what's going on besties today we're going to be tackling the t7 mathematics portion of the exam and we're going to be talking about mean median mode range and probability let's get started let's start off by talking about mean median and mode so what our mean is the average number of a data set our median is the middle number of a data set and our mode is going to be the most frequent number of a data set so when you are taking your te they are going to give you a list of numbers that is going to be your data set that you have to compare in order to find these values what I highly recommend that you do when you're taking your te's is to place those numbers in numerical order from least to greatest trust me it's going to save you a whole lot of time as you're doing your calculations so I've already done that here for you in the data set that was provided so starting with our mean remember that's the average we have to figure out the average of all these numbers our very first step that we want to do is we want to add all of our numbers together so based on our data set if we were to add all of these numbers together we're going to get 127 but we're not done yet we have another step that's very important once we figure out what the overall sum is of all these numbers we need to divide it by the total numbers that are within our data set so if we were to add these numbers together 1 2 3 4 5 6 7 7 8 9 10 we have 10 numbers so we're going to take 127 which was our sum and divide it by 10 which was the total number of numerical digits that were provided to us that's going to give us our number of 12.7 or 13 if the t's ask you to round it up so the mean of this data set is 12.7 or 13 next let's move on to our median so the median is our middle number of AAP ass set so when you think of median think about when you're driving down a highway right there's always something in the middle that is always protecting you from being able to go into the oncoming traffic lane right we don't want that we don't want accident and when we think of median we think of middle think of that middle median that's down our Highway roads so again organizing those numbers are going to help save you so much time on the te's we're going to have to start kind of Crossing off numbers from each end until we figure out what our median numbers are so we would cross have 8 and 18 10 and 16 10 and 15 10 and 14 and look at that we have an even amount of numbers so we actually have two middle numbers so when we have an even amount of numbers we need to find the average between the two in order to determine what our median is so if we were to add 2 and 14 that's going to give us 26 you're going to divide that by two and that is going to give us 13 so the median for this data set is 13 now it's way easier if you have an odd number because of course whatever that number is in the middle would be your median but when you have even numbers make sure you're adding them together and dividing them by two and then lastly we have mode so when I think of mode I think of most m o mode M most mode is most it's the most frequent number in our data set so again lining them up save you a whole bunch of time so if we look here we have 18 we've got 3 10 112 2 14s 11 15 116 and 118 what number do we see the most well the number that we see the most is 10 So based on the numbers that we have here our mode most number is going to be 10 as it appears three times in our data set our first practice question is you are a nurse in a post-operative unit test with assessing the pain levels of 10 patients who have recently undergone surgery you are using a standard 0 to 10 Pain Scale where zero indicates no pain and 10 signifies the worst possible pain throughout the day you record the following pain scores for patients after knee surgery so as we can see here we have our patients rooms which really isn't something that we have to worry so much about when we're trying to find mean median mode but we we have pain scores these are the ones that going to be important so let's start off with finding our mean again we need to put these in numerical order it's going to save us a whole bunch of time later trust me so let's see the lowest number we have here is three so we've got three we've got another three here and we have another three here next we have four we've got another four here and another four here after that we have five and we have another five here that's all of our fives then we have six and our last number is seven perfect now we got them in order we're going to save ourselves a whole bunch of time so with me and remember that's our average so we're going to add all of these numbers together so if we were to add three + 3+ 4 so on and so forth it's going to give us a total of 44 but next we're not done we have to divide it by the total number that's within our data set so we have 1 2 3 4 5 6 7 8 9 10 we have a total of 10 numbers so we're going to divide 44 by 10 and that is going to give us our mean of 4.4 next let's consider our median so remember we're driving down the highway we get a median in the middle of us add is our middle number so again I've already written the numbers out here for you in numerical order so now we got to find our middle so when I'm doing this what I'm doing is I'm just marking off the end so 37 3 six three five four five look at that remember we have an even number of data points so we have four + 4 well 4 + 4 is equal to 8 / 2 is equal to 4 so our median number here is 4 our last data point that we have to find is mode remember m o and mode m o and most we're looking for the most frequent number of our data set so I have it written out here but I'll just visualize it for you we have three threes we have three fours we have two FS 1 six and 1 7 well look at that we