Stat 100

[BarTopDancer]

I got number 3 wrong.

Suppose studies have shown that approximately 20% (.2) of Americans regularly engage in exercise (at least 3 times per week).

a. If one American is randomly selected, what is the probability that he or she does not regularly engage in exercise? Set up calculation.

1 - .20 = .80

The probability that a randomly selected American does not engage in regular exercise is 80%

b. If three Americans are randomly selected, what is the probability that all three regularly engage in exercise? Set up the calculation.

.8 x .8 x .8 = .51

(-1)
(.2) x (.2) x (.2) = .008

c. What assumption is being made in part b above?

The assumption being made in part B is that if you remove one person from the equation it will not change the outcome of the probability.

(-1) sure the probability will change!!!

The answer is that we are assuming independence of the events. Otherwise, we could not just multiply the individual event probabilities together to arrive at the “joint probability” of all 3 events happening

d. If two Americans are randomly selected, what is the probability that the first American regularly engages in exercise while the second American does not regularly engage in exercise? Set up the calculation.

.80 x .20 = .16
1 - .20 = .80

(-.5)
Correct order is important and will be important on the exam. The answer is :
(.20) x (.80) = .16



Make sure that the values are in the correct order.