Tuesday, August 26, 2008

Jounal Assignment for September holidays

Chapter 10: Measures of Central Tendency

  1. Consider a company that has nine employees with salaries of $35 000 a year, and their supervisor makes $150 000 a year. If you want to describe the typical / most common salary in the company, which measure will you use – Mean, median or mode? Explain your answer.
  2. If you are the recruiting officer for the company and want to make a good impression on prospective employees, what would you tell them?
  3. Hence, does the mean always give a realistic picture of a certain situation? Explain why.

YOUR TASK

- Answer the above questions based on your knowledge of mean, median and mode.

- This assignment is graded (upon 10 marks)

- Marks will be awarded on 1) Content & Knowledge, 2) Effort & participation

- Click on the 'comments' link below to do your assignment. Remember to write in your register number as well as your name!

last but not least, do enjoy your class outing as well as your sept hols! =)

41 comments:

Anonymous said...

1)I would use mode because I want to find the most common salary in the company as the mode of the set of data is the value that occurs most often.I would not use median because because the median of a data set is the middle value when the data are arranged in order from the smallest to largest.Its calculation involves just one or two middle values and does not take into consideration the rest of the data.Since I want to describe the most common salary ,just taking one or two middle values is not accurate as some of the most common salary are in front or behind the one or two middle salaries.I would not use mean because its calculation involves all the data and it is affected by extreme values.Hence,in some cases,it may give a misleading values.For example,the mean of the data set(35000,35000,35000,35000,35000,35000,35000,35000,35000,150000)is 46500.In this case,only one value is above average.But some may think that the most common value is 46500.But actually there is a very large value in the data compared to other values so it is not accurate.
2)I would tell them that I would see their performance in their work over the next few months since the most common salary is 35000.
3)Mean does not always give a realistic picture of a certain situation.The reason is because it is affected by extreme values.Hence,in some cases,it may give a misleading average value.

Anonymous said...

1) Use Mode. It is to describe the most COMMON salary in the company, not the Mean nor the Median. Although the answer would be the same even if we use the Median, that might not be correct/accurate in other cases. Common means most people have it which is actually the Mode. Mean is the average (affected by extreme values) while Median is the centre value when all of the values are arranged in ascending order.
2)I would tell them the Mean of our workers' salary. To make a good impression on prospective employees, we have to tell them the highest possible salary so that they would be prompted to join the company and the Mean in this case is of the highest value as both the Mode and Median are $35'000 while the Mean is $46'500. (And also, most people think that the most accurate measure is the Mean/average so they would think that once they join the company, they would get an annual salary of around $46500.)
3)No. The Mean is the average and the average is affected by extreme values (as shown in the above case) and would, in some case, give inaccurate/misleading value.

&&note that this is done by reg. 30, shu xian =)

Anonymous said...

1. I would take the mode. The mode is the most number of common data. It is not affected by extreme values, and is easy to find as it does not take into consideration of other data. The mean involves all the data, hence usually more representative of the centre of a distribution. It is also affected by extreme values and in this case, the extreme value is the salary of the supervisor so it will give a misleading average value, thus I did not use it. Although the calculation of median is not affected by extreme values like the mode, it only involves one or two of the middle values. Therefore, I would choose mode as my answer.
2. I would tell them the mean of our workers’ salaries, which is the centre of the distribution. However, I would also tell them the amount of the workers’ salary and the amount of the supervisor’s salary truthfully, to encourage them to work harder and after their efforts are recognised, they could get a pay rise of $100 000 a year.
3. No, it does not always give a realistic picture of a certain situation. Its calculation involves all the data given , usually representing more of the centre of a distribution. Another point to mention, two means can be combined to form an overall mean, which will cause the mean to change either to be higher or lower. Lastly, I want to mention that it is also affected by the extreme values in a data, hence in some cases like the one above, it would give a misleading average value.

Anonymous said...

1. I would use mode, as most of the employees have $35000 as their main salary, and only the supervisor has $150 000 as his main salary, and if we were to use mean, it would be incorrect, as for example:
(35000 x9) + 150000
10
= 46500
The mean of their salaries is very far away from the employee’s salary, which makes it quite inaccurate.
2. I would tell them that the mode of the salaries is $35000 and that this money is very good.
3. No. sometimes if you have an extreme value, the total mean would be very inaccurate. We should use the different methods of data analysis according to the different situations.

i am shuwei(28) ^^

Anonymous said...

omg... i dunno how do xPPP omgomgomg mus try!!! er bwhahahaha im jeannie(24) btw here goes xDDDD

1. I would use mode as the question already gave a huge clue *most common salary*. Mode of a set of data is the value that occurs the most often. This means that i can prove that the most typical salary in the company is $35000.
Mean is the sum of data divided by the total number of data. Although this can be useful in certain circumstances, this mean also has disadvantages, such that extreme values can affect the result. The employees have salaries of $35000 each and the supervisor has $150000, this extreme value will raise the result: [(35000 x 9)+ 150000]/10= 46500. This means that the mean has changed the 'most common salary'.
Even if we use the median, the answer would be the same. But a sudden change in a few of the employees' salaries may cause the whole result to change IF we were to use median. Therefore, Mode is still the best measure to be used.

2. To give a good impression, of course we have to show how generous my company is-- By giving the employees HIGH PAY!!!!!
Hence, i would tell them the MEAN salary, which is definitely higher than the median and mode salary. *Unlike my above friends' answers, I WILL NOT TELL THE PEOPLE THE DETAILS!!!! I WILL ONLY PUT IT IN SMALL PRINT (for legal purposes)AT THE VERY BOTTOM OF THE AGREEMENT! BWHAHAHAHA!*

3. As what i have mentioned above, of course the mean does not always give a realistic picture of a certain situation. As can be seen by this situation, the actual amount is $35000, while the mean is $45600!! THE DIFFERENCE IS HUGE!!!! When someone hears that the pay is $45600, of course people are eager to join!! but the exact amount is only $35000! this becomes a lie to people! Giving an UN-realistic picture of the situation. *TADAAA!!*

THE END!!!!!!!!!!!! xDDDDDDDD
PLEASE JOIN MY COMPANY WITH HIGH PAY OF $45600!!!!!!!!!!

Recruiting officer: Jeannie (24)

Anonymous said...

1. I would use the measure of central tendency - mode.
Reason:
The mode of a set of data is the value that occurs most frequently.
In this question, the question is asking for the most common salary.
Mean is mostly used when finding the average of the values. The mean of a data set is the sum of all the data values divided by the total number of data.
Calculation of median involves only just one or two middle values and will not be affected by the extreme values.