actually have two numbers that appear most often within our data set when this happens it means we have two modes there's no additional addition subtraction multiplication nothing else we need to do we just know that our mode is going to be both three and four all right let's do some practice a small bakery tracked its daily cookie sales over one week and recorded the following numbers 12 15 11 12 14 16 and 12 calculate the mean median and mode of the bakery's daily cookie sales now what I did over here in the corner is I already put them in order for you but it's very important remember when you're doing this put them in order save yourself some time so mean we're finding the average what's the average of all these numbers so we have 11 + 12 + 12 + 12 + 14 + 15 + 16 so when we add all these numbers together we're going to get 91 but again we're not done we have to divide these numbers by the total numbers within our data set so we have 1 2 3 4 five 6 7 so we have a total of seven numbers in our data set so 91 / 7 is equal to 13 cookies next let's figure out our median so again our median is is our middle number driving down the highway we're looking at that median so again I'm going to go ahead and write out all of these numbers here and now we have to find that middle number so I'm going to start Crossing some numbers out so 11 and 16 12 and 15 12 and 14 and then look at that remember we have an odd number so automatically our middle number is our median so our median is 12 cookies and then lastly let's talk about our mode so again we're going to write this out all of our numbers here to figure out what our mode is so we have remember mode is most most frequent number we have 111 we have 3 12s 114 115 and 116 so our mode of this data set is going to be 12 cookies because this is the number that appears most within our problem next let's explore how we determine the range of a data set where is the range defined it's the difference between the highest and lowest numbers of our set in our initial example we have a jumble data set so we're going to need to organize that to figure out what our highest number is and our lowest number is so if we organize this we can see that our lowest number is two and our highest number is 10 so as we said as we're calculating range we want to subract our highest number from our lowest number so 10 - 2 is equal to 8 so our correct answer for range with this data set is eight pretty simple right it's going to be really easy when you're taking it on the t's let's take a look at our other data set so again it's just a bunch of jumble numbers so we want to make sure that we put that in numerical order from least to greatest save ourselves some times later right when we organize this we have 73 as our least number and 96 as our greatest number so we're going to just subtract the two 96 - 7 73 is equal to 23 so our range for this data set is 23 so let's take a look at a practice question to tie all this together so the question States a student scores on five math quizzes are as follows 82 76 91 85 and 88 what is the range of the students quiz scores so very first thing we want to do is we want to sort our data set so we have our lowest number is 76 then we have 882 and what I like to do is I like to just cross out as I'm moving through so that way you're not double counting numbers 76 82 the next number we have is 85 then we have 88 and our last number is 91 now that we've sorted our data set we know what our greatest number is and our least number is so our greatest number is 91 our least number is 76 and we are going to calculate our range we do this by subtracting ing our greatest number 91 by 76 and that is going to give us 15 so the range of this equation is 15 and look at that b is our correct answer so something else that's important when you're taking your t's especially when it comes to range is going to be distribution of that range when it comes to shape so first up we have symmetry so a distribution is considered symmetric if you can draw a vertical line through the center of its graph representing two halves that approximately mirror each other for our first example let's consider a continuous distribution graph here it appears that the mean is equal to the median that makes sense right our median would be right down the middle our mean is going to be equal to our median and as you can see we have an even Distribution on our left side as we do our right side something else that's important to note here is our graphs Peaks they indicate the frequency and provide us with our modes as you can see here we can observe two prominent Peaks suggesting that we actually have two modes this leads us to identify this kind of distribution as bodal meaning that we have two modes or two frequencies it's also symmetric and we have coning mean and median values so in this particular example this is going to be bodal B meaning two two modes in our next scenario we have a mound shape or it's also commonly referred to as the bell shape because of the way that the distribution looks drawing a central line down the middle through our graph it reveals again that symmetry with equal distribution on both sides of that line thus again it means that our mean is going to align with our median once more the notable feature here that we didn't see in our previous example was that we only have one one highest peak so this is going to illustrate in this scenario that our mean is going to be equal to our median which is also going to be equal to our mode so we would call this kind of distribution uni modal uni meaning one one mode it's also symmetric so we can call this again unimodal I hope you now see the difference between bodal and unimodal examples next let's talk about uniform or rectangular symmetry so this distribution