Question 1190589
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Question 1, part A


Given data table
<table border = "1" cellpadding = "5"><tr><td>X</td><td>1</td><td>2</td><td>3</td><td>4</td><td>5</td><td>6</td><td>7</td><td>8</td></tr><tr><td>P(X)</td><td>0.2</td><td>0.119</td><td>0.128</td><td>0.079</td><td>0.058</td><td>0.03</td><td>0.036</td><td>0.35</td></tr></table>


Form a new row consisting of X*P(X)<table border = "1" cellpadding = "5"><tr><td>X</td><td>1</td><td>2</td><td>3</td><td>4</td><td>5</td><td>6</td><td>7</td><td>8</td></tr><tr><td>P(X)</td><td>0.2</td><td>0.119</td><td>0.128</td><td>0.079</td><td>0.058</td><td>0.03</td><td>0.036</td><td>0.35</td></tr><tr><td>X*P(X)</td><td>0.2</td><td>0.238</td><td>0.384</td><td>0.316</td><td>0.29</td><td>0.18</td><td>0.252</td><td>2.8</td></tr></table>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.<table border = "1" cellpadding = "5"><tr><td>X</td><td>1</td><td>2</td><td>3</td><td>4</td><td>5</td><td>6</td><td>7</td><td>8</td></tr><tr><td>P(X)</td><td>0.2</td><td>0.119</td><td>0.128</td><td>0.079</td><td>0.058</td><td>0.03</td><td>0.036</td><td>0.35</td></tr><tr><td>X*P(X)</td><td>0.2</td><td>0.238</td><td>0.384</td><td>0.316</td><td>0.29</td><td>0.18</td><td>0.252</td><td>2.8</td></tr><tr><td>(X-mu)^2</td><td>13.3956</td><td>7.0756</td><td>2.7556</td><td>0.4356</td><td>0.1156</td><td>1.7956</td><td>5.4756</td><td>11.1556</td></tr></table>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.<table border = "1" cellpadding = "5"><tr><td>X</td><td>1</td><td>2</td><td>3</td><td>4</td><td>5</td><td>6</td><td>7</td><td>8</td></tr><tr><td>P(X)</td><td>0.2</td><td>0.119</td><td>0.128</td><td>0.079</td><td>0.058</td><td>0.03</td><td>0.036</td><td>0.35</td></tr><tr><td>X*P(X)</td><td>0.2</td><td>0.238</td><td>0.384</td><td>0.316</td><td>0.29</td><td>0.18</td><td>0.252</td><td>2.8</td></tr><tr><td>(X-mu)^2</td><td>13.3956</td><td>7.0756</td><td>2.7556</td><td>0.4356</td><td>0.1156</td><td>1.7956</td><td>5.4756</td><td>11.1556</td></tr><tr><td>(X-mu)^2*P(X)</td><td>2.67912</td><td>0.8419964</td><td>0.3527168</td><td>0.0344124</td><td>0.0067048</td><td>0.053868</td><td>0.1971216</td><td>3.90446</td></tr></table>



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.<table border = "1" cellpadding = "5"><tr><td>X</td><td>1</td><td>2</td><td>3</td><td>4</td><td>5</td><td>6</td><td>7</td><td>8</td></tr><tr><td>Y</td><td>2.75</td><td>5.5</td><td>8.25</td><td>11</td><td>13.75</td><td>16.5</td><td>19.25</td><td>22</td></tr><tr><td>P(Y)</td><td>0.2</td><td>0.119</td><td>0.128</td><td>0.079</td><td>0.058</td><td>0.03</td><td>0.036</td><td>0.35</td></tr></table>


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<table border = "1" cellpadding = "5"><tr><td>X</td><td>1</td><td>2</td><td>3</td><td>4</td><td>5</td><td>6</td><td>7</td><td>8</td></tr><tr><td>Y</td><td>2.75</td><td>5.5</td><td>8.25</td><td>11</td><td>13.75</td><td>16.5</td><td>19.25</td><td>22</td></tr><tr><td>P(Y)</td><td>0.2</td><td>0.119</td><td>0.128</td><td>0.079</td><td>0.058</td><td>0.03</td><td>0.036</td><td>0.35</td></tr><tr><td>Y*P(Y)</td><td>0.55</td><td>0.6545</td><td>1.056</td><td>0.869</td><td>0.7975</td><td>0.495</td><td>0.693</td><td>7.7</td></tr></table>
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<table border = "1" cellpadding = "5"><tr><td>X</td><td>1</td><td>2</td><td>3</td><td>4</td><td>5</td><td>6</td><td>7</td><td>8</td></tr><tr><td>Y</td><td>2.75</td><td>5.5</td><td>8.25</td><td>11</td><td>13.75</td><td>16.5</td><td>19.25</td><td>22</td></tr><tr><td>P(Y)</td><td>0.2</td><td>0.119</td><td>0.128</td><td>0.079</td><td>0.058</td><td>0.03</td><td>0.036</td><td>0.35</td></tr><tr><td>Y*P(Y)</td><td>0.55</td><td>0.6545</td><td>1.056</td><td>0.869</td><td>0.7975</td><td>0.495</td><td>0.693</td><td>7.7</td></tr><tr><td>(Y-mu)^2</td><td>101.304225</td><td>53.509225</td><td>20.839225</td><td>3.294225</td><td>0.874225</td><td>13.579225</td><td>41.409225</td><td>84.364225</td></tr><tr><td>(Y-mu)^2*P(Y)</td><td>20.260845</td><td>6.367597775</td><td>2.6674208</td><td>0.260243775</td><td>0.05070505</td><td>0.40737675</td><td>1.4907321</td><td>29.52747875</td></tr></table>
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
<a href = "https://www.mathportal.org/calculators/statistics-calculator/probability-distributions-calculator.php">https://www.mathportal.org/calculators/statistics-calculator/probability-distributions-calculator.php</a>
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