In conclusion, the other 2 measures do not really meet the requirement of the question. Hence, mode is the most suitable one.


2. To make a good impression on the prospective employees, I would tell them the mean values for the salary which is a more attractive offer compared to the other 2 measures -- median and mode.
Calculation:
Mean: ($35000 x 9) +$150000/10=$46500
Mode: $35000
Median: $35000


3. No. Mean does not always give a realistic picture of a certain situation.
From the above example:
Mean will be affected by extreme values.
9 employees’ salary = $35000 each
Total: $35000 x 9 = $315000
Mean = $35000
After adding their supervisor’s salary,
Total: $315000+$150000=$465000
Mean= $465000/10= $46500
The mean value is affected by the extreme value ($150000), hence the new mean value ($46500) increases after the addition of the supervisor’s salary.
Thus, it gives misleading and inaccurate values when calculation of extreme value is involved, as shown in this case.


Irene Tan Mei Chin (26)
2A

JENNIFER:D said...

1. I would use the mode.
The mode of a set of data is the value that occurs most often and what the question is asking for is the most common salary.
If we use the mean, it may be affected by extreme values and may eventually give a misleading average value.
If we use the median, it does not take in into consideration the rest of the data and two medians cannot be combined to give an overall median.

2.I'd tell them the mean values of the salary as it is more attractive than the other two measures and gives them a good impression on our company.

3.No, the mean does not always give a accurate value as it may be affected by extreme values which might lead to misleading average values

Anonymous said...

ERM,this is ANGELA btw.shocked ma haha
1.)I would use mode.BECAUSE I
IF you use mean,i would be not so accurate.The workers mostly earm quite little,and the supervisor earn so much.If you want to fnd out the most common salary,the supervisor's salary will pull up.And it will be not fair to the poor little workers.

-->35000/9 = 3888.89(without supervisor)

35000+150000/10 = 18500(with supervisor)

SEE,the difference is so big,and threfore is not accurate to use mean in this case.
If you use median,it would be not accurate as well,it will not be affected if a big amount is added in because we only choose the middle or the middle 2.But for this case we can use median.

Conclusion:I think mode is the best because it accurately shows the most people having that amount of salary.

2)I tell them i will increase the mode salary i their performance is good.=D

3.)Of cause not.Mean will be affected by an extreme value.So if you add an extreme value,it will give an unaccurate value of average.

DONE.

Anonymous said...

1. I would use mode.

Definition of mean : Average
Definition of median : the middle value when the data is arranged from smallest to largest.
Definition of mode : value that occurs most often.

Aim : To find the common salary in the company. The keyword is common.

It would not be very accurate if we used mean and median to find the common salary.

Reason 1 : As mentioned, mean is the AVERAGE of the data set. The nine employees earn $35 000 a year, and their supervisor makes $150 000 a year. There is such a huge difference between them. If i used mean, the overall result would not be fair, as the result will change depending on the vaules placed in it.
Eg: the mean salary of the company is $46500. if i add 2 more supervisor in the company, the overall result will be $63750.
This is not the common salary in the company.
Hence, im not using mean.

Reason 2 : Median is also unfair as it only chosed the middle value. Using median in this situation would be like this :
$3500, $3500, $3500, $3500, $3500, $3500, $3500, $3500, $3500, $150 000.
The median would be ($3500 + $3500)/2 = $3500
Although its the same ans as Mode, but it would also be affected if i add some other values to it. The answer depends on the middle value.
Eg: i add 9 more supervisor. The data value would be : $3500, $3500, $3500, $3500, $3500, $3500, $3500, $3500, $3500, $150 000, $150 000, $150 000, $150 000, $150 000, $150 000, $150 000, $150 000, $150 000,
The median would be : ($3500 + $150000)/2 = $76750.
This also proves that median would also be affected if i add in more values.

Reason 3 : Using mode would be the best. If i want to add any other supervisor, i would not change the common salary, unless the overall supervisor exceeds the number of workers.

Thus, using mode would be the best answer to describe the situation.

2. I would use mean.

Reason 1 : If i now include the supervisor in the company, and using median, one person working in this company will earn $3500 (which is same as mode, and very little.)

Reason 2 : If i used mean, including the supervisor, one person working in this company will earn $46500.

Can you see the difference ? Mean has a bigger value.
Therefore using mean would be gd to trick the recuits in the company.

3. No. Mean does not.
As i said, different values will affect the results if i uses mean.
Without the supervisor salary, the 9 workers' salary are $315000
WITH the supervisor salary, it is $465000
lol. big difference.
Mean only gives us the average of the situation, not the specific ones, and Mislead us, which is inaccurate.
Thus, it does not always give a realistic picture of a certain situation.

Anonymous said...

1) I would choose to use mode.
This is because it the mode will give the most suitable answer. Mode helps us to find the most common value, which is exactly what we are looking for in this case. If we use mean, we would get the average among the entire company instead. If we use median, thoug the result we get will be the same in this case, it will change if more values are added. Median is used to find the middle value. Thus, mode will be the most suitable method :)

2) I would tell them the mean salary of the company as it is higher than the median and mode salary of the company, thus it will be more attractive. But i will tell them the values of respective workers as well so that they would be motivated to work harder as the difference in the salary between an employee and a supervisor is very huge. This would make them strive to do a good job consistently :)

3) NOPE!
The mean is the average of values. If there are extremes in a range of numbers, the mean would not be accurate. The mean is easily affected and definitely does not give a realistic picture of situations with extreme values. In this case, the mean salary would give the prospective workers a false impression.

JIALIN (22)

a desperate soul... said...

Before answering the questions let us see the meanings of mean, median and mode respectively.
Mean= average of the no
Median=middle
(workings depending on the total amount of no on whether it is even or odd)
eg.
1,2,3
In this case, total amount of no is odd so we can just find the middle position
Median=2
However let us look at another case
1,2,3,4
The total amount of no is even so we need to do extra steps
Median positions= 4/2 and 4/2+1
=2nd and 3rd positions
which are =2 and 3
Median = 2+3/2
= 2.5
We can see that different workings are needed for different cases in Median.
Mode= the most common no

