Question

. A nationwide survey of college seniors by the University of Michigan revealed that almost 70% disapprove of daily pot smoking, according to a report in Parade. Suppose 12 seniors are selected at random and asked their opinion. Our goal is to investigate the number of pot smokers. (a) Verify that this is an example of a binomial experiment. (b) What is the probability that exactly one of the 12 seniors disapproves daily pot smoking

Answer 1

Answer:

a) Trials are independent, and there is a fixed number of trials, which means that the binomial probability distribution can be used.

b) 0.000015 = 0.0015% probability that exactly one of the 12 seniors disapproves daily pot smoking

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

a)

For each senior, there are only two possible outcomes. Either they disapprove daily pot smoking, or they do not. The seniors are selected at random, which means that the probability of one disapproving daily pot smoking is independent of other seniors, that is, the trials are independent. There is also a fixed number of trials.

b)

70% disapprove of daily pot smoking, according to a report in Parade, which means that $p = 0.7$

12 seniors means that $n = 12$

The probability is P(X = 1).

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

$P(X = 1) = C_{12,1}.(0.7)^{1}.(0.3)^{11} = 0.000015$

0.000015 = 0.0015% probability that exactly one of the 12 seniors disapproves daily pot smoking