The Downtown Parking Authority of Tampa, Florida, reported the following information for a sample of 228 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 36 6
3 53 9
4 40 13
5 22 14
6 11 16
7 9 18
8 36 22
228
A. Convert the information on the number of hours parked to a probability distribution. Is this a discrete or a continuous probability distribution?
B. Find the mean and the standard deviation of the number of hours parked. How would you answer the question: How long is a typical customer parked?
C. Find the mean and the standard deviation of the amount charged.

Step-by-step explanation: Probability Distribution is an equation or table linking each outcome of an experiment with its probability of ocurrence. For this case, since the experiment is performed a high number of times and in a long run, the relative frequency of the event is its probability. Therefore:

A. To convert to a probability distribution, find the probability through the frequency by doing:

Hour 1

P(X) = [tex]\frac{21}{228}[/tex] = 0.09

Hour 2

P(X) = [tex]\frac{36}{228}[/tex] = 0.16

Hour 3

P(X) = [tex]\frac{53}{228}[/tex] = 0.23

Hour 4

P(X) = [tex]\frac{40}{228}[/tex] = 0.17

Hour 5

P(X) = [tex]\frac{22}{228}[/tex] = 0.09

Hour 6

P(X) = [tex]\frac{11}{228}[/tex] = 0.05

Hour 7

P(X) = [tex]\frac{9}{228}[/tex] = 0.04

Hour 8

P(X) = [tex]\frac{36}{228}[/tex] = 0.16

The table will be:

hours probability

1 0.09

2 0.16

3 0.23

4 0.17

5 0.09

6 0.05

7 0.04

8 0.16

This is a discrete distribution because it lists all the possible values that the discrete variable can be and its associated probabilities.

B. Mean for a probability distribution is calculated as:

E(X) = ∑[[tex]x_{i}[/tex].P([tex]x_{i}[/tex])]

E(X) = 1*0.09 + 2*0.16+3*0.23+4*0.17+5*0.09+6*0.05+7*0.04+8*0.16

E(X) = 4.12

Standard Deviation is:

σ = √∑{[x - E(x)]² . P(x)}

σ = [tex]\sqrt{(1-4.12)^{2}*0.09 + (2-4.12)^{2}*0.16 + ... + (7-4.12)^{2}*0.04 + (8-4.12)^{2}*0.16}[/tex]

σ = [tex]\sqrt{4.87}[/tex]

σ = 2.21

The average number of hours parked is approximately 4h with a standard deviation of approximately 2 hours, which means that a typical costumer parks between 2 to 6 hours.

C. Mean for a sample is given by: μ = ∑[tex]\frac{x_{i}}{n}[/tex] , which is this case is:

μ = [tex]\frac{4+6+9+13+14+16+18+22}{8}[/tex]

μ = 12.75

Standard Deviation of a sample: s = √[tex]\frac{1}{n-1}[/tex]∑([tex]x_{i}[/tex] - μ)²

s = [tex]\sqrt{ \frac{(4-12.75)^{2} + (6-12.74)^{2} + ... + (18-12.75)^{2} + (22-12.75)^{2} }{8-1}}[/tex]

s = 6.11

The average amount charged is 12.75±6.11.