Question

The Downtown Parking Authority of Tampa, Florida, reported the following information for a sample of 236 customers on the number of hours cars are parked and the amount they are charged.

Number of Hours		Frequency		Amount Charged
1		25		$	3
2		42			7
3		53			9
4		40			12
5		20			14
6		11			17
7		9			19
8		36			22
		236

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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)   For Probability of each number of hours, we divide the corresponding          
   frequency by total sum of frequency          
   We get          

Number of Hours (X)	Frequency	P(X)
1	25	0.106
2	42	0.178
3	53	0.225
4	40	0.169
5	20	0.085
6	11	0.047
7	9	0.038
8	36	0.153
Sum	236

a-2)   Since the number of hours are countable and whole numbers,      
   this is a discrete distribution.      
   Answer :      
   Discrete     

b-1)   For finding mean, we find x.P(x) for each X and sum it      

Number of Hours (X)	Frequency	P(X)	X.P(X)
1	25	0.106	0.106
2	42	0.178	0.356
3	53	0.225	0.674
4	40	0.169	0.678
5	20	0.085	0.424
6	11	0.047	0.280
7	9	0.038	0.267
8	36	0.153	1.220
Sum	236		4.004

Mean = ∑X.P(X) = 4.004          
Mean = 4.004    hours

For finding standard deviation, we also find x2.P(x) for each X and sum it          

Number of Hours (X)	Frequency	P(X)	X.P(X)	X2.P(X)
1	25	0.106	0.106	0.106
2	42	0.178	0.356	0.712
3	53	0.225	0.674	2.021
4	40	0.169	0.678	2.712
5	20	0.085	0.424	2.119
6	11	0.047	0.280	1.678
7	9	0.038	0.267	1.869
8	36	0.153	1.220	9.763
Sum	236		4.004	20.979

    Variance = ∑X².P(X) - (Mean)²              
       = 20.979 - (4.004)²          
       = 4.946984          
   Standard Deviation = sqrt(Variance)              
       = sqrt(4.946984)          
       = 2.224          
   Standard Deviation = 2.224        hours     
                  
b-2)   Since the mean = expected number of hours, = 4.004              
   a typical customer is parked for 4.004 hours              
                  
c)   We find the mean and standard deviation of amount charged              
   in the same way as above              

Amount Charged (X)	Frequency	P(X)	X.P(X)	X2.P(X)
3	25	0.106	0.318	0.953
7	42	0.178	1.246	8.720
9	53	0.225	2.021	18.191
12	40	0.169	2.034	24.407
14	20	0.085	1.186	16.610
17	11	0.047	0.792	13.470
19	9	0.038	0.725	13.767
22	36	0.153	3.356	73.831
Sum	236		11.678	169.949

Mean = ∑X.P(X) = $11.678          
Mean = $ 11.678         
          
Variance = ∑X².P(X) - (Mean)²          
   = 169.949 - (11.678)²    
   = 33.57332      
Standard Deviation = sqrt(Variance)          
   = sqrt(33.57332)      
   = 5.794      
Standard Deviation = $ 5.794