*Homework discussion:* Measures of central tendency

Hello, everyone. Below are some submitted answers to the homework covering material from the Measures of central tendency lecture, as well as my commentary. Have a listen and feel free to ask questions in the comments section below.

As always, there are many ways to say the same thing and many ways to say correct things!

*With all lecture audio, please be careful about your computer volume and your use of headphones or earbuds.*

Note the particular responses I’ve underlined below. These responses are the ones that stood out as most correct and least problematic. How does your response differ from these (if yours wasn’t selected)? What part of your response could be improved to make it more perfect? (It’s often just one word!) Don’t be discouraged. Responses often are largely correct but ultimately include irrelevant or contradictory or just vague elements that downgrade them. These homeworks are for your benefit to help you understand how to think critically and clearly and to prepare you for the exams and how I’ll grade you, as well. Just imagine if I was deducting points for every little bit that wasn’t perfect! We’re here to learn this summer, so let’s learn from yours and the mistakes of your classmates. Feel free to comment below.

**Join the discussion in the Comments section. What answers are better than others and why? Feel free to specify Question # and Response # when you post (Q#R#)!**

**1. Given that µ = 13.74, Mdn = 13, and Mo = 13, what else can you tell me about our first distribution? That is, how might we describe this distribution’s ‘shape’, given the values of the three measures of central tendency I just provided? Why?**

In lieu of audio, please see some comments posted below in red!