occurs when all categories or classes within a distribution have frequencies that are equal or very close to equal observing such a distribution in our first example we notice that it appears symmetric with an even split of data on each side of our median so the Symmetry suggests that our mean is likely equal to the median in this distribution as well however the concept of mode is going to be somewhat unique when we're looking at this kind of graph in this case we have no mode as we do not have any Peaks as we would have seen in our previous examples in addition the frequencies don't have to be exactly the same in all classes there can be slight variations as we see here in our second example the distribution maintains symmetry implying that our mean is still equal to our median although the mode is less defined in a uniform distribution due to the equal or nearly equal frequencies of all classes one could point to multiple modes or consider the entire distribution mode like for its uniformity so we actually call this roughly uniform while it's not perfectly symmetric a roughly uniform distribution comes close showcasing an even spread across all of the classes lastly we're going to talk about skewness everybody's favorite the most confusing right well what I want you to do when you're taking your test is I want you to look down at your toes and I want to notice the way that your toes trend on each side of your foot in a left skewed distribution the bulk of the data is off to our left side so that is where we are going to find our mean at the point of the slope we're going to find our mode right that's our most frequent number if you look at your toes on the left side of your foot you're great toe is at its highest and each toe that follows is going to get slightly shorter giving you kind of like an arch this is the same thing that's happening with a left skewed graph so in this skewed landscape the mean or the average is going to be positioned a little bit more to the left the median which divides the data set into two is going to be found slightly closer to our mode but still between the mode and the mean this Arrangement means that half of the data points are going to fall on the left of the median with the remainder distributed to the right conversely our right skew distribution is going to be in the opposite direction here the data is going to accumulate mostly on the left side before it starts trending off onto the right just like we pictured before looking at our toes the same thing is happening with our right foot our greatest toe is going to be higher on the left side and as each toe follows you're going to see see that same skewness take place that same Trend so when you're trying to figure out the difference between your left and right skewed when you're looking at crafts I want you to look at your feet is it your right foot that it's mirroring or is it your left foot that is mirroring again the mode when it comes to right skewed is going to remain the highest point right here our median is going to be slightly positioned to the right very close to our mode and then our mean which is the average of our numbers are going to pull a little bit further to the right indicating the skewness very important for you know to your teas hopefully now you got it let's take a look at our practice question a researcher is analyzing the distribution of a data set representing a number of daily online orders received by two different products over a month the data shows two distinct Peaks so that is going to be our keyword here in our practice question what is most like the shape of our distribution well we can automatically eliminate right and left skewed as we know we only really have one Peak so we're only left with uniform and bodal well uniform as we discussed doesn't have any Peaks at all right it's uniform it's one straight line across so we can automatically eliminate that and if we're thinking about two distinct Peaks we talked about that bodal two modes that is our two distinct Peaks so out of all the answers that we have available to us the one that makes sense is C bodal our next question States a teacher records the test scores of students in a class and notices that the frequencies of scores is evenly distributed across all positive values which of the following best describes the shape of the distribution of these test scores so remember when we're seeing even distribution we're looking at one thing in particular so let's take a look at our example well right skewed and left skewed we can automatically eliminate that because as we know we don't have an even distribution it kind of skews off either to the right or skews off a little bit to the left next we have uniform and bodal so as we said before in our previous practice question B modal means we're going to have two peaks right two modes so that is obviously not going to be the right answer based on everything that we have available to us knowing that the data set is evenly distributed the only only answer that's going to make sense is a uniform let's take a look at our last question before we move on to probability so the question States in a study of household income in a certain region it was found that most of the data points cluster at the lower end of the income range with fewer households earning significantly higher income this results in a long taale to the rights of the distribution how is this distribution best described so if we're talking about this we know that the cluster points for the lower end of the incoming range right is going to be most of the data with fewer households earning a significantly higher so we know that we've got some households right having higher data and then as we start to see income households it's going to start to kind of long tail