Question

The Downtown Parking Authority of Tampa, Florida, reported the following information for a sample of 250 customers on the number of hours cars are parked and the amount they are charged. Number of Hours 1 2 3 4 Frequency 20 38 53 45 40 13 5 36 250 Amount Charged $ 3.00 6.00 9.00 12.00 14.00 16.00 18.00 20.00 5 CONO

Answer 1

a-1 Hours probability Pi 0,152 3 0 212 4 0,180 0,16o 0.052 6 T 8 0,020 0,144 there for probability Pi = thi 10.09 20 250 P2 = 38 Joils 250 Similarly do it for all for P3 to Pg

b-1 X: No. of hours parked I. Mean a Elx) = {xipi - 1.10.08) + 210.152)+.. +810.149) - (4.149 Standard deviation Levy) - [E1013 C. Elx²=EX:² Pi 1² 10.08172260.1527t annet 82 (0.144) -- 21.544 Elx2) - (F(x)]² 21.549 - 14,144)2 2.090756 y (2.0908 Answers. Mean = 4.144 Standard deviation = 2,0908

brz is more A typical customer parked often that is expected value of hours and frequency of parking customer. . Mean is expected value here : 4,144 hours typical customer parked car Answer: The typical customer hours 4.144 parked for a Nowe here xi = Amound of charged X XI 3 6 9 9 36 81 12 16 1५५ 196 256 324 400 1446 18 20 98 2 Total

Mean= 98 112.25 8 Standard deviation (s) 2 2 ) De (1946-1902) 8-1 5.922113 23 15 5.92 Mean -- 12.25 standard deviation = 5.92