*Actual answers from students this term:*

*Actual answers from students this term:*

- 1. For the first graph we can tell that the numbers are distributed equally. I can’t tell that. I think you mean that there is a normal distribution, but that is not the same thing as saying the numbers (by which you mean observations, or scores) are distributed equally. What does ‘equally’ mean here? We can see this because of the bell curve shape. Another way we can see this is because the median, mean, and mode are all relatively close (if not all) at 13. This forms a bell graph on the graph and ultimately that the numbers are distributes normally.
- 1. According to the given values and the shape of the graph, it can be seen that there is normal distribution in this data set. This is because the mean, mode, and median all have roughly the same values and fall in the middle of the the graph with a normal bell-shaped curve.
- 1. I would describe the distribution as perfectly symmetric because the mean, median and mode are all about the same and that information makes it have a normal curve. I might quibble about ‘perfectly’, but it certainly appears symmetrical enough!
- 1. The graph is very similar to the Normal Distribution graph because the mode and median were the same, and the mean was almost the same as well. It shows the normal curve.
- 1. The graph shows that the central tendency is symmetrical with the data. What does it mean to say ‘the central tendency is symmetrical with the data’? The graph or the distribution is symmetrical. Meaning that the mean, median, and mode are all in the middle of the graph and it slopes off to both sides relatively equal. Slope has a very specific meaning that we’ll see when discussion correlation and regression. I wouldn’t use it hear, especially because of the ‘curve’ of our distribution and that we’re not dealing with a ‘line’.
- 1. The distribution of this bar graph, and the value of both the mean (13.74), Mdn (13) and the mode MO (13) being so close arithmetically, is a fine indicator of the resulting information’s validity and reliability. There is a very symmetrical distribution in this graph with no extreme asymmetrical right or left distributions. What does ‘with no extreme asymmetrical right or left distributions’ mean? There is only one distribution. It’s your data. Even with a few high (25) and low (0? Were they watching or sleeping?) outliers, the majority of the populations answers fall within the middle range, and the proximity of the three numeric values for mean, Mdn and Mo “provide a very good indicator of what is true with respect to central tendency” (Ramey, 2015).
- 1. Since the mean, median, and mode are very close in value, I would expect the distribution to resemble a normal bell-shaped curve.
- 1. The first curve is very close to a normal distribution curve which means that the mean, median and mode are all the same number or very close to the same number, i.e. 13 in this example. I don’t think the second sentence adds much. This occurs because the number that occurs the most also happens to be the middle number of the set of data, as well as the mean of the data.
- 1. Based on the information given, the shape of the distribution given would be a normal curve. Although the mean, median, and mode aren’t exactly the same shape it is still close enough. Shape characterizes a distribution, not measures of central tendency like mean, mode, and median. Be careful as shape is really about variability in a way. It can’t be skewed negatively, because the mode doesn’t have the highest value, followed by the other two. It also can’t be skewed positively because the median and the mode are of equal value instead of the mode having the lowest value. Think about skew in terms of the pull of mean toward extreme, infrequent scores. Modes don’t care about infrequent scores, extreme or not. Medians are only concerned with adding a score to a distribution’s ends.
- 1. The distribution appears symmetrical, or nearly symmetrical. At 13, the mode and median are equal. Because the mean is slightly higher, at 13.74, the distribution may be slightly skewed positively to the right. I would not think that this difference is very significant, but it is technically not the same value. This is not a normal curve, because the scores are not evenly distributed with 50% below the mean. I know what you mean by significant, but don’t use the term. Reserve it for ‘statistically significant’. Otherwise just use ‘important’ or ‘big’, etc. I don’t understand the last sentence. If we’re dealing with a normal distribution the mean is equivalent to the mean (or roughly in real life) and so 50% of a distribution (or near enough in real life) is below the mean.
- 1. This graph is unimodal and resembles the shape of a bell curve. It’s mostly symmetrical and does not show much of a skew even though there are some outliers in the data. We know that the graph is not skewed also because the three measures of central tendency were either very close or equal to each other. In this case, basically very close to 13.
- 1. Given that the mean is 13.74, the median is 13 and the mode is 13, it can be said that the first distribution is a normal distribution. This distribution’s shape would be considered a bell-shape curve because in this case the mean, median and mode are all approximately the same number.
- 1. It is consider a normal curve in which the mean, mode and median are all roughly the same, making it a symmetrical and unimodal distribution. -unimodal: one value, 13, having a larger frequency (approx 23) more than any others
- 1. The first distribution could be described as ‘normal’ distribution since every score is generally equal (that is all ~13). No, not every ‘score’ is equal. If that was the case, we’d be dealing with a constant, not a variable’s distribution. You mean every measure of central tendency. The next sentence doesn’t erase the first sentence, so be careful in phrasing. In this instance, the three measures of central tendency are similar (not unique)—mean, median and mode are all roughly the same and would be represented as normal or similar frequency.
