Question

The Downtown Parking Authority of Tampa, Florida, reported the following information for a sample of 231 customers on the number of hours cars are parked and the amount they are charged.

Number of Hours		Frequency		Amount Charged
1		22		$	3
2		38			6
3		51			8
4		45			12
5		20			14
6		14			16
7		5			18
8		36			22
		231

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?

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.)

c. 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)

dividing each value with 231 below is the probability distribution of number of hours

Hours	probability
1	0.095
2	0.165
3	0.221
4	0.195
5	0.087
6	0.061
7	0.022
8	0.156

a-2)

discrete

b-1)

x	P(x)	xP(x)	x2*P(x)
1	0.095	0.095	0.095
2	0.165	0.330	0.660
3	0.221	0.663	1.989
4	0.195	0.780	3.120
5	0.087	0.435	2.175
6	0.061	0.366	2.196
7	0.022	0.154	1.078
8	0.156	1.248	9.984
		4.071	21.297
E(x) =μ=	ΣxP(x) =		4.071
E(x2) =	Σx2P(x) =		21.297
Var(x)=σ2 =	E(x2)-(E(x))2=		4.724
std deviation=	σ= √σ2 =		2.173

mean =4.071

standard deviation =2.173

b-2)

typical customer parked =mean value =4.071

c)

x	P(x)	xP(x)	x2*P(x)
3	0.095	0.285	0.855
6	0.165	0.990	5.940
8	0.221	1.768	14.144
12	0.195	2.340	28.080
14	0.087	1.218	17.052
16	0.061	0.976	15.616
18	0.022	0.396	7.128
22	0.156	3.432	75.504
		11.405	164.319
E(x) =μ=	ΣxP(x) =		11.405
E(x2) =	Σx2P(x) =		164.319
Var(x)=σ2 =	E(x2)-(E(x))2=		34.245
std deviation=	σ= √σ2 =		5.852

mean =11.405

standard deviation =5.852