SOLUTION: Greetings, I solved part of this problem, just need a bit of help with the second part. Thank you, I greatly appreciate it! When parking a car in a downtown parking lot, dri

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Question 1190589: Greetings,
I solved part of this problem, just need a bit of help with the second part. Thank you, I greatly appreciate it!
When parking a car in a downtown parking lot, drivers pay according to the number of hours or fraction thereof. The probability distribution of the number of hours cars are parked has been estimated as follows:

X 1 2 3 4 5 6 7 8
P(X) 0.2 0.119 0.128 0.079 0.058 0.03 0.036 0.35
(I solved these 2 answers)
A. Mean = 4.66
B. Standard Dev. = 2.8408
The cost of parking is 2.75 dollars per hour. Calculate the mean and standard deviation of the amount of revenue each car generates.
(I solved this)
A. Mean = 12.815
B. Standard Deviation = ????
I can't seem to figure of the standard deviation.
I appreciate your help.

Answer by math_tutor2020(3816) (Show Source):
You can put this solution on YOUR website!

Question 1, part A

Given data table

X 1 2 3 4 5 6 7 8
P(X) 0.2 0.119 0.128 0.079 0.058 0.03 0.036 0.35

Form a new row consisting of X*P(X)

X 1 2 3 4 5 6 7 8
P(X) 0.2 0.119 0.128 0.079 0.058 0.03 0.036 0.35
X*P(X) 0.2 0.238 0.384 0.316 0.29 0.18 0.252 2.8

For example, 8*0.35 = 2.8 in the far right column.

Add up everything in the X*P(X) row
0.2+0.238+0.384+0.316+0.29+0.18+0.252+2.8 = 4.66

Your answer of 4.66 is perfectly correct. Nice work.

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Question 1, part B

Form a new row that involves subtracting each X value from the mean mu = 4.66
Afterward, square the result
The expression is of the form (X-mu)^2 which is what I'll title this new row.

X 1 2 3 4 5 6 7 8
P(X) 0.2 0.119 0.128 0.079 0.058 0.03 0.036 0.35
X*P(X) 0.2 0.238 0.384 0.316 0.29 0.18 0.252 2.8
(X-mu)^2 13.3956 7.0756 2.7556 0.4356 0.1156 1.7956 5.4756 11.1556

Example: In column 1 we have
(X-mu)^2 = (1-4.66)^2 = 13.3956

Next, multiply each (X-mu)^2 value with its corresponding P(X) value.
I'll make a new row for that.

X 1 2 3 4 5 6 7 8
P(X) 0.2 0.119 0.128 0.079 0.058 0.03 0.036 0.35
X*P(X) 0.2 0.238 0.384 0.316 0.29 0.18 0.252 2.8
(X-mu)^2 13.3956 7.0756 2.7556 0.4356 0.1156 1.7956 5.4756 11.1556
(X-mu)^2*P(X) 2.67912 0.8419964 0.3527168 0.0344124 0.0067048 0.053868 0.1971216 3.90446

Then add up everything in the (X-mu)^2*P(X) row
2.67912+0.8419964+0.3527168+0.0344124+0.0067048+0.053868+0.1971216+3.90446 = 8.0704

The last step is to apply the square root
sqrt(8.0704) = 2.8408

The standard deviation is approximately 2.8408

You're on a roll with correct answers.

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Question 2, part A

X = number of hours parked
Y = revenue
To find the revenue, we multiply 2.75 by the X value
Y = 2.75X

For example, if a car is parked for 4 hours, then
Y = 2.75*X = 2.75*4 = 11
meaning they pull in $11 in revenue

Let's form a table of Y values with their corresponding P(Y) probabilities.
The probabilities will be the same as before. The only thing that changes is the introduction of the Y row.

X 1 2 3 4 5 6 7 8
Y 2.75 5.5 8.25 11 13.75 16.5 19.25 22
P(Y) 0.2 0.119 0.128 0.079 0.058 0.03 0.036 0.35

We follow the same steps as part A in the previous question. I'll skip a few steps, but basically this is what you should have as a final table

X 1 2 3 4 5 6 7 8
Y 2.75 5.5 8.25 11 13.75 16.5 19.25 22
P(Y) 0.2 0.119 0.128 0.079 0.058 0.03 0.036 0.35
Y*P(Y) 0.55 0.6545 1.056 0.869 0.7975 0.495 0.693 7.7

Adding everything in that bottom row gets you 12.815

You have the correct answer once again.

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Question 2, part B

We follow the same steps as part B from the previous question
This time use mu = 12.815 which was found in the previous section.

This is what the table should look like when all is said and done

X 1 2 3 4 5 6 7 8
Y 2.75 5.5 8.25 11 13.75 16.5 19.25 22
P(Y) 0.2 0.119 0.128 0.079 0.058 0.03 0.036 0.35
Y*P(Y) 0.55 0.6545 1.056 0.869 0.7975 0.495 0.693 7.7
(Y-mu)^2 101.304225 53.509225 20.839225 3.294225 0.874225 13.579225 41.409225 84.364225
(Y-mu)^2*P(Y) 20.260845 6.367597775 2.6674208 0.260243775 0.05070505 0.40737675 1.4907321 29.52747875

Add up everything in the bottom row and you should get 61.0324

Apply the square root to get
sqrt(61.0324) = 7.8123
The value is approximate.

This is a useful calculator to check your work
https://www.mathportal.org/calculators/statistics-calculator/probability-distributions-calculator.php