Answer to Question #345223 in Statistics and Probability for harry

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. 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) anywhere from 7 to 9;

(b) at most 5;

(c) not less than 8

Answer 1

Let X be the random variable denoting number of college seniors who disapprove of smoking pot daily from the sample of 12.

So, clearly X ~ Binomial (12, 0.7).

The probability that the number of

people who disapprove of smoking pot daily

is

a) Anywhere from 7 to 9

Required probability is P(X=7) + P(X=8) + P(X=9)=

$=\frac{12!}{7!5!}*0.7^7*0.3^5+\frac{12!}{8!4!}*0.7^8*0.3^4+\frac{12!}{9!3!}*0.7^9*0.3^3=\\=0.63$

b) At most 5

Required probability is P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)=

$=0.3^{12}+\frac{12!}{11!}*0.7*0.3^{11}+\frac{12!}{10!2!}*0.7^2*0.3^{10}+\frac{12!}{3!9!}*0.7^3*0.3^9+\frac{12!}{4!8!}*0.7^4*0.3^8+\frac{12!}{7!5!}*0.7^5*0.3^7=\\=0.0386$

c) Not less than 8

Required probability is P(X=12) + P(X=11) + P(X=10) + P(X=9) +

P(X=8)

$=0.7^{12}+\frac{12!}{11!}*0.7^{11}*0.3+\frac{12!}{2!10!}*0.7^{10}*0.3^2+\frac{12!}{9!3!}*0.7^9*0.3^3+\frac{12!}{8!4!}*0.7^8*0.3^4=\\=0.724$