Question

According to the 2016 National Survey on Drug Use and Health (NSDUH), 57.2% of full-time college students ages 18 to 22 drank alcohol in the past month. What is the probability that in a sample of 20 full-time college students in this age bracket that 8 have drank alcohol in the past month?

Suppose that the probability a lost microchipped dog is returned to its owner is 0.53. What is the probability that in a sample of 30 lost microchipped dogs that less than 12 will be returned to their owner?

According to the Centers for Disease Control approximately 16% of Americans (18 and over) smoke cigarettes. In a sample of 50 adult Americans

What is the mean number that smoke cigarettes? Blank 1

What is the standard deviation? Blank 2 Round to 1 decimal place.

Would be unusual if 12 were found to be smokers in this sample? Blank 3 Choose yes or no.

Answer 1

Question)

According to the 2016 National Survey on Drug Use and Health (NSDUH), 57.2% of full-time college students ages 18 to 22 drank alcohol in the past month. What is the probability that in a sample of 20 full-time college students in this age bracket that 8 have drank alcohol in the past month?

Answer)

As here there are fixed number of trials and the probability of each and every trial is same and independent of each other

Here we need to use the binomial formula

P(r) = ncr*(p^r)*(1-p)^n-r

Ncr = n!/(r!*(n-r)!)

N! = N*n-1*n-2*n-3*n-4*n-5........till 1

For example 5! = 5*4*3*2*1

Special case is 0! = 1

P = probability of single trial = 0.572 (57.2%)

N = number of trials = 20

R = desired success = 8

After substitution

Required probability is

P(8) = 0.05454570858

Question )

Suppose that the probability a lost microchipped dog is returned to its owner is 0.53. What is the probability that in a sample of 30 lost microchipped dogs that less than 12 will be returned to their owner?

Answer)

N = 30

P = 0.57

R = less than 12

P(0) + P(1) + P(2) + P(3) + P(4).....+P(11) = 0.05361922324

Question)

According to the Centers for Disease Control approximately 16% of Americans (18 and over) smoke cigarettes. In a sample of 50 adult Americans

What is the mean number that smoke cigarettes? Blank 1

What is the standard deviation?

N = 50

P = 0.16 (16%)

Standard deviation is given by √n*p*(1-p) = 2.6

-)

Would be unusual if 12 were found to be smokers in this sample? Blank 3 Choose yes or no.

We say something unusual when it is 2 standard deviation away from mean

Mean = n*p

N*p = 50*0.16 = 8

8 + (2*2.6) = 13.2

As 12 is not beyond 13.2

Answer is no