Let us move on to answer the questions.
Qn1) Now the questions ask us to find the most common salary of the company staff. We can derive the answer from the meanings of the mean, median and mode. As we can see the meaning of mode fits perfectly to find the answer to this questions since mode actually finds out the most common salary in this case.
(n)35000=9
(n)150000=1
Mode=$35000
So we can see from this case that $35000 is the most common salary which answer the question perfectly.
Reason for not choosing mean and median.
Mean) We are not finding the average of the salary so mean will not be useful in this case.
Median) The answer for Median may be the same as mode as shown
Let us arrange the no in ascending order:
35000,35000,35000,35000,35000,
35000,35000,35000,35000,150000
The median positions should be
10/2 and 10/2 + 1
since the total amount of no is even
which are the 5th and 6th positions
which are 35000 and 35000
so the median is
35000x2/2
= 35000
so the median is $35000 but since we are not finding the middle positions of the salaries this will not be useful in cases like this in the future.
Qn2) To make a good impression on employees, we need tell them about the good future prospects of the company so whatever workings that come out is the highest we should follow that since it will make them more willing to work hard since their future in this company will be good.
So let us start on the workings
Mean:
($35000+$35000+$35000+$35000+$35000+
$35000+$35000+$35000+$35000+150000)/10
=465000/10
=46500
Median:
Let us arrange the no in ascending order:
35000,35000,35000,35000,35000,
35000,35000,35000,35000,150000
The median positions should be
10/2 and 10/2 + 1
since the total amount of no is even
which are the 5th and 6th positions
which are 35000 and 35000
so the median is
35000x2/2
= 35000
Mode:
(n)35000=9
(n)150000=1
Mode=$35000
As we can see from the workings, actually, the mean is the highest so we can tell the employees that staying in the company would secure their future since their annual salary is $46500 which is $3875 a month which is actually quite high for an employee. This male them work harder as the more hardworking workers are paid more.
Qn3)
Mean does not give a realistic picture of a certain situation. As defined, mean shows the average of values. If the values are all low the mean would be very low as well which means that it will not attract a lot of employees in qn 2. However we must also take note that mean, median and mode are not all useful in all situations. Mean, median and mode should be used in what they are situated for. In the case of the mean, it is useful in some cases like our budget everyday if we know the average we spend everyday, we can actually make an effort to spend less and save more.

a desperate soul... said...

Real name is
Ng Yee Hang(37)
sorri forgot to put my name thr

Anonymous said...

1) I would use mode, as it is the most common value in a set of data and this makes it easier to find the most common salary in the company. It does not take into consideration the bulk of the data and it is not affected by extreme values. Therefore, it is useful as a measure of opinion and popularity and it will be more accurate as it will not be affected by the supervisor’s high pay.

I will not use Mean as it is obtained by dividing the sum of all the data by the total number of data. It involves all the data and therefore it is affected by extreme values. It will be the most reliable representative of numerical data provided only if there are no extreme values.

Also I will not use median as it is only a better representation of average than the mean when there are extreme values. Even though it is not affected by extreme values as its calculation uses only one or two middle values and it does not use the total quantity represented by the data.

Therefore I feel that mode is the most suitable method of all.

2) To give a good impression I would tell them the mean values of the salaries as it is higher than the values of mode and median
MEAN=([$35000X9]+$150000) ÷9=$46500
MODE & MEDIAN=$35000
This would of course leave a good impression on them that we are a high paying company.

3) NO, it does not.
It uses all the values in the data and therefore is affected by extreme values.
So unless there’s no extreme value the data will not be accurate.

JODIC[9]

Anonymous said...

1. The mode. It tells us the typical salary which is $35,000. Mode is a set of data that occurs most often, with 9 of the employees earning $35,000 a year, it is the salary that occurs most often as the 9 employees occupy the majority of the number of employees in the company, while the only supervisor earns $150,000 a year. Thus the mode will be able to describe the most common salary. The mean is the sum of data divided by the total number of data, it therefore tells us the average salary of the employees when the salaries are all added together and divided by the no. of employees. The mean would be :
$35,000 x 9 =$315,000
$315,000 + $150,000 = $465,000
Mean =$465000/10
=$46,500
It would not be able to describe the most common salary as the mean will be affected by the extreme value: $150,000 and it has totally changed the most common salary that is earned when it uses all the salaries. The median can also be used in this situation as it uses the middle values and $35,000 occupies the middle positions in the salaries. However it is not wise to use it in all situations, thus I will use the mode.

2. I would tell the mean salary to make a good impression on the prospective employees. The mean would be:
$35,000 x 9 =$315,000
$315,000 + $150,000 = $465,000
Mean =$465000/10
=$46,500
However using the median salary would be:
$35,000, $35,000, $35,000, $35,000, $35,000, $35,000, $35,000, $35,000, $35,000,$150,000
Median= ($35,000+$35,000)/2
=$35,000
And using the mode would be:
Mode:$35,000
Therefore the mean salary would be the highest possible salary that can be used to attract prospective employees and I will also tell them that a raise can be given in addition to the high salary that they would earn: $46,500 if they were to work hard. But how the salary :$46,500 can be driven would be printed in small at the bottom of the agreement. :p lalala
3. No. The mean adds up all the salaries of the workers and it is affected by extreme values such as in this situation, the extreme value is:$150,000. Therefore, resulting the mean salary: $46,500 which is a lie to attract people to join my company when they can only earn the likely salary:$35,000. Thus, the mean does not always give a realistic picture of situation.

Wanying said...

1.I would use mode as it is most appropriate and is not affected by extreme values.
Since, the question asks us to describe the typical / most common salary in the company and the mode of a set of data is the value that occurs most often.
I would not use the median and the mean as both are affected by extreme values.
The median involves just one or two middle values and does not take into consideration the rest of the data.

2. I would tell them the mean of the workers’ salary, because mean is affected by extreme values, so the people would be misleaded, thinking that they will have high salary if they work in the company
Mean = 9($3500) + $150 000
-----------------
10
= $46500

Which is definitely much greater than the median and mode, with them both being $3500 only.
This would of course leave a good impression on them that we are a high paying company, and we can recruit that officer within a shorter period of time.

3. So, mean, as the average of a set of data, does not always give a realistic picture of a certain situation as it is badly affected by extreme values.
Mean should be only use in survey results, in advanced statistical work, etc.
However, it can be useful to data with no extreme values as it is usually more representative of the centre of a distribution.As seen already in the above answer, the huge difference between the mean and the( median and mode).

Koh Wan Ying(11)
2A

Anonymous said...

1. I would use mode. It is not affected by extreme values, and of course, it is the most appropriate from what i see. It is the most common value in a set of data and this makes it easier to find the most common salary in the company. Therefore, it will be more accurate as it is not affected by the supervisor's high salary.
Meen and median will not be my choice as both of them will be affected by extreme values.

