First graders who misbehave in school may be more likely to be regular smokers as young adults according to a new study presented in the July 2004 issue of the American Journal of Epidemiology. After following a group of U.S. first graders for 15 years, it was found that among those kids who had tried smoking and misbehaved, 68% were daily smokers. (Give your answers correct to three decimal places.)

Question

First graders who misbehave in school may be more likely to be regular smokers as young adults according to a new study presented in the July 2004 issue of the American Journal of Epidemiology. After following a group of U.S. first graders for 15 years, it was found that among those kids who had tried smoking and misbehaved, 68% were daily smokers. (Give your answers correct to three decimal places.)

(a) What is the probability that exactly two of the next three randomly selected young adults who misbehaved in early grades and have tried smoking are daily smokers?
(b) What is the probability that exactly 8 of the next 12 randomly selected young adults who misbehaved in early grades and have tried smoking are daily smokers?
(c) What is the probability that exactly 20 of the next 30 randomly selected young adults who misbehaved in early grades and have tried smoking are daily smokers?

Answer 1

a, ,68^2

b. 12!/4!8! (.68)^8 (.32)^4

Answer 2

To solve these probability problems, we need to use the binomial probability formula. The formula is:

P(x) = nCx * p^x * q^(n-x)

Where:
- P(x) represents the probability of getting exactly x successes,
- n is the total number of trials,
- x represents the number of successful outcomes,
- p is the probability of success in a single trial, and
- q is the probability of failure in a single trial (1 - p).

Now let's solve the given problems:

(a) What is the probability that exactly two of the next three randomly selected young adults who misbehaved in early grades and have tried smoking are daily smokers?

From the given information, we know that the probability of success (p) is 0.68, and the number of trials (n) is 3.

Using the binomial probability formula, we have:
P(x=2) = 3C2 * 0.68^2 * (1 - 0.68)^(3 - 2)

Calculating this, we get:
P(x=2) = 3 * 0.68^2 * (1 - 0.68)^(3 - 2)
= 3 * 0.4624 * 0.32
≈ 0.445

So, the probability that exactly two of the next three randomly selected young adults who misbehaved in early grades and have tried smoking are daily smokers is approximately 0.445.

(b) What is the probability that exactly 8 of the next 12 randomly selected young adults who misbehaved in early grades and have tried smoking are daily smokers?

Here, the probability of success (p) is still 0.68, and the number of trials (n) is 12.

Using the binomial probability formula, we have:
P(x=8) = 12C8 * 0.68^8 * (1 - 0.68)^(12 - 8)

Calculating this, we get:
P(x=8) = 495 * 0.068719476736 * 0.150255706624
≈ 0.205

So, the probability that exactly 8 of the next 12 randomly selected young adults who misbehaved in early grades and have tried smoking are daily smokers is approximately 0.205.

(c) What is the probability that exactly 20 of the next 30 randomly selected young adults who misbehaved in early grades and have tried smoking are daily smokers?

Here, the probability of success (p) is still 0.68, and the number of trials (n) is 30.

Using the binomial probability formula, we have:
P(x=20) = 30C20 * 0.68^20 * (1 - 0.68)^(30 - 20)

Calculating this, we get:
P(x=20) = 30 * 0.008792670265014 * 0.02799214976336
≈ 0.021

So, the probability that exactly 20 of the next 30 randomly selected young adults who misbehaved in early grades and have tried smoking are daily smokers is approximately 0.021.