) P(7≤ X ≤9) = 0.628

c) P(X≤5) = 0.0386

Explanation:

This can be solved using the binomial distribution formula:

P(X=x) = ⁿCₓ pˣ qⁿ⁻ˣ

Where p = probability of success

q = probability of failure = 1-p

n = number of trials

x = number of successful trials

We have p = 70% = 0.7

n = 12

a) Find the probability that the number who disapprove of smoking pot daily is no less than 6 i.e. P(X≥6). This can be calculated as:

P(X≥6) = 1 - P(X<6)

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

= 1 - [¹²C₀ (0.7)⁰(0.3)¹² + ¹²C₁ (0.7)¹(0.3)¹¹ + ¹²C₂ (0.7)²(0.3)¹⁰ + ¹²C₃ (0.7)³(0.3)⁹ + ¹²C₄ (0.7)⁴(0.3)⁸ + ¹²C₅ (0.7)⁵(0.3)⁷]

= 1 - (0.000000531 + 0.0000148 + 0.0001909 + 0.00148 + 0.00779 + 0.02911)

= 1 - 0.0386

P(X≥6) = 0.9614

b) P(7≤ X ≤9) = P(X=7) + P(X=8) + P(X=9)

= ¹²C₇ (0.7)⁷(0.3)¹²⁻⁷ + ¹²C₈ (0.7)⁸(0.3)¹²⁻⁸ + ¹²C₉ (0.7)⁹(0.3)¹²⁻⁹

= 0.158 + 0.231 + 0.239

P(7≤ X ≤9) = 0.628

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

= ¹²C₀ (0.7)⁰(0.3)¹² + ¹²C₁ (0.7)¹(0.3)¹¹ + ¹²C₂ (0.7)²(0.3)¹⁰ + ¹²C₃ (0.7)³(0.3)⁹ + ¹²C₄ (0.7)⁴(0.3)⁸ + ¹²C₅ (0.7)⁵(0.3)⁷

P(X≤5) = 0.0386