off to the right base on what the question is saying long tail off to the right so based on what we have available to us we have uniform well that doesn't make any sense right because that would mean we would have no modes it would just be all evenly distributed so we know that that is incorrect we have bodal well we don't have bodal because we know we don't have two peaks we have to have two modes it tells us that our results are going to long tail off to the right of the distribution so we can automatically eliminate that next we have left skewed well look at your toes based on what we just drew is your left foot the weight that it Trends down exactly with what we're looking for here no right we would have to have larger numbers to the right of the graph with the smaller numbers trailing down to the left so we can automatically eliminate that leaving us with our only correct answer which is going to be d a right skewed graph now let's talk about probability in most mathematical scenarios outcomes are going to be certain right if we say 1 + 1 is going to give us two if we multiply 2 by three is going to give give us six however sometimes in math you're often going to be presented with situations that are far from predictable for this we use the probability equation which is the total number of favorable outcomes divided by the total number of possible outcomes let's take a look at the simple Act of flipping a coin we can't be certain if it's going to land on heads or tails and while we can't predict the outcome of each individual coin toss we do understand some basic princip princi about it for example in a Fair coin toss the likelihood of landing on heads is going to be equal to the chance of it landing on Tails since there's only two outcomes heads or tails each is going to be equally as likely we would expect that half the flips would be heads and half the flips would be Tails meaning that the probability of flipping heads is 50% or 1/2 and the probability of flipping tails is also going to be 50% or 1/2 let's dive deeper into this concept using the probability line which is essentially a number line ranging from 0 to 1 on this scale a probability of zero indicates the impossibility of an event occurring and the probability of one guarantees that this vent will most likely occur marking it as complete certainty this is the reason why the probability line is confined from 0 to one think about it no event can have a probability lower than zero as nothing is less likely than impossible and similarly no events can have a probability higher than one since nothing can be more certain than certainty itself when we talk about a probability of 1/2 such as would the case with our coin toss it means that the event has an equal chance of occurring as it does not occurring if the probability Falls below 1/2 the event is considered unlikely whereas if the probability is above 1/2 it's going to indicate the event is likely to happen in addition the t's will commonly practice expressing probabilities in various formats including fractions decimals and percentages they're all going to be used interchangeably for example a probability of zero can be viewed as 0% chance of an event happening similarly a probability of 1/2 can equate to a 50% chance of something happening and a probability of one corresponds to 100% chance of an event from occurring now let's take a look at an example that's a bit more complicated than a coin toss now let's take a look at dice so a standard dice has six faces each numbered one to six when you roll the dice each face has an equal opportunity of Landing face up similar to the 50/50 chance that we see with a coin flip either being heads or tails however that's where the comparison is going to stop because unlike a coin which only has two outcomes the role of a dice is going to have six possible outcomes the certainty of landing on one of the six faces is represented by a probability of one or 100% however since only one face can be on top after a roll the probability must be distributed equally among all six outcomes so we divide that by six doing the math 1 / 6 is going to give us six which can be translated into 0.167 or 16.7% this value should be positioned in our probability line to indicate the likelihood of rolling a specific number say maybe a three while that suggests that the three is unlikely it is equally probable as to Rolling any other number on our dice I'd like to highlight an important aspect of probability remember that probability of Landing heads when you flip a coin is 1/2 and the chance of getting tails is also 1/2 when it comes to Rolling a dice the probability of rolling a one is indeed six now that you tally the probabilities of all outcomes of flipping a coin the sum is going to be 2 over two which is going to simplify to one likewise summing up all the probabilities for each face when it comes to a dice is going to be 6 over 6 which is also going to equate to one the cumulative probability of all potential outcomes for a given event always is going to equal one or 100% this reflects the certainty that one of the possible outcomes will occur and our last example let's make it a bit more complicated by introducing a spinner if our spinner only had six sections of equal size the probability is going to mirror that of rolling a dice but to make things more interesting let's increase the number of sections now our spinner has 16 equal segments so what's the likelihood of landing on a number say 10 similar to dividing a die 100% chance among the six faces we're going to evenly distribute that percentage across the Spinner's 16 sections therefore the chance that that spinner is going to land on 10 is 116th or 6.25% which again we can plot on our probability line it's now evident that landing on a 10 with