2. I would tell them the mean of the workers' salaries. Mean will be affected by extreme values, therefore many people may be tricked in this case, thinking that they will receive very high salaries if they were to work in the company. Adding on, the mean values of the salaries is higher than the values of mode and median, thus it is a better choice.
Mean: ($35000 x 9) +$150000/10=$46500
Mode: $35000
Median: $35000

3. From what i have mentioned above, of course the mean does not always give a realistic picture of a certain situation. From what can be seen by this situation, the actual amount is $35000, while the mean is $45600. The difference is obviously ALOT. Mean will be affected by an extreme value. So if added an extreme value,it will give an unaccurate value of average.
Therefore, the mean does not always give a realistic picture of a certain situation.

Lim Qi Min [14]

Anonymous said...

1) I would use the mode because mode represent the most common data in the table. Mean is the average of all the data and might not be accurate if there are extreme values which may cause unaccurate value of average. Median is the middle data of the whole thing and does not represent much because it only shows the middle data and does not show anything about the most common salary. Therefore if we use mode to represent the most common salary, it would be the most appropiate. Mode gives us the data which is the most common and thus mode can show us the most common salary in the company.

2)I would show the workers the mean value. As mentioned above, extreme values may affect the whole average amount and if we use mode and median the answer would be $35000 but however if we use the mean value, it would be $45600 because the supervisor salary has affected the whole average value. Therefore if we are a worker, we will be tempted by the $45600 more than the $35000.

3)Mean does not always show us a realistic picture of a certain situation. As stated above, if extreme values come to play, it would affect the whole data and might cause the data to be not accurate and false. Thus it may not be suitable in all situation and might give us the wrong data if not used correctly!

Lau Kok Ting (36)(2A)

Anonymous said...

1. I will use mode as I want to let people know the most common salary. Mode is the value that has the largest number of observations. Mean is the average, which will not be accurate as the values differ. As for median, it is the middle value but not necessary the most common salary.
2. I will use mean as the supervisor gets more salary compared to the other employees. Adding in the supervisor’s salary will increase the average pay in the company.
If I were to use mean, the average pay will be $46500 per year. For median and mode, the pay is both only $35000.
3. Mean does not always give a realistic picture of a certain situation. As the values may differ greatly, the biggest or smallest value will affect the overall results.

yiwen {27}

Anonymous said...

1)The mode is the most frequently occurring value in the data set.For example, in the data set {1,2,3,4,4}, the mode is equal to 4. The mode can be very useful for dealing with categorical data.
For example, if 9 workers are receiving $35000, and the supervisor is receiving a higher salary, the mode would show the most common salary. To show another example, if a sandwich shop is selling strawberry sandwiches and chocolate sandwiches, the mode would show the most popular sandwich.

2)The mean refers to the average, simply the sum of the values divided by the total number of items in the set. The mean is influenced by other extreme numbers of the data set, therefore increasing the average salary of the workers in the company. New recruits would be attracted by the higher average salary of the company.

3)Mean does not always give a realistic picture of a certain situation. Although it is used often, it is greatly influenced by other extreme numbers. An example is the average income.When presented with an "average" one may be led to believe that most people's incomes are near this number. This "average" income is higher than most people's incomes, because high income outliers skew the result higher.

Zeng Fan Jun(41)

-lihUi` said...

1. MODE.
If by mean:
total no of employees10.
total amount of salary (35 000 x 9) + 150 000
= 465 000
Mean of 1 employee’s salary = total amt. of salary/total no. of employees
=465 000/ 10
=$46 500
If by median:
{35 000,150 000}
Median of 1 employee’s salary= (35 000 + 150 000)/2
= $92 500
If by mode:
{35 000, 35 000, 35 000, 35 000, 35 000, 35 000, 35 000, 35 000, 35 000, 150 000}
salary that appears most = mode of 1 employee’s salary
= $35 000
Hence, we can see that by using the mode, it gives us a more typical salary. Interviewers would be able to find out a more accurate amount they would be able to receive when they enter this company. However, if we use the mean, the average salary we get is $46 500, which is not very accurate. This is because the supervisor has an extreme salary of $150 000, and upsets the balance of the mean. If we use by median, it is also not accurate as the result it $92 500, which is too far away from a normal employee’s salary, which is only $35 000.

2. If I were the recruiting officer, I would use the mode of the employee’s salary AND the salary of the supervisor. I think that being truthful is more important, as when we are in the working industry, trust counts. X) so yeah, I would tell the interviewers the mode of the salary. :D

3. Nope. Just like the example of the company, the mean is not always accurate. Sometimes, an extreme value may pop out of nowhere, making the mean of the whole amount/thing turn upside down and inaccurate. But sometimes, the mean may be useful when the values are about the same. For example, the mean of the amount of sweets in a huge bag. Thus, the mean does not always give a realistic picture of a certain situation.

Tan Li Hui [25]
{paiseh a bit messy hurhhs >P}

Anonymous said...

1)I would use mode as mode refers to The mode of a distribution with a discrete random variable is the value of the term that occurs the most often. From the first situation, there are nine workers who have the same salary as compared to the supervisor who is the only one who has a different salary from the nine workers. Hence,as the nine workers' salary occurs the most often,using mode would be the most efficient way as the mode would show the most common salary.

2)If i were the recruiting officer, i would use the mean as meaning of mean means the most common expression for the mean of a statistical distribution with a discrete random variable is the mathematical average of all the terms. So, by using the mean salary of the workers, it would attract more recruits as there will be a 'higher' amount of salary offered though it isn't the case. By using this method, it can increase the amount of recruits to come for interviews.

3)No, A mean does not always give a realistic picture of a certain situation as there may be a very extreme number appearing in the equation, like for example,there might be nine workers with a the same salary but there may be a worker who has double the salary. In these type of circumstances, mean will be inaccurate as it can be influenced by extreme numbers. Mean also does not mean that the amount you will get would be near the average. It can be lower than expected as mean is affected greatly by higher amounts as the amounts that are added up and divided can be very extreme. Hence, mean does not give a realistic picture of a certain situation.

- eLiZa - said...
This comment has been removed by the author.
Anonymous said...

1. I would use mode. We are wanting to find the most common salary in the company after all. "The mode of a set of data is the value that occurs the most often." That is what our textbooks say. Hence, this could help us find the most common salary in the company. The median of a data set is the middle value when the data are arranged in order from the smallest to the largest and the mean of a data set is the sum of all the data values divided by the total number of data. How can we use those to find the most common salary in the company? That would be totally wrong.

Using mode it would be like this :

(35000, 35000, 35000, 35000, 150000, 35000, 35000, 35000, 35000,35000)


Mode: 35000

This would be correct as we are finding the most common salary in the company.


