Howdy!  I'm Professor Curtis of Aspire Mountain Academy here with more statistics homework help.  Today we're going to learn how to find “significant values” using a normal distribution.  Here's our problem statement: Suppose that the sitting back-to-knee length for a group of adults has a normal distribution for the population mean of 24.1 inches and a standard deviation for the population of 1.2 inches.  These data are often used in the design of different seats, including aircraft seats, train seats, theater seats, and classroom seats.  Instead of using 5% for identifying “significant values,” use the criteria that a value x is significantly high if the probability of x or greater is less than 1% and a value is significantly low if the probability of x or less is less than one 1%.  Find the back-to-knee length separating “significant values” from those that are “not significant.”  Using these criteria, is a back-to-knee length of 26.5 inches significant?

Here we have the first part of our problem, which says, “Find the back-to-knee length separating significant values from those that are not significant.”  To help us with this, we're going to use StatCrunch.  I could alternatively go old school and use the z-tables to find my answer, but that would require me to convert between random variables and z-scores.  I'm going to use StatCrunch because it's much simpler; it does all the conversion for you here inside StatCrunch.

I'm going to go to Stat –> Calculators –> Normal.  Notice the default settings for the Normal calculator are those for the standard normal distribution.  The normal distribution that we have here in our problem statement is not a standard normal distribution, because we have a mean value that's not zero and a standard deviation that is not one.  What I'm going to do is put those population parameters here into these fields in StatCrunch.  This will allow StatCrunch to adjust everything for me and do the conversions that I need for me.

Now the next step is to identify the criteria that we're using for separating out “significant” and “not significant” values.  Notice this is statistician talk.  What we're talking about are values that are usual that we would expect to see, and then there's values outside that range of usual values which are unusual that we would not expect to see.  So “significant” values are what a statistician would call values that are in the tails of our distribution — in other words, those that are not usual, the ones we wouldn't expect to see — whereas the “not significant” ones are the ones that are within that range of usual values.

The criteria that we’re given is that a value x is significantly high or low if the probability of being in that unusual range is less than 1%.  That means we have 1% in each of the tails of our distribution.  That area of 1% in each of the tails of our distribution is a total of 2%, which means 98% of the area under the distribution curve must be within that range of usual values that we would expect, or values that a statistician would call “not significant.”

So if I put down here in my probability field 0.98, and I select the Between option up top, I compute an outcome: the random variables values that I'm looking for.  I'm asked to round to one decimal place.  Fantastic!

Notice again we have 1% in the tail on the left.  This is for values that are significantly low.  And the probability of values being significantly high over here on the right is also 1%.  So 98% is here in the middle, and that's what I've calculated with the Between option from my Normal calculator.

Now the second part asks, “Using these criteria, is a back-to-knee length of 26.5 inches significantly high?”  The way you judge this is by looking at whether or not the value you're asked to evaluate is inside the range of usual values, or what a statistician would call values that are “not significant.”  26.5 is within the range of usual values in this problem; it's between 21.3 inches and 26.9 inches.  Therefore, it's in that range of usual values, or what a statistician would call “not significant.”