- 1. The data is slightly skewed to the left, because the mean is 13.74, which is higher than the mode and median, which are both 13. Also, the mean value is to the left of the peak. It is skewed to the right (not really, but a reasonable thing to speculate on). The mean is pulled to the extreme, infrequent high scores to the right. Some high score is making the calculated mean higher, because the mean has to take into account all scores (sigma x, remember). Counterintuitively, I know, skew is not about the bulge. It is about the elongated tail.
- 1. This would be an almost perfect normal distribution that should be almost perfectly symmetrical, because the Mo, Mdn, and M (is there a trick to making X-bar?) but it might be skewed just a bit to the right because M is 13.74 while the other two values are 13.
- 1. We’d expect a distribution shape that is neither skewed negatively or positively because of the closeness in value of the mean, median and mode.
- 1. The shape of the first distribution is normally distributed because the three measures of central tendency (mean, median, and mode) are all nearly the same number, which is 13.
- 1. Due to the fact that the mean, median and mode are very close to equal, it is evident that the distribution is very close to symmetrical. Even though the mean is very slightly higher, the median and mode are equal (instead of the median being higher than the mode) so a positive skew is not really notable. From the fact that the shape of the distribution is roughly symmetrical, we can call the sample distribution roughly Normal.
- 1. The first distribution has a slightly positively skewed distribution, because the mean is greater than the mean or median. If it was equal to then it would be perfectly symmetrical.
- 1. The first distribution is symmetrical. This means that the data is not skewed either positively or negatively. The data falls into a bell graph shape meaning that it is perfectly symmetrical rising from the left side to an upmost peak then declining on the right side until a plateau. I’m not sure of your use of plateau. The tails are asymptotic which means they go left and right and get closer and closer to the
*x*-axis without ever reaching it. - 1. The shape of the distribution is almost perfect, a normal curve. There are a few outlying answers but overall the shape is symmetrical. Symmetry has nothing to do with ‘outlying answers’, so your sentence is odd. I can have or not have extreme scores in a distribution and it be symmetrical; it’s about mirror images of the left and right half, is all. It shows that most people observed the same thing, around 13 punches. That is why the mean, median and mode are 13.
- 1. This set of data is almost perfectly symmetrical because the median and mode are the same number (13) and the mean is almost the same (13.74). If it was perfectly symmetrical the mean, median and mode would have all been the same number.
- 1. The data represent a bell-curve, which shows the distribution of the scores. The central tendency of these scores proves that there is variability in the data. The mean of a distribution of values like 14, 14, 14, 14 = 14. So the mean doesn’t prove variability. Also, don’t use the word prove! Everything in science is about probability, as mentioned before and soon to come. Proofs exist in geometry. The few outliers affected the mean, which is why in this case the median would be most accurate, it would not be affected by the outliers. I like what you want to say here, but say more. What outliers, in what direction, for example?
- 1. Overall the distribution shape would look similar to a perfect normal curve because the mean, median, and mode are virtually the same value. The mean is slightly higher than the median and mode indicating the distribution is marginally skewed in the positive direction.
- 1. From the first distribution you can see a bell shaped curve with a majority of participants reporting around the 13 and 14 mark and a majority of the individuals falling between 5 and 20 punches.
- 1. Well 13.74 is the mean where they added all the numbers together and then divided by the total number of people who watched. The number in the middle would be 13 and the number said the most or the mode would be 13 too. This would show that over all the number 13 would be that was most common number used or the number that would most likely be the answer on how many times he punched. It also looks like it is a little skewed to the left. To the right! (see discussion above) The outlier of 20-25 and the zero could have been confused on what counted as a punch.
- 1. The first distribution would be a bell-shaped curve. Most participants counted between 5 and 20 punches, with a few outliers. Within that, most participants observed between 13 and 14 punches. The measure of central tendency is an almost perfect symmetrical distribution, with the three measures being very close to each other. The distribution overall is almost perfectly symmetrical distribution. ‘The measure of central tendency is an almost perfect symmetrical distribution’ doesn’t make sense. If you mean the
*three*measures of central tendency are almost identical, that is true, but we would just characterize them as nearly identical and would not call them ‘symmetrical’. - 1. The distributing is mostly symmetrical since the values seem to be highest (you mean most frequently occurring!) in the middle and gradually become lower the farther they are from the middle; however, it is slightly positively skewed since the mean is greater than the median and the mode. I might only want you to use the actual calculated measures of central tendency in your answer, as well.
- 1. For the first distribution, it can be concluded from the data that the graph is a normal distribution. Although it has several small outliers, the graph is almost a perfectly symmetrical distribution. Why?