Using mean it would be like this :

(35000, 35000, 35000, 35000, 150000, 35000, 35000, 35000, 35000,35000)

Total number of data: 10

Sum of all the data values:

35000 + 35000 + 35000 + 35000 + 35000 + 35000 + 35000 + 35000 + 35000 + 150000

= 465000

Mean = sum of all the data values/total number of data

= 465000/10
= 46500

This is definitely not the most common salary in the company. Therefore, mean is wrong.


Using median it would be like this:

(35000, 35000, 35000, 35000, 150000, 35000, 35000, 35000, 35000,35000)

150000 + 35000 = 185000

Median = 185000/2
= 92500

This is definitely not the most common salary in the company. Therefore, median is also wrong.


From all this, i can conclude that mode is the best solution for finding the most common salary in the company.


2. I would tell them that i don't judge people by their looks but only their ability. Also, if they are graduated from a well known university but slack their days in the company and hope for a promotion. I would rather give it to a person with only an 'O' level certificate but is very serious about its job. I would also hope they will do their best in job and if they have any enquirers they could ask me and always treat me like a friend instead of a boss.


3. Mean does not always give a realistic picture of a certain situation. It will be affected by extreme values. Like in this company for example, actually the salary of the employees should be only $35000 however the mean is $46500. Only because there is an extreme value of the supervisor's salary which is so different from the employees. If the employee really joined this company, they would have been 'cheated' about $11500 a month for salary. Who would want to join a company like that?


ELIZA (3) :D

Jia Yu said...

1. I would use the mode method as it is not affected by extreme values and the question also tell us to find the common salary, not average. Thus, mode will be a better choice to use.
If we use mean, it will be like this:
(35000 x 9) + 150000 = 465000
Mean of 1 employee salary = 465000/10
= $46500
If we use median, it will be like this:
{35000, 150000}
Median of 1 employee salary = (35000 + 150000)/2
= $92000
If we use mode, it will be like this:
{35000, 35000, 35000, 35000, 35000, 35000, 35000, 35000, 35000, 150000}
Mode= $35000
Therefore, we use mode as it give the interviewers a more accurate amount of money they would get if they get into the company. However, if we use mean, the average salary is $46500, which is not as accurate as the amount get from using mode method. If we use median, the amount we get will be $92500, which is very far away from an average employee salary.

2. To make a good impression on prospective employees, i would tell them the mode of the salary as I think that they have the rights to know the facts. I would not tell them the median or mean of the salary as this will be consider dishonest. Telling them the facts will also let them leave a good impression on us.

3. No. The mean adds up all the different salaries of the employees and is affected by extreme values. It is also not accurate. Thus, the mean does not always give a realistc picture of a certain situation.

jiayu(15)

Anonymous said...

1. I would use mode to describe the most common salary as it is the value that occurs most often in a set of data which is equivalent to the most common salary. I do not use mean as it will be affected by extreme values such as the supervisor salary. The median would also not be very accurate in finding the most common salary as it is the middle value of the data from the smallest to the largest and it may not always be equivalent to the most common value to appear in a data set.

2. I would use mean as extreme values has the greatest impact on it while extreme values will have lesser impact on median and hardly any impact on mode. For example, using the values given, the mode salary of the employees would be
($35 000 X 9workers)/9workers = $35 000. The median and mode would also be $35 000 as all the values are the same. When the salary of the supervisor is included, the mean salary would increased to $46 500 while there would not be any change to the mode and median of the workers salary. As the mean would give the impression of the company giving its workers high salary, it would be possible to make a good impressioon on prospective employees.

3.The mean does not always give a realistic picture of a certain situation as it is affected greatly by extreme values.

Toh Kai Bin(39)

Anonymous said...

Adelyn Sim (1)

1) I would use the measure mode. The mode of the a set of data is the value that occurs most often. If i were to use mean, the average pay a year would be $46500, but for most of the employees, they only received $35000 a year, which has a great difference of $11500. For median, although the result is the same as the mode,
Mode: $35000
Median: $35000

it might not always when the digits of their salary changes. Therefore, the best measure is MODE.

2)If i were the recruiting officer, of course i would use MEAN. The mean of a data set is the sum of all the data vaules divided by the total number of data. By using mean, a higher salary would be achieved and this "higher" pay would attract more people to come and interview for this job. Although this is an act of dishonesty, it cant be blamed as the world is realistic :D

3)No, mean does not give a realistic picture of many situations. For example, the results of 5 students are 49, 59, 70, 71 and 99. When the mean is calculated, the result is 69.6, which is 71 when rounded off. This is unfair as one of the student got 99, a distinction but the average marks is only an A2. as for the student who fail, due to the help of the student who achieved distinction, the average score had been pulled up by 21. From this example we can learn not to use mean if we want a realistic picture.

Adelyn Sim (1)

Anonymous said...

Q1. I would use mode. The mode of a set of data is the value that occurs most frequently. Hence in the case, ($35 000, $35 000, $35 000, $35 000, $35 000, $35 000, $35 000, $35 000, $35 000, $150 000) $35 000 would be the modal number since it occurs nine times. This gives an accurate picture of the most common salary of the company.

On the other hand, using mean in this case would be inaccurate considering the fact that there is one extreme value above the average. Mean involves calculating all the data and this may give misleading values in certain situations since it is affected by extreme values.

Also, I would not use median in this case. Median is the value exactly in the middle of a set of ordered numbers (ascending or descending). Although the median in this case is the same value as the mode, it only takes one or two data into consideration and not the rest of the data. Hence, median is not feasible in this situation whereby we want to find to find the most common salary.

Most common means the value that occurs the most times, therefore, we should use mode.

Q2. I would tell them the mean of the current employees’ salaries. When there is an extreme value of $150 000, the mean would be increased to the value of $ 46 500, instead of the typical salary, $ 35 000. Hence, I tell them that the mean of the workers’ salaries, the highest possible value of all three measures of central tendency. It would leave an impression that on them that the most common salary would be $ 46 500. This would entice them to join the company as they would have the false impression that they would most likely get an annual salary of $46 500.

Q3. No, it does not. It is often misunderstood by people that the most accurate measure of central tendency is mean. However, this is not always the case. In a certain situation whereby there are extreme values, the mean would either be pulled up or down. In turn, this will not give a realistic picture of the most common values of the situation.

Anonymous said...

1) I would use mode to describe the most common/ typical salary in the company. Mode is the most frequent value that appeared in the data set. One example of a mode in a data set {1, 2, 3, 4, 5, 5}, the mode will be 5 as it occured the most. In this case, it is {$35000, $35000, $35000, $35000, $35000, $35000, $35000, $35000, $35000, $150000}. So apparently the mode will be $35000.

