QA14

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course Mth 152

5:16pm July 24

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.

014. mean vs median

Self-critique:

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Question: `q001. Note that there are 8 questions in this assignment.

{}{} What is the average, or mean value, of the numbers 5, 7, 9, 9, 10, 12, 13, and 15? On the average how 'far' is each number from this mean value?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

80/8=10 average

5+3+1+1+0+2+3+5=20

20/8=2.5 difference

confidence rating #$&*

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Given Solution: To get the mean value of the numbers, we first note that there are eight numbers. Then we had the numbers and divide by eight. We obtain 5 + 7 + 9 + 9 + 10 + 12 + 13 + 15 = 80. Dividing by 8 we obtain

mean = 80 / 8 = 10.

The difference between 5 and the mean 10 is 5; the difference between 7 and the mean 10 is 3; the difference between 9 and 10 is 1; the differences between 12, 13 and 15 and the mean 10 are 2, 3 and 5. So we have differences 5, 3, 1, 1, 0, 2, 3 and 5 between the mean and the numbers in the list. The average difference between the mean and the numbers in the list is therefore

ave difference = ( 5 + 3 + 1 + 1 + 0 + 2 + 3 + 5 ) / 8 = 20 / 8 = 2.5.

Self-critique:

OK

Self-critique rating #$&*3

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Question: `q002 What is the middle number among the numbers 13, 12, 5, 7, 9, 15, 9, 10, 8?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

5,7,8,9,9,10,12,13,15

9-middle number

confidence rating #$&*

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Given Solution: It is easier to answer this question if we place the numbers in ascending order. Listed in ascending order the numbers are 5, 7, 8, 9, 9, 10, 12, 13, and 15.

We see that there are 9 numbers in the list. If we remove the first 4 and the last 4 we are left with the middle number. So we remove the numbers 5, 7, 8, 9 and the numbers 10, 12, 13, and 15, which leaves the second '9' as the middle number.

Self-critique:OK

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Self-critique rating #$&*3

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Question: `q003. On a list of 9 numbers, which number will be the one in the middle? Note that the middle number is called the 'median'.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Between the first 4 and last 4 numbers.

confidence rating #$&*

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Given Solution: If the 9 numbers are put in order, then we can find the middle number by throwing out the first four and the last four numbers on the list. We are left with the fifth number on the list.

In general if we have an odd number n of number in an ordered list, we throw out the first (n-1) / 2 and the last (n-1) / 2 numbers, leaving us with the middle number, which is number (n-1)/2 + 1 on the list.

So for example if we had 179 numbers on the list, we would throw out the first (179 - 1) / 2 = 178/2 = 89 numbers on the list and the last 89 numbers on the list, leaving us with the 90th number on the list. Note that 90 = (179 - 1) / 2 + 1, illustrating y the middle number in number (n-1)/2 + 1 on the list.

Self-critique: 3

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Question: `q004. What is the median (the middle number) among the numbers 5, 7, 9, 9, 10, 12, 13, and 15?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(8-1)/2=3.5

Median is between 9 and 10 so 9.5 is the median

confidence rating #$&*

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Given Solution: There are 8 numbers on this list.

If we remove the smallest then the largest our list becomes 7, 9, 9, 10, 12, 13.

If we remove the smallest and the largest from this list we obtain 9, 9, 10, 12.

Removing the smallest and the largest from this list we are left with 9 and 10.

We are left with two numbers in the middle; we don't have a single 'middle number'. So we do the next-most-sensible thing and average the two numbers to get 9.5. We say that 9.5 is the middle, or median, number.

Self-critique:OK

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Question: `q005. We saw that for the numbers 5, 7, 9, 9, 10, 12, 13, and 15, on the average each number is 2.5 units from the average. Are the numbers in the list 48, 48, 49, 50, 51, 53, 54, 55 closer or further that this, on the average, from their mean?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Mean-51

3+3+2+1+0+2+3+4=18/8=2.25

confidence rating #$&*

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Given Solution: The mean of the numbers 48, 48, 49, 50, 51, 53, 54, and 55 is (48 + 48 + 49 + 50 + 51 + 53 + 54 + 55) / 8 = 408 / 8 = 51.

48 is 3 units away from the mean 51, 49 is 2 units away from the mean 51, 50 is 1 unit away from the mean 51, and the remaining numbers are 2, 3 and 4 units away from the mean of 51. So on the average the distance of the numbers from the mean is (3 + 3 + 2 + 1 + 0 + 2 + 3 + 4) / 8 = 18 / 8 = 2.25.

