The activities of the Downtown Parking Authority can be represented on a probability distribution table
a-1 Convert the table to a probability distribution
This is calculated as:
[tex]\mathbf{Probability = \frac{f}{Total}}[/tex]
So, we have:
[tex]\mathbf{P(1) = \frac{25}{234} = 0.108}[/tex]
[tex]\mathbf{P(2) = \frac{38}{234} = 0.162}[/tex]
[tex]\mathbf{P(3) = \frac{51}{234} = 0.218}[/tex]
And so on.
So, the table is:
[tex]\mathbf{\left[\begin{array}{ccc}x&f&P(x)\\1&25&0.108\\2&38&0.162\\3 & 51 & 0.218 & 4 & 45 & 0.192 & 5 & 20 & 0.085 & 6 & 14 &0.06 & 7 & 5 & 0.021 & 8 & 36 & 0.154 \end{array}\right] }[/tex]
a-2 The distribution type
The number of hours is from 1 to 8, and all are given in whole numbers.
Discrete probability distributions are represented by whole numbers.
Hence, the distribution is a discrete continuous probability distribution
b-1. The mean and the standard deviation of the number of hours parked.
The mean is calculated using:
[tex]\mathbf{\bar x = \frac{\sum fx}{\sum f}}[/tex]
So, we have:
[tex]\mathbf{\bar x = \frac{25 \times 1 + 38 \times 2 + 51 \times 3 + 45 \times 4 + 20 \times 5 + 14 \times 6 + 5 \times 7 + 36 \times 8}{234}}[/tex]
[tex]\mathbf{\bar x = \frac{941}{234}}[/tex]
[tex]\mathbf{\bar x = 4.02}[/tex]
The standard deviation is calculated as:
[tex]\mathbf{\sigma = \sqrt{\frac{\sum f(x - \bar x)^2}{\sum f}}}[/tex]
So, we have:
[tex]\mathbf{\sigma = \sqrt{\frac{25 \times (1 -4.02)^2+ 38 \times (2 -4.02)^2 + 51 \times (3 -4.02)^2 +.............+ 36 \times (8 -4.02)^2}{234}}}[/tex]
[tex]\mathbf{\sigma = \sqrt{\frac{1124.8936}{234}}}[/tex]
[tex]\mathbf{\sigma = \sqrt{4.80723760684}}[/tex]
[tex]\mathbf{\sigma = 2.19}[/tex]
Hence, the mean is 4.02 and the standard deviation is 2.19
b-2. How long is a typical customer parked?
This is the mean of the distribution.
In (a), we have:
[tex]\mathbf{\bar x = 4.02}[/tex]
Hence, a typical customer is parked for 4.02 hours
c-1. The probability that a car would be parked for more than 6 hours
This is calculated as:
[tex]\mathbf{P(x > 6) = P(7) + P(8)}[/tex]
So, we have:
[tex]\mathbf{P(x > 6) = 0.021 + 0.154}[/tex]
[tex]\mathbf{P(x > 6) = 0.175}[/tex]
Hence, the probability that a car would be parked for more than 6 hours is 0.175
c -2. The probability that a car would be parked for 3 hours or less?
This is calculated as:
[tex]\mathbf{P(x \le 3) = P(1) + P(2) + P(3)}[/tex]
So, we have:
[tex]\mathbf{P(x \le 3) = 0.108 + 0.162 + 0.218}[/tex]
[tex]\mathbf{P(x \le 3) = 0.488}[/tex]
Hence, the probability that a car would be parked for 3 hours or less is 0.488
Read more about probabilities at:
https://brainly.com/question/11234923