However if we were to calculate using mean, which is to find the average, it will be inaccurate as the $150000 salary is only for the supervisor, not for the employees. The $150000 acts as a extreme value and therefore the mean will not be an accurate answer to the question.

The median is basically sorting out data set in ascending order and taking the data point in the middle of the sequence. However to find the most common, this is not applicable where only two data (or one in other cases) are used.

2) For the recruiting of officers, I will tell them the average salaries that is given to concurrent working members, which is using mean. Since there is an extreme value, it definitely helps. Both median and mode are $35000. However, when using mean, the average salary is $46500. This gives them a false impression that they are all going to get approximately $46500 per month. Although this may not be possible, but there is a reason behind it.

3)No. The mean does not always give a correct picture of a situation. In other situations, there will be more than one extreme value, which will clearly affect the result. The average does not always tell the most common salary.

Ang Teng Da (33)

Anonymous said...

1. Use mode. Mode is the value that occurs the most number of times . Hence, it is the most common occuring.Since there are 9 workers and only 1 supervisor, we use the workers as they have the common salaries. (i.e {35000,35000,35000,35000,35000,35000,35000,35000,35000,150000}
The mode would be 35000.
{LAUREEEEEEEN:D}12
2. I will present a range of the salaries which the employees in the company are earning. The lowest being 35,000 per yr and highest of 150,000 per yr depending on the person's ability and performance.Also , i could also mean , to attract more people to join my company as there is an extreme value of $150000.
This will pull up the income of other workers too.
(35000 x 9)+150000=465000
465000/10=46500
Of course, the workers would want more money (: therefore , they will join my company .

3. In this context, the mean is defined as ( total salary/ total number of employees). Hence, it only presents an aggregate of the salaries amongst employees. It does not present a realistic amount which the employee can receive based on the position he/ she will be holding.

Anonymous said...

1)I will use mode. The mode of a set of data is the value that occurs most frequently. Thus, the most common salary in the company is $35000.
However, if mean is used, the extreme values could affect the result since the mean of a set of data is obtained by dividing the sum of all the data by the total number of data. As each of the employees have salaries of $35000 while their supervisor makes $150000, the result will be different due to the extreme values.
($35000x9) + $150000 = $465000
$465000/10=$46500
If median is used, the result will be $35000 which would be the same as mode. The value exactly in the middle of a set of ordered numbers (ascending or descending) is the median. Thus it will be unfair to use median since only the middle value is selected.

2)I would tell them the mean values of the salary so that the prospective employees would want to join the company as they will be tempted for the high amount of money given to them.
($35000x9)+$150000=$465000
$465000/10=$46500
If median is used, the result as compared to mean would be smaller.
$35000, $35000, $35000, $35000, $35000, $35000, $35000, $35000, $35000, $150 000.
The median would be ($35000 + $35000)/2 = $35000
If mode is used, the result would still be smaller than the result we used using mean since the salary $35000 occurs most frequently.

3)In my opinion, I do not think that mean would always give a realistic picture of a certain situation since the mean of a set of data is obtained by dividing the sum of all the data by the total number of data. Hence, if there are any extremes values in the data, the result will be affected and would be inaccurate. Thus, the employees would be misled. As you can see, the typical salary is $35000 if we use mode to calculate but if we use mean to calculate, the salary would be $46500. There is a total difference of $ 11500 since the extreme value in the data is the supervisor’s salary ($150000).

Gan She Hwa(6) 2A (:

Anonymous said...

1.I would choose to use mode as it is not affected by the extreme values and that it is more accurate in the sense that the extreme value would not so called affect the average and give an answer that is actually too far-fetched. :)

2.If i were recruiting people to work in my company,i would use the method of mean to calculate the average pay in the company as there is an exteme value of $150000 of pay in the company.Therefore,if i calculate the mean,the value would be much higher and people would then be attracted to work in my company as there is a higher average pay.

3.No,the mean does not always give a realistic picture as when in such a case,where there is an extreme value,the average would seem much higher.But in actual fact,when i take out the extreme value,you would notice that the given averege is actually much lower.Therefore,the mean could sometimes mislead us in thinking that the average is that high.

Lye Jie Yi(16) 2A :)

Anonymous said...

1)Mean, median and mode are measures of central tendency.
I would choose mode over median and mean.

Reason for not choosing mean:
Mean of a set of data is obtained by dividing the sum of all the data by the total number of data.
Mean= sum of data/ number of data
By using mean, we will be finding the average of amount of salary a person in the company will receive.
Mean = ((35000x9+150000)/10
= 465000/10
= 46500

Reason for not choosing median:
Median is the value that is exactly in the middle of a set of ordered numbers (ascending or descending).
By using median we would be finding the middle number of the set of numbers instead of the number that appears the most.
We would start by arranging the values from smallest to largest
35000, 35000, 35000, 35000, 35000, 35000, 35000, 35000, 35000, 150000
Middle position= 4th and 5th
Median= (35000+35000)/2
= 70000/2
= 35000

Reason for choosing mode:
Mode is the value that occurs most frequently in a set of data.
This has fulfilled the requirement as stated in the question " if you want to describe the typical/ most common salary.
35000, 35000, 35000, 35000, 35000, 35000, 35000, 35000, 35000, 150000
Due to 35000 appearing the most, the mode is 35000.

2)Now that the economics is not very well, people would be more concerned about their pay. By telling them the mean pay of all the workers in the company would be something good. The mean of the pays of all the workers in the company is higher than both the median and mode of the salaries of all the workers in the salary. This can be shown by the following using the above situation.
35000, 35000, 35000, 35000, 35000, 35000, 35000, 35000, 35000, 150000
Middle position=4th and 5th
Median= (35000+35000)/ 2
= 70000/2
= 35000
Mode = 35000 (35000 appears the most)
Mean= sum of data/ number of data
= ((35000x9+150000)/10
= 465000/10
= 46500
The mean is so much higher than the mode and median. The recruits would be more attracted by the mean than the other two.Telling them the mean would give them a good impression in prospective employees.
However trust within the workers is still very important. I would tell them the salary that they would be receiving truthfully. Knowing these, they would want to work their way up to the higher salary’s position. With trust and a goal to work towards to, they would try their ultimate best to do things to the best. With such efficient way, the company would also be able to produce wonderful results.