**2. Report to me your calculated mean, median, and mode for this second distribution. Take great care to describe in words your use of proper notation and ‘spell out’ your steps so that you can show all of your formulae and work.**

Many answers were pretty good. Here are some points on a few of them. In lieu of audio, please see comments posted below in red!

*Actual answers from students this term:*

*Actual answers from students this term:*

- 2. The second chart was referred to as the “second distribution” and varies from the first, so my calculations were based on the sample equation, not population. The calculated mean is 14.44. To find that, sigma x was calculated (11+12+12+12+13+13+13+13+14+14+15+16+16+16+16+17+18+19) and that was divided by 18 (n=18) which equals 14.44. The median was 14. n is an even number, the two middle scores are both 14. So (14+14)/2=14. The mode is 13 and 16. Both numbers appear most frequently — four times each.
- 2. Mean=14.4 Two decimal places for all calculations. Thanks! Median= Missing! Mode= 13 and 16. In order to find the mode, I just had to find which number(s) occurred most frequently. For this data set, there were two modes: 13 and 16. To find the median, I put the data in rank from lowest to highest. Although N is an even number, Tell me why
*N*as opposed to*n*. the two remaining numbers were 14; therefore, the median for this data set is 14. Lastly, in order to find the mean for this data, I took the list ranked from lowest to highest that was used to find the median and added the figures together; it totaled 260. N=18, so i was able to take 260 divided by 18 and discover that the mean is 14.4. - 2. The median of this data is 14. Correct answer, but it’s accidental, and so ultimately wrong! The number of observations is incorrect. Double and triple-check. No partial credit will be given. Consider what would happen in a real life lab. Every datum matters. Be crazy vigilant. There are 17 values, so taking 17 +1 and then multiplying it by .5 equaled 9. This means the median is the 9th value, or in this case, 14, with there being 8 values before and after it. There are two modes, 13 and 16. I looked at the graph and saw that the maximum frequency was 4 and that this frequency was shared by two numbers, 13 and 16. The mean is 13.53. The scores were all added together to equal 230 and then divided by the number of values, 17, to end up with the quotient 13.53. The
*n*(whether you mean*n*or*N*) is wrong and so your mean is too, unfortunately. Make sure to indicate what formula you’re using, so conceptually you’re indicating whether you think this is a sample or a population. - 2. The mean is the average of all the number so you add them together then divide by the total number. List of numbers: 11,12,12,12,13,13,13,13,14,14,15,16,16,16,16,17,18,19 In the power point for this week the it was also said the equation for it is X={(sum)x}/n. This meaning the same thing as the sum of all the numbers divided by the total amount of numbers you added. So the mean would be (all the numbers added together) 272/18, which equals 15.1 Your sum (sigma
*x*) is off and so your mean is ultimately wrong. Median would be the value in the middle (on a scale highest to lowest), which that is 14. The mode is the numbers that occur the most or show up in the line the most. What’s ‘the line’? If the number occur the same amount of times as in and both the most they both would be the mode. For this example the mode would be 13 and 16 - 2. Mean is calculate by taking the number of counts and adding them together and then dividing that number by the different punches so 9. Ex: 1+3+4+2+1+4+1+1+1 = 18/9 = 2 for the count. but for the number of punches the mean would be 15 because 135/9. Your sigma
*x*and your*n*are incorrect. In the graph, several observations occurred multiple times. The mean is the average*x*, not the average number of times these*x*s occurred. The Median will be the number in the middle when they’re organized lowest to highest. For the count this would be 1,1,1,1,1,2,3,4,4 so in this case would be 1. For the number of punches it would be 15 as it falls directly in the center. The mode on the other hand is the number that occurs the most often. For the count this would be 1 as it appears most frequently in the data. - 2. First, I calculated the mean. In this situation, the data represents the entire population so I would use the formula for the population mean. Tell me why it’s the population. The mean is represented by “mu” with a subscript x. So μx equals the sum of the scores (Σx) divided by total number of observations in the population (N). The sum of the scores is 260. There were 18 total observations so N = 18. 260 divided by 18 is 14.4. Two decimal places please. So the mean of the population 14.4 punches. I calculated the median next. There are 18 scores, so the median score is at the 9.5th location. The scores at this location are 14 and 14. So the median score is 14. Finally the mode. In this distribution, two scores tie for the highest frequency. The score 13 and the score 16 both occur most often. Therefore, the distribution is bimodal with a mode of 13 and 16.
