Question

The frequency distribution shown in the following table lists the number of hours per day a randomly selected sample of teenagers spent watching television. Where possible, determine what percent of the teenagers spent the following number of hours watching television. (Round your answers to one decimal place. If not possible, enter IMPOSSIBLE.)

Hours per day	Number of Teenagers
0 ≤ x < 1	17
1 ≤ x < 2	31
2 ≤ x < 3	24
3 ≤ x < 4	37
4 ≤ x < 5	27
5 ≤ x < 6	11
6 ≤ x < 7	15

(a) less than 4 hours
%

(b) at least 5 hours
%

(c) at least 1 hour
%

(d) less than 2 hours
%

(e) at least 2 hours but less than 4 hours
%

(f) more than 3.5 hours
%

Answer 1

The table for probability distribution is ,

Hours per day	Number of Teenagers	Probability
0x<1	17	17/162=0.1049
1x<2	31	31/162=0.1914
2x<3	24	24/162=0.1481
3x<4	37	37/162=0.2284
4x<5	27	27/162=0.1667
5x<6	11	11/162=0.0679
6x<7	15	15/162=0.0926
Sum	162	1

(a) P(less than 4 hours)=P(X=3)+P(X=2)+P(X=1)+P(X=0)= 0.2284+0.1481+0.1914+0.1049=0.673=67.3%

(b) P(at least 5 hours)=P(X=5)+P(X=6)=0.0679+0.0926=0.161=16.1%

(c) P(at least 1 hour)=1-P(\ess than 1 hour)=1-P(X=0)=1-0.1049=0.895=89.5%

(d) P(less than 2 hours)=P(X=1)+P(X=0)=0.1914+0.1049=0.296=29.6%

(e) P(at least 2 hours but less than 4 hours)=P(2X<4)=0.1481+0.2284=0.377=37.7%

(f) P(more than 3.5 hours)=P(X=4)+P(X=5)+P(X=6)=0.1667+0.0679+0.0926=0.327=32.7%