3)Hence, mean does not always give a realistic picture of a certain situation. As stated above, the mean salary of all the workers is $46500 but this does not mean that the workers would have a salary of $46500 or more than that. Mean can only show the average salary a person receive in the company thus not being able to show a realistic picture of the situation. Mean can be affected by extreme values. The mean is greatly pulled up by the $150000 in the above situation. With the existence if any extreme values, the mean would be greatly affected since mean is obtained by adding up all the values and dividing the result with the number of values.

Done by: Sim Hui Ping (21)

Anonymous said...

Answer 1) I would choose to use the method of Mode to describe the most common salary in the company. This is because Mode calculates the number that occurs most frequently in the set, which in this case is the most common salary in the company. Whereas Median calculates the number that is in the middle of the set that is ordered from lowest to highest, so that information will not be useful in this case. Mean is also useless in this case as it calculates the average value of a set of numbers, and therefore it will be affected by the extreme value, in this case, the supervisor’s salary.
Answer 2) I would tell the prospective employees that the Mean value of the company’s salary is very high and this would impress them as this shows that no matter what job you get in the company, you will still be very highly paid. So the prospective employees feel safe to join the company without getting a lowly-paid job.
Answer3) I do not think that the Mean always give a realistic picture as in such cases, there is an extreme value, this in turn will cause the average to seem much higher than it actually is. So therefore it will not be as accurate. So the mean could sometimes be misleading and causes us to think that the average of the set of numbers is high.

Dan Jun Yuan(35)

Anonymous said...

1. For question 1, I would choose mode rather than mean and median because the mode of a set of a data is the value that occurs most often.

I choose mode because $35 000 occurs most often and we are suppose to find the most common salary in the company.

If I use mode, it will be like this:
$35 000 $35 000 $35 000 $35 000 $35 000 $35 000 $35 000 $35 000 $35 000 $150 000
Most common: $35 000

I did not choose mean because it is affected by extreme values.

Therefore, in this case, it may give a misleading average value.
If I use mean, it will be like this:
[( 9 x $35 000 ) x $150 000] / 10 = $ 46500

The above answer does not show the typical and most common salary in the company.

I did not choose median because it only involves just one or two middle values and does not take into consideration the rest of the data.

If I use median, it will be like this:
$35 000 $35 000 $35 000 $35 000 $35 000 $35 000 $35 000 $35 000 $35 000 $150 000
Middle position: 5th and 6th position
Median: ( $35 000 + $35 000) / 2 ) = $35 000

Although the answer is still the most common salary, it is not the best way in compared with mode.

2. For this case, if I am the recruiting officer for the company and want to make a good impression on prospective employees, I will tell them the mean salary as it is the highest among mode, mean and median.

If I use mean, it will be like this:
[( 9 x $35 000 ) x $150 000] / 10 = $ 46500

If I use median, it will be like this:
$35 000 $35 000 $35 000 $35 000 $35 000 $35 000 $35 000 $35 000 $35 000 $150 000
Middle position: 5th and 6th position
Median: ( $35 000 + $35 000) / 2 ) = $35 000

If I use mode, it will be like this:
$35 000 $35 000 $35 000 $35 000 $35 000 $35 000 $35 000 $35 000 $35 000 $150 000
Most common: $35 000

Therefore, by choosing mean, people will think that by joining our company, they can get a higher pay, $ 46500 rather than $35 000. Thus, they will be attracted to the pay, and join the company willingly.
However, although I will probably use mean, it is still not the best way as people will feel betrayed and will not do their best for the company after they joined as it not a good representation of the average pay.

3. Hence, after I come across these two questions, I realise that mean does not always give a realistic picture of a certain situation.
Although, mean calculation involves all the data and it is usually more representative of the centre of a distribution, it is affected by extreme values thus resulting in a misleading average values.
In the above case ( question 2 ), the mean salary is $ 46 500. However, it does mean that any worker that work in this company will get $46 500 thus it does not give a realistic picture of this situation.
However, in some cases, if there is no extreme values, mean is usually a good way to represent the centre of a centre of a distribution.
Thus, mean has its advantage and disadvantage.

Yvonne Thung ( 29 )

Anonymous said...

XinLu (20)
1) I would use the mode as in this case as it had asked me to describe the typical/most common salary in the company . Mode of the set of data is the value that occurs most often.It does not take into consideration of other data, therefore , it represents more of the centre of a distribution which the question had asked us to .

However , we cannot use median here even though the answer would be the same . If there are some changes in some employees' salaries , it might cause the final result to change too . Therefore , it is inaccurate to use the median .

ALso , mean is frequently used in finding the average of values , which is
mean= total sum of data values / total number of data .
Mean is the average of all the date values and is affected by extreme values too .

Therefore , Mode is most suitable to be used here .

2) Firstly , I will tell them the mean of the company s' salaries . Then , be truthful to them and tell them their actual salary , if not , once they joined the company and found out that their actual salaries were much lower than the MEAN one , they would be disappointed . We can tell them the difference in pay of a normal staff and a supervisor , to give them MOTIVATION to work hard and perform well to becomea supervisor to get a much higher salary . This way , the company will thrive if all workers worked hard and the workers themselves will get a higher pay and benefit from it too .

3) No , the mode does not always give a realistic picture of a certain situation . As shown from the example a supervisor's yearly revenue can be more than several normal employees yearly revenue , therefore it does not give a realistic picture of a certain situation . The final results involves all the data given to us , representing the centre of a distribution instead . It is also affected by the extreme values as shown above and would then give inaccuratate/misleading value .

Anonymous said...

Joel yeo ying yao (40)

1)I would choose to use the mode method as it shows the value that occurs the most frequently in a data set.
The median method would not be accurate as the results only shows the centre of the data.
The mean method is also not accurate as it shows the average of their total income.

2)I would tell them the mean of the company's salary as it is affected positvely even if there is only one extreme high salary compared to the other workers.
The median and the mode would not be affected as greatly as compared to the one calculated in the mean method if there is at least one extreme high salary.
Thus, the prospective employees would be attracted to work in the company due to the 'high' pay.

3)No. Even though the mean may seem realistic as it shows the average value of the data set, it is affected if there is an extreme value in the data set. However, in cases whereby there is no extreme value in the data set, it would be more realistic and useful.

Joel yeo ying yao (40)

Anonymous said...

Dear All,

Just leaving some comments to clear a few misconceptions that you all may have about mean, mediand and mode. I will quote some mistakes that your classmates have made in their responses. Pls don't take it personally, instead use it as an opportunity to learn!

