Question

5. (3 points) The General Social Survey asked 1000 people how many hours per day la- they were able to relax. Consider these 1000 people to be a population, and let distribution for the number of hours of relaxation time r. Number of Hours1 a be the number of hours of relaxation for a person at random from this popu tion. The results are presented in the following table. Complete the probability 6 8 3 Frequency | 54 | 96 | 166 245 175 135 85 54 Probability 1000 1000 of 044 Relaxation Time .054 (a) (2 points) Find the probability that a person relaxes at most 3 hours. (b) (2 points) Find the probability that a person relaxes at least 3 hours. (c) (2 points) Find the probability that a person relaxes from 3 to 5 hours. (d) (2 points) Find (e) (2 points) Find ơ

Answer 1

5.

First of all we complete the table :

Number of hours	1	2	3	4	5	6	7	8
Frequency	54	96	166	245	175	135	85	44
Probability of Relaxation time	0.054	0.096	0.166	0.245	0.175	0.135	0.085	0.044

Let X be the Number of hours of Relaxation time.

(a)

Probability that a person relaxes at most 3 hours = P(X=1) + P(X=2) + P(X=3)

= 0.054 + 0.096 + 0.166 = 0.316

(b)

Probability that a person relaxes at least 3 hours = P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) + P(X=8)

= 0.166 + 0.245 + 0.175 + 0.135 + 0.085 + 0.044 = 0.85

(c)

Probability that a person relaxes from 3 to 5 hours = P(X=3) + P(X=4) + P(X=5)

= 0.166 + 0.245 + 0.175 = 0.586

(d)

$\mu = \sum_{x=1}^{x=8}xP(X=x) =1*0.054 + 2*0.096 + ...........+ 8*0.044 =4.356$

(e)

$\sigma = \sqrt{ \sum_{x=1}^{x=8}x^2P(X=x) - \left (\sum_{x=1}^{x=8}xP(X=x)\right )^2}$

$\sum_{x=1}^{x=8}x^2P(X=x) =1^2*0.054 + 2^2*0.096 + ...........+ 8^2*0.044 =22.068$

and we already calculated :

$\sum_{x=1}^{x=8}xP(X=x) =1*0.054 + 2*0.096 + ...........+ 8*0.044 =4.356$

$\sigma = \sqrt{ 22.068 - \left (4.356\right )^2} = 1.7588$