Question

choose a person aged 19 to 25 years at random and ask, \in the past seven days, how many times did you go to an exercise or fitness center to work out?\ call the response y for short. based on a large sample survey, here is a probability model for the answer you will get.

days	probability
1	0.05
2	0.07
3	0.08
4	0.05
5	0.04
6	0.01
7	???

(a) write the event \went to exercise/fitness center to work out on all 7 days\ in terms of y. then find its probability.

• $p(y < 7) = 0.98$
• $p(y = 7) = 0.02$
• $p(y = 7) = 0.02$
• $p(y < 7) = 0.98$
• $p(y < 7) = 0.02$
• $p(y = 7) = 0.98$
• $p(y < 7) = 0.02$

(b) whats the probability that a randomly selected person says they went to an exercise/fitness center to work out on at most 2 days?
(include 2 decimal places.)

Explanation:

Step1: Identify the event

The event "at most 2 days" means \( Y = 0 \), \( Y = 1 \), or \( Y = 2 \). We need to find \( P(Y \leq 2) \), which is \( P(Y = 0) + P(Y = 1) + P(Y = 2) \).

Step2: Get the probabilities from the table

From the table:

• \( P(Y = 0) = 0.68 \)
• \( P(Y = 1) = 0.05 \)
• \( P(Y = 2) = 0.07 \)

Step3: Calculate the sum

Add the three probabilities: \( 0.68 + 0.05 + 0.07 = 0.80 \)

Answer:

0.80