This list of numbers is a bit closer, on the average, then the first list.

Self-critique:OK

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Self-critique rating #$&*3

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Question: `q006. On a 1-10 rating of a movie, one group gave the ratings 1, 8, 8, 9, 9, 10 while another gave the ratings 7, 7, 8, 8, 9, 10. Find the mean (average) and the median (middle value) of each group's ratings.

Which group would you say liked the movie better?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

First group-

1+8+8+9+9+10=45/6=7.5 average 8.5 median

7+7+8+8+9+10=49/6=8.16 average 8 median

The 2nd group like the movie better because their average is higher

confidence rating #$&*

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Given Solution: The mean of the first list is (1 + 8 + 8 + 9 + 9 + 10) / 6 = 45 / 6 = 7.5. The median is obtained a throwing out the first 2 numbers on the list and the last 2 numbers. This leaves the middle two, which are 8 and 9. The median is therefore 8.5.

The mean of the numbers on the second list is (7 + 7 + 8 + 8 + 9 + 10) / 6 = 49 / 6 = 8 .16. The median of this list is found by removing the first 210 the last 2 numbers on the list, leaving the middle two numbers 8 and 8. The median is therefore 8.

The first group had the higher median and the lower mean, while the second group had the lower median but the higher mean. Since everyone except one person in the first group scored the movie as 8 or higher, and since everyone in both groups except this one individual scored the movie 7 or higher, it might be reasonable to think that the one anomalous score of 1 is likely the result of something besides the quality of the movie. We might also note that this score is much further from the mean that any of the other scores, giving it significantly more effect on the mean than any other score. We might therefore choose to use the median, which limits the otherwise excessive weight given to this unusually low score when we calculate the mean. In this case we would say that the first group liked the movie better.

Self-critique:

I got the median as being 8.5 on 2nd group, I understand though. I refigured it and I got 8.166 on the second try.

But I don’t understand why we use the median to determine who liked the movie better rather than using the average. I’ve reread but it may not be clicking with me.

The 1 rating has a lot of influence on the mean, much more than would a rating of 10, since the mean is much closer to 10 than to 1. However the 1 has no more influence on the median than would any other number below 8. If we wish to avoid the undue influence of anomalous ratings, the median is the better measure.

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Question: `q007. Suppose that in a certain office that ten employees make $700 per pay period, while five make $800 per pay period and the other two make $1000 per pay period. What is the mean pay per period in the office? What is the median?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

700*10=7000

5*800=4000

2*1000=2000

700+700+700+700+700+700+700+700+700+700+800+800+800+800+800+1000+1000=13000/17=764.70 mean 700-median

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Given Solution: There are a total of 10 + 5 + 2 = 17 employees in the office. The total pay per pay period is 10 * $700 + 5 * $800 + 2 * $1000 = $13,000. The mean pay per period is therefore $13,000 / 17 = $823 approx..

The median pay is obtained by 'throwing out' the lowest 8 and the highest 8 in an ordered list, leaving the ninth salary. Since 10 people make $700 per period, this leaves $700 as the median.

STUDENT QUESTION:

Is it typical to use the median value if there are ‘oddball’ scores in a group?

INSTRUCTOR RESPONSE

A few 'oddball' scores have little effect on the median, but can have a great effect on the mean.

Other factors can also be important depending on the situation, but if a lot of 'oddball' scores, or 'outliers', are expected the median is often the better indication of average behavior than the mean.

Self-critique:

I got 13000/17=764.70 as the average not 823?

your calculation is correct

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Question: `q008. In the preceding problem ten employees make $700 per pay period, while five make $800 per pay period and the other two make $1000 per pay period; we just found that the mean pay per period was $823. On the average, how much to the individual salaries differ from the mean?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

823-700=123

823-800=23

1000-823=177

123*10+5*23+2*177= 1699/17=99.94

confidence rating #$&*

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Given Solution: The mean was found in the preceding problem to be $765. The deviation of $700 from the mean is therefore $65, the deviation of $800 from the mean is $35 and the deviation of $1000 from the mean is $135.

Since $700 is paid to 10 employees, $800 to five and $1000 to two, the total deviation is 10 *$65 + 5 * $35 + 2 * $235 = $1295. The mean deviation is therefore $1295 / 17 = $76.18 , approx..

I did the problem from $823.00. I found the answer the same way. I don’t know if I’m confused or if theres a typing error because I got 765 for the average for the original problem(previous problem) but the average was 823 in the problem but then in the answer it was 765. Let me know if I’ve missed this completely

"

Your original calculation was correct.

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

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