- 2. Considering the fact that there are only 17 observations here, Uh-oh! Recount. while the previous distribution had 72, I will treat this data set as a sample rather than as the population. a. Mean: In order to find the Xbar of the sample, I took the sum of the observations (11+12+12+12+13+13+13+13+14+14+15+16+16+16+16+17+18+19=260) and divided this by n, the total amount of observations in the sample (18). Apparently you did! From this, I found xbar to be ~14.444. b. Median: In order to find the mdn, I computed the position at which the median should exist and then listed all of the observations in numerical order to find the right number at the aforementioned position. By calculating half of n+1, I found the position of the median to be 9.5th location (i.e. halfway between the 9th and 10th positions). Finally to find the median, I took the average of the two numbers at the 9th and 10th positions (mdn=(14+14)/2) and thus found the median to be 14. c. Mode: In this instance, the mode was very simple to find since this is ungrouped data presented in a bar graph. I simply had to find the observation(s) with the highest bar or bars (i.e. the observation(s) with the highest frequency). This distribution is bimodal; Mo1 is 13 and Mo2 is 16. Both modes have a frequency of four. Great, but the last sentence was unnecessary.
- 2. Mean: ((11*1) + (12*3) + (13*4) + (14*2) + (15*1) + (16*4) + (17*1) + (18*1) + (19*1)) / 18 = 14.4. Median: .5(18 + 1) = 9.5th location Median = 14 [….] the sum of all of the data is 232, and N = 14, making the mean 16.57 for this data set. I understand you’re weighting each observation by the number of times it occurs in the distribution, but stick to sigma x and list each one because your sum is off and that makes your mean incorrect.
- 2. The calculated mean is 13.94, the modes are 11 and 13, the median is 14. The mean is calculated by adding up all the data found (11+12+12+13…. I think something is lost in the ellipsis because your sum and thus your mean is off+18+19) then divided by the amount observed (18) which was 13.94. Stick to sigma x and list each observation. The time you saved isn’t worth being wrong and receiving no credit for the exam. The modes are 11 and 13 because those are the amounts that occurred the most which can be found from the graph. The median is found by finding the middle number in the set of data when set from lowest value to highest value.
- 2. The mean is equal to the sum of all the scores divided by the number of scores. To calculate the mean, I added all the scores together (11+12+12+12+13+13+13+13+4+14+14+15+16+16+16+16+17+18+19)=274, then I divided that number by the number of scores in the sample (19): 274/19= 14.42 µ=274/19=14.42 You’re off because you have a phantom “4” after “13” and then use that
*n*of 19 instead of 18. Be careful. There’ll be no partial credit or close enough on the exam! The median is the value that lies at the midpoint of the scores when the scores are in numerical order. To find the median, I wrote all of the scores in order from lowest to highest and then identified the middle value. 11,12,12,12,13,13,13,13,14,14,14,14,15,16,16,16,16,17,18,19 There are nine scores on both sides of the value 14, therefore, 14 is the median. The mode is the value that has the most scores. Since the graph is available, I just looked at the two bars with the greatest frequency to determine the mode. In this case there are two modes, 13 &16, because both of these values have the same frequency (4). Use formulae so I know if this is a sample or a population.

*Please feel free to ask a question in the comments section below. They can be anonymous, but signing with your initials would be helpful for me!
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These reviews are extremely helpful, but I do have one question. If our answers have any red next to them, does this mean they were viewed as incorrect and we did not get credit for them? I’m just a little confused on how these assignments are being graded!

Red is just my commentary on how to improve the answer or wording. For the exam they could be marked off in part or full depending. I intend HW to be a chance to try things out and learn from mistakes rather than be punished. Answers won’t be perfect at first. HW ‘grading’ for the course is about participation and being willing and open to learning from me and your peers. It’s all or none. Turning something credible in (on time) is what matters. Does that make sense?

Yes! Thank you!

I really like the homework review as well! However, I have a question about how homework is graded. Is it done just on a completion basis? Or is every answer that is correct receives a point? Also, when do you put the homework points into blackboard?

Actually you can ignore my “how is it graded question” as I just noticed that was answered in the question above. If only I would read things before I would speak, then I would be going places!

i’ll be putting the HW grades on blackboard soon. thanks for checking!

I appreciate the comments on answers they really helped!

So glad!

I agree. This method is quite helpful. Thank you.

Again very helpful! Especially the red commentary. I too was confused by that at first, but you already answered that question. Thanks!