Misconception 1:

Mean
-->35000/9 = 3888.89(without supervisor)

35000+150000/10 = 18500(with supervisor)

Clarification: Each worker earns $35000 a year, so the mean is not calculated by $35000/9.
Instead, the mean salary (without including the supervisor) should be ($35000 X 9)/9 = $35000.
Mean salary (including supervisor) should be
($35000 x 9 + 150000)/10 = $46 500

Misconception 2:

Using median it would be like this:

(35000, 35000, 35000, 35000, 150000, 35000, 35000, 35000, 35000,35000)

150000 + 35000 = 185000

Median = 185000/2
= 92500

Clarification:

Before you take the average of the middle 2 values to obtain the median, first you need to arrange the data values in ASCENDING ORDER!!
This is very important, pls don't forget!!
Hence median should be found this way instead:
(35000, 35000, 35000, 35000, 35000, 35000, 35000, 35000, 35000,150000)

Median = (35000 + 35000)/2 = 35000

In this case, the median HAPPENS TO give the same value as the mode. However this is COINCIDENTAL.
The MOST COMMON salary refers to the MODE. So median cannot be used for Qn 1.

Hope it helps! =)

Anonymous said...

1) Since we are finding the most common salary in the company, I would use the MODE as the mode of a set data is the value that occurs most often and it is not affected by extreme values. Mean is affected by extreme, hence in some cases; it may give a misleading average value. Even though median is not really affected by extreme values, its calculation involves just one or two middle values and does not take into consideration the rest of the data.

2) I would tell them the mean of the entire worker’s salary. The mean of the salary is the highest possible salary and a high salary will tempt the prospective employees to come and join the company. The MEAN is $46500 while the MEDIAN and the MODE is $35000.

3) No, it does not always give a realistic picture of a certain situation. Its calculation involves all the data. Hence, it is usually more representative of the centre of a distribution. It is affected by extreme values. Hence, it may give a misleading average value.

Eugenia Teh Shu Wen (4)
2A

Anonymous said...

1)The mode.
Mode is frequency of a data. With 9 employees earning $35,000 a year, it is the salary that occurs most often since the 9 employees occupy the majority of the number while the only supervisor earns $150,000 a year, which is more than 4 times as compared with the salary the other 0 employees received. Thus the mode will be able to describe the most common salary.
The mean is the sum of data divided by the total number of data, therefore it tells us the average salary of the employees when the salaries are all added together and divided by the total muber of employees and the superviser:
$35,000(9) + $150,000=$465,000
Mean =$465,000/10
=$46,500
The above calculation shows that using the mean method is unable to describe the most common salary as the mean will be affected by the extreme value, $150,000 and it has exceed the most common salary that is earned by $110,000 when using the mean method.
The median method can also be used in this situation as it uses the middle values and $35,000 occupies the middle positions in the salaries. However it is not wise to use this method in all situations. Hence, I will use the mode.

2)To make a good impression on the prospective employees, i will tell them the mean values as it is more attractive as compared to the other 2 methods' values. The mean would be:
$35,000(9) + $150,000=$465,000
Mean =$465,000/10
=$46,500
The median salary would be:
$35,000, $35,000, $35,000, $35,000, $35,000, $35,000, $35,000, $35,000, $35,000,$150,000
Median= 5th + 6th position/2
=($35,000+$35,000)/2
=$35,000
The mode salary would be: $35,000
Therefore the mean salary would be able to attract the prospective employees since it is the highest value among the 3 methods used.

3. No.
It does not always give a realistic picture of a certain situation. (E.g.the situation given.) The mean adds up all the salaries of the workers and it is affected by extreme values which is the supervisor's salary, and therefore, resulting in $46,500 which is not the most typical salary in the company.

Jaslyn Tay

✖jIng__ru♥♥ said...

ms chum!!!! i just check and found out that the comment wasnt submited... and i hav to do it again....

1. I would use mode to describe the typical salary in the company. The nine employees earn $35,000 each, which cause it to be a mode for all the amount of salary as it is the highest repetition of the same value.

i wont use mean because it is not the most accurate factor to describe about the values.

i also wont use median because it is not regarding median here.

2. to the employees, i would tell them about the mean as it would cause the value of the salary to be higher. therefore the amount of salary would appear to be more attractive than the actual amount.

3.no. it does not always give us a clear impression of the real value. As shown in the above situation, we can see that having the value calculated by mean method would give us a misleading idea, thinking that the salary is that high. thus this method does not always give a realistic picture of a certain situation sometimes.

sorry for late submitting of this assignment. i only know that it wasnt posted today.

Wong Jing Ru(31)

Anonymous said...

1. I would choose to use mode. In the case, the question stated to find the most common type of salary.
Mode is is the value that occurs most often in a data set.
It cannot be affected by the extreme value, in this case is $150 000 and there is only 1 supervisor, compared to the other 9 employees having the same amount of salary. For this scenario, the value that occurs the most time is actually the salary of the 9 employees with $35 000.
Compared to mode, median is not the most ideal way to describe the most common type of salary as it gives the average of the total sum of the salaries of 1 supervisor and 9 employees.
Mean= [35 000(9) + 150 000] /10 = $46 500
And in this case, it is not accurate as both the supervisor and the 9 employees do not receive this amount of salary. The employees receive lesser than $46 500 while the supervisor still receive higher amount of salary. Thus this proves that mean can be affected by extreme values.
median, is more accurate the mean but not as accurate as mode.
S= {35 000, 35 000, 35 000, 35 000, 35 000, 35 000, 35 000, 35 000, 35 000, 150 000}
There is even number of elements. Thus we need to add the 5th and 6th salaries up and divide it by 2. Since the middle position is $35 000, the mean we get is still $35 000. But if the 2 numbers are different, we might not get the most accurate reading.
*thus I would choose mode*

2. As for this, I will choose to use mean. As the supervisor’s salary is $150 000, it has a great difference between with any of the employee’s salary, $35 000. Because of the difference in salary of $115 000, it is not possible to use mode as it shows the employees salary of $35 000. It will not be attractive enough to attract the prospective employees. In this case, using median will not help as the calculated answer will still be $35 000. By using mean, the average salary will increase due to the extreme value, $150 000.
Total salary= 35 000(9) + 150 000
= $465 000
Mean= 465 000/ 10
= $46 500.
So mean provides a higher salary which is not too high as $150 000 and not too low as $35 000.

3. Mean does not always give a realistic picture of certain situation. Mean can be affected by the extreme values if there is any. In the above situation, the mean calculated was $46 500 and it is higher than the original employee’s salary. Thus it gives people a misleading idea that the salary is high, but the actual fact, it is not. But in cases where there is no extreme value, mean can be realistic and helpful.

Sorry Ms Chum, I enter the comment box for weeks.. Sorry for the late submission. (:

Done by: Chen Liyun (02)