MTH152-Q15

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course MTH 152

Time: 4:22PM Date: 7/29

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution.

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

015. ``q Query 15

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Question: `q Query problem 13.2.10 .3, .4, .3, .8, .7, .9, .2, .1, .5, .9, .6

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Your solution:

- The first thing I do is make the numbers a little easier to read, and list them in order as set ‘A’.

- Set ‘A’ = (.1, .2, .3, .3, .4, .5, .6, .7, .8, .9, .9) = 11 #’s.

- The mean is the sum of all elements / number of elements = .518

- The median follows the equation (n+1)/2 where n= number of elements in the set, so (11+1)/2 = 6, being the sixth number in the set, which is .5 This is proven by 5 numbers being on either side of .5 (in addition, ‘n’ = an odd number here, so there is one median. For a set with an even number, there would be 2 where I believe the two are divided together or just left separate)

- The mode is essentially any numbers listed more than one time. How many times a number repeats and then the value of that number.

- In this case, the mode is 2, for values .3 and (.9)

confidence rating #$&*: 3

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Given Solution:

`aThe numbers, in order, are .1, .2, .3, .3, .4, .5, .6, .7, .8, .9, .9

The mean, obtained by adding the 11 numbers then dividing by 11, is .518.

The median occurs at position (n + 1 ) / 2 = 6 in the ordered list. This number is .5. Note that there are five numbers before .5 and five numbers after .5.

The maximum number of times a number repeats in this distribution is 2. So there are two modes (and we say that the distribution is bimodal). The modes are .3 and .9. **

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Self-critique (if necessary):

- Easy stuff, only hard part is not going cross-eyed with all the commas and decimals. Expect extra spaces in my sets of numbers to preserve sanity.

- Additional Inquiry = consider my definition of “mode” and provide any comments/insight as necessary.

@&

I'm all for sanity.

The mode is the number that's listed the most times. If there's a tie then all such numbers are modes.

*@

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Self-critique Rating: 3

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Question: `q Query problem 13.2.24 more effect from extreme value

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Your solution:

- My text reads “In general, which measure, mean or median, is affected less by the presence of an extreme value in the data?”

- To which I reply, “Assuming the ‘extreme value’ relates to the variable ‘n’ denoting the total number of elements in the set in question, I theorize that the median is not affected as much as the mean, because the median is reflected by position and not necessarily numerical value.”

confidence rating #$&*: 3

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Given Solution:

`aThe mean is drastically affected by the error; correcting the error changes the mean by about 3 units.

The median number, however, simply shifts 1 position, changing from 2.28 to 2.39. **

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `q Query problem 13.2.30 Salaries 6 @$19k, 8 @ 23k, 2 @ 34.5k, 7 @ 56.9k, 1 @ 145.5k.

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Your solution:

- I experiment with combining the number of employees / salaries as follows

- (6*$19k, 8 * $23k, 2 *$34.5k, 7 * $56.9k, 1 * 145.5k )

- Taking the product of number of employees * salary

- (114, 184, 69, 398.3, 145.5) = 841.8

- Taking the quotient of the total salary and the average number of employees in question,

- (6, 8, 2, 7, 1) = 24 = 841.8 / 24 = 35.075

- Following the constraint listed in the text to round to the nearest hundred dollars, the mean salary is $35.1K

confidence rating #$&*:

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Given Solution:

`aIF THERE ARE 28 EMPLOYEES:

The totals paid for each salary level are:

6 * $19,500 = $117,000

8 * $23,000 = $184,000

4 * $28,300 = $113,200

2 * $34,500 = $69,000

7 * $36,900 = $258,300

1 * $145,500 = $145,500

The grand total paid in salaries to the 28 employees is therefore $887,000, giving an average of $887,000 / 28 = $31,700.

The median occurs at position (n + 1) / 2 = (28 + 1) / 2 = 14.5. Since the 14 th salaray on a list ordered from least to greatest is $23,000 and the 15 th is $28300 the median is ($23000 +$28300) / 2 = $25,650.

The mode is 23,000, since this salary occurs more frequently than any other.

IF THERE ARE 24 EMPLOYEES:

The totals paid for each salary level are:

$19,000 * 6 = $114,000

$23,000 * 8 = $184,000

$34,500 * 2 = $69,000

$56,900 * 7 = $398,300

$145,500 * 1 = $145,500

Adding these gives a ‘grand total’, which is divided by the number 24 of employees to obtain the mean $37,950.

The median occurs at position (n + 1) / 2 = (24 + 1) / 2 = 12.5. Since the $23000 salary covers positions 7 thru 14 in an ordered lise of salaries the median is $23,000.

The mode is 23,000, since this salary occurs more frequently than any other.

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Self-critique (if necessary):

- My mathematical instincts are becoming sharper. I came very close this time, construed only in execution.

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Self-critique Rating:

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Question: `q Query problem 13.2.51 mean, med, mode of 0, 1, 3, 14, 14, 15, 16, 16, 17, 17, 18, 18, 18, 19, 20

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Your solution:

- Nice how the numbers are organized thusly.

- Mean is 206/15 = 13.73

- Median is (n+1)/2 = (15+1)/2 = 8th position, the second 16.

- Mode is 3, 18 occurred three times. (Also 14, 16, and 17 occurred twice)

confidence rating #$&*:

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Given Solution:

`aThe mean is 13.73, obtained by adding together all the numbers and dividing by n = 15.

The median is in position (n+1) / 2 = (15+1)/2 = 8 on the ordered list; the 8 th number is 16.

The mode is 18, which is the only number occurring as many as 3 times. **

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Self-critique (if necessary): Okay.

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Self-critique Rating:

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Question: `q Query Add comments on any surprises or insights you experienced as a result of this assignment

- Interesting to put the (n+1)/2 position of median equation (hereafter known as POM equation) to use.

- Surprising to see how cross-eyed one gets when staring at numerous decimals in a set.

- I believe my method of creating “sets” of numbers is handy, especially when running into errors like accidently combining two different sets of numbers with each other. Hopefully this is better prevented by denoting each set as set ‘A, B, C’ and so on.

"

Self-critique (if necessary):

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Self-critique rating:

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Question: `q Query Add comments on any surprises or insights you experienced as a result of this assignment

- Interesting to put the (n+1)/2 position of median equation (hereafter known as POM equation) to use.

- Surprising to see how cross-eyed one gets when staring at numerous decimals in a set.

- I believe my method of creating “sets” of numbers is handy, especially when running into errors like accidently combining two different sets of numbers with each other. Hopefully this is better prevented by denoting each set as set ‘A, B, C’ and so on.

"

Self-critique (if necessary):

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Self-critique rating:

#*&!

&#Good work. See my notes and let me know if you have questions. &#