Question

The Downtown Parking Authority of Tampa, Florida, reported the following information for a sample of 229 customers on the number of hours cars are parked and the amount they are charged.

Number of Hours		Frequency		Amount Charged
1		21		$	4
2		38			5
3		50			10
4		45			13
5		18			14
6		16			16
7		5			18
8		36			20
		total 229

a-1. Convert the information on the number of hours parked to a probability distribution. (Round your answers to 3 decimal places.)

a-2. Is this a discrete or a continuous probability distribution? Discrete Continuous

b-1. Find the mean and the standard deviation of the number of hours parked. (Do not round the intermediate calculations. Round your final answers to 3 decimal places.)

b-2. How long is a typical customer parked? (Do not round the intermediate calculations. Round your final answer to 3 decimal places.)

Find the mean and the standard deviation of the amount charged. (Do not round the intermediate calculations. Round your final answers to 3 decimal places.)

Answer 1

a-1)

The probability for the number of car parked for each hour is obtained by using formula,

Number of Hours		Probability
1	21/229=	0.092
2	38/229=	0.166
3	50/229=	0.218
4	45/229=	0.197
5	18/229=	0.079
6	16/229=	0.070
7	5/229=	0.022
8	36/229=	0.157

Number of Hours	Probability
1	0.092
2	0.166
3	0.218
4	0.197
5	0.079
6	0.070
7	0.022
8	0.157

a-2)

This is a discrete probability distribution because the variable of interest, "number of hour car parked" is discrete in nature which takes value 1, 2, 3, 4, 5, 6, 7 and 8.

b-1)

The mean of the number of hours car parked or the expected value is obtained using the formula,

The standard deviation, is obtained using the formula,

Number of Hours, x	mean	(x-mean)	(x-mean)^2	(x-mean)^2 P(x)
1	4.087	-3.087	9.532	0.874
2	4.087	-2.087	4.357	0.723
3	4.087	-1.087	1.182	0.258
4	4.087	-0.087	0.008	0.001
5	4.087	0.913	0.833	0.065
6	4.087	1.913	3.658	0.256
7	4.087	2.913	8.484	0.185
8	4.087	3.913	15.309	2.407
			SUM	4.770

b-2)

The time such that how long a typical customer parked? is obtained as follow,

The mean of the amount chared is obtained as follow,

Number of Hours	Amount Charged, y	Probability, p	y*p
1	4	0.04	0.16
2	5	0.05	0.25
3	10	0.1	1
4	13	0.13	1.69
5	14	0.14	1.96
6	16	0.16	2.56
7	18	0.18	3.24
8	20	0.2	4
Total	100	1	Sum=14.86

The standard deviation, is obtained as follow,

Amount Charged, y	mean	(y-mean)	(y-mean)^2	(y-mean)^2 P(y)
4	14.860	-10.860	117.940	4.718
5	14.860	-9.860	97.220	4.861
10	14.860	-4.860	23.620	2.362
13	14.860	-1.860	3.460	0.450
14	14.860	-0.860	0.740	0.104
16	14.860	1.140	1.300	0.208
18	14.860	3.140	9.860	1.775
20	14.860	5.140	26.420	5.284
			SUM	19.760