our spinner is going to be less likely probable than rolling a three when we have a die this is logically given with the number of outcomes based on our spinner and our dice now let's add one more layer to this scenario how do we determine the probability of a spinner landing on a specific color with our spinner Now featuring five blue tiles and 11 gray ones let's delve into the probability of landing on one of those blue tiles drawing parallels to previous examples like the coin flip resulting in 1/2 and the dice resulting in one six we notice a common theme the numerator from all of these fractions were one this is because we were focusing on One Singular outcome such as rolling heads on a coin flip or rolling three on a dice however in our current scenario the numerator of this fraction is going to change to five because we are reflecting the five blue tiles that must meet our criteria for a successful outcome the denominator is still going to remain the total number of outcomes which is 16 representing the total number of tiles that we find on our spinner thus the probability of the spinner landing on a blue tile is calculated as 5 over 16 also translated to 31.25% well 31.25% might still be considered not highly probable it's significantly more likely that we're going to land on a blue tile then we were to land on one specific number with our spinner hang tight with me as we're going to work through this probability question I promise we're going to get through it the question States a bag contains five red marbles three blue marbles and two green marbles if two marbles are drawn one at a time from the bag at random with that replacement what is the probability that both marbles drawn are going to be red so we're looking for red marbles we know that we're drawing marbles one at a time from the bag at random and we're not replacing them without replacement so we start off by asking ourselves how many marbles do we have because based on probability our equation is going to be the number of favorable outcomes that's our top number our numerator and underneath that is going to be the total possible number of outcomes so that's what we're going to figure out first the total possible number of outcomes so we know that we have five red marbles we have three blue marbles and we have two green marbles so if we add all these together we know that we have a total of 10 marbles so we know that the denominator of our equation is going to be 10 okay so now we need to figure out what we're drawing each time we draw we know that we have five red marbles so the possibility of us getting one of those red marbles is going to be five out of 10 five is our number of favorable outcomes because we have five red marbles and 10 is a total number of possibilities that can occur because we have 10 marbles in the bag so say we pulled out a red marble the first time we're not going to replace it right without replacement that's what it states so now how many marbles do we have in the bag well now we have nine right so now our denominator is going to be nine because we have only nine possible outcomes since we've already drawn one of the red marbles well we've already drawn one of the red marbles how many red marbles do we now have in the bag 5 - 1 is equal to 4 so our next fraction is going to be 4 out of nine there's four number of favorable outcomes cuz we have four red marbles left and we have nine total number of possible outcomes because we only have nine marbles left so based on our two drawings our first should be 5 out of 10 and our second should be four out of n now we are going to multiply these two possibilities together okay so we know that we have 5 out of 10 we are going to multiply that by 4 over 9 because we have the word and so whenever we are looking for two things together we use the word and that means we are going to multiply our outcomes something very important that you're going to have to know for the t's is when we are looking for two outcomes together we use usually the word and and anytime we have the word and we're going to multiply the outcomes together because we're going to have two outcomes that we're going to need to know together whereas if you start to see the word either or or we are going to to add the outcomes together so anytime that you are trying to figure out two things like and in this case we're trying to figure out the probability of both marbles right and two marbles we are going to multiply outcomes whereas if it was asking or whether we were going to get a red or a blue then we would add the outcomes together so very important when you were taking your te's exam so in this particular case we are trying to figure out both marbles right that's our and so we are going to multiply so we are going to multiply our top together 5 * 4 is equal to 20 and we're going to multiply our denominators together so 9 * 10 is equal to 90 but if we take a look at our examples here for our answers we don't have anything that says 20 over 90 right we're going to have to simplify our fraction you can either do the math if you want to but what I do is I just remove the corresponding zeros from the end of each equation it just makes it a lot easier here and that is going to give me 2 over 9 so the probability that we are going to pull a red marble both times well that replacement is going to be 2 over 9 do we have that yes we absolutely do the correct answer is going to be b 2 over9 I hope that this video was helpful in making mean median mode probability and range make sense as always if you have any additional questions I'd love to hear from you leave them down below go over to nurse store.com where there's a a ton of additional resources available to you to help you Ace those exams and as always I'll catch you in the next video bye