Question

The Downtown Parking Authority of Tampa, Florida, reported the following information for a sample of 232 customers on the number of hours cars are parked and the amount they are charged.

Number of Hours	Frequency			Amount Charged
1		20		$	3
2		37			7
3		49			10
4		39			17
5		33			21
6		17			25
7		3			27
8		34			31
		232

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

Hours	Probability
1
2
3
4
5
6
7
8

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

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

Mean
Standard deviation

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

The typical customer is parked for

hours

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

Mean
Standard deviation

Answer 1

a-1)

The probability is for each number of hour is obtained by dividing the frequency by total frequency,

Number of Hours	Frequency	Probability
1	20	20/232=0.086
2	37	37/232=0.159
3	49	49/232=0.211
4	39	39/232=0.168
5	33	33/232=0.142
6	17	17/232=0.073
7	3	3/232=0.013
8	34	34/232=0.147
Total	232

a-2)

Answer: discrete

Explanation: Since we calculating the probability for discrete data (frequency), this is a discrete probability distribution.

b-1)

The mean for the the number of hours parked is obtained using the formula,

Number of Hours, x	Frequency, f
1	20	20
2	37	74
3	49	147
4	39	156
5	33	165
6	17	102
7	3	21
8	34	272
Total	232	957

The standard deviation for the number of hours parked is obtained using the formula,

Number of Hours, x	Frequency, f
1	20	-3.125	9.766	195.313
2	37	-2.125	4.516	167.078
3	49	-1.125	1.266	62.016
4	39	-0.125	0.016	0.609
5	33	0.875	0.766	25.266
6	17	1.875	3.516	59.766
7	3	2.875	8.266	24.797
8	34	3.875	15.016	510.531
Total	232			1045.375

b-2)

A typical customer parked with mean value = 4.125 hours.

c)

The mean for the amount charged is,

Amount Charged	Frequency, f
3	20	60
7	37	259
10	49	490
17	39	663
21	33	693
25	17	425
27	3	81
31	34	1054
Total	232	3725

The standard deviation for the amount charged is,

Amount Charged	Frequency, f
3	20	-13.056	170.460	3409.201
7	37	-9.05603	82.012	3034.435
10	49	-6.05603	36.676	1797.102
17	39	0.943966	0.891	34.752
21	33	4.943966	24.443	806.612
25	17	8.943966	79.995	1359.907
27	3	10.94397	119.770	359.311
31	34	14.94397	223.322	7592.952
Total	232			18394.272