Question

Please help

The Downtown Parking Authority of Tampa, Florida, reported the following information for a sample of 234 customers on the number of hours cars are parked and the amount they are charged. Amount Charged $ 4 Frequency 25 38 Number of Hours 1 6 2 14 20 17 14 6 19 5 22 36 234 Click here for the Excel Data File a-1. Convert the information on the number of hours parked to a probability distribution. (Round your answers to 3 decimal places.) Answer is complete and correct. Probability Hours 0.107 1 2 0.162 3 0.218 0.192 4 0.085 6 0.060 7 0.021 0.154 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.) Answer is complete but not entirely correct Mean Standard deviation 4.020 21.140 b-2. How long is a typical customer parked? (Do not round the intermediate calculations. Round your final answer to 3 decimal places.) hours The typical customer is parked for 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 is complete but not entirely correct. 653.000 Mean Standard deviation 10.883

Answer 1

Part a1)

Number frequenc Realative of hours y frequenc |(x) (f) Y 1 25 0.108 0.162 2 38 3 51 0.218 4 45 0.192 0.085 5 20 6 14 0.06 7 0.021 5 36 0.154 |ТOTAL 234 1

PART a2)

Yes, it is an example of a discrete series.

Part b1) The standard deviation is computed using the formula-

ΣίΧ ΣiΧ2 2 (mean) σ - Σί

Number frequenc Realative of hours y (x- MEAN) frequenc Amount f x x2 (X) (f) charged CX B Y 0.108 25 1 25 4 25 -3.02 38 0.162 -2.02 152 2 6 76 3 -1.02 459 51 0.218 153 4 45 0.192 -0.02 720 12 180 5 20 0.085 0.98 500 14 100 6 0.06 17 1.98 504 14 84 7 0.021 5 19 35 2.98 245 8 36 0.154 22 288 3.98 2304 ТОTAL 234 1 941 3.84 4909 Mean 4.02 standard deviation 2.195

Part b2)

Number frequenc of Number y hours of hours (f parked 1 25 25 76 2 38 153 3 51 180 4 45 5 20 100 6 14 84 7 5 35 36 8 288 234 941 Ln

The maximum number of frequency is 51 corresponding to 3 hours. So, a typical person park for 3 hours.

Part c)

frequenc Amount Number y charged |fx x2 of hours (f) (X) fxX 1 25 100 400 2 38 6 228 1368 51 8 408 3264 540 45 12 6480 3920 5 20 14 280 14 6 17 238 4046 7 5 19 95 1805 36 22 792 17424 TOTAL 234 2681 38707 11.457 mean standard deviation 5.844 LO O0