Open Query 15

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course mth 152

2:25 11/9

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

1.02, 2.31, 4.35, 5.64, 7.68, 8.97

mean= 1.02+2.31+4.35+5.64+7.68+8.97/6=4.99

median= 4.35+5.64/2=4.99

modes=0

.1, .2, .3, .3, .4, .5, .6, .7, .8, .9, .9

.1+.2+.3+.3+.4+.5+.6+.7.+.8+.9+.9/11=5.7/11=.5 is the mean

.5 is the median number

the set is bimodal because .9 and .3 both occur twice

confidence rating #$&*:

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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):

I did not understand what you were wanting until I looked at the answer because the set you gave is different from the problem in the book for 13.2.10.

I did the problem from the book first

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Self-critique Rating:

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The exteme value has a much larger effect on the mean than the median.

confidence rating #$&*:

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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:

6*$19500= $117000

8*$23000= $184000

4*$28300= $113200

2*$34500= $69000

7*$36900= $258300

1*$145500= $145500

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$818000/28=$29214.28

the mean salary to the nearest hundred dollars is $29,200

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):

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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:

A)0+1+3+14+14+15+16+16+17+17+18+18+18+19+20/15=207/15=13.8 is the mean

B) the eigth number in the set of fifteen is the middle number which makes 16 the median

C) 18 occurs more than any other number making it the mode

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):

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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.

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Self-critique (if necessary):

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Self-critique rating:

&#Very good responses. Let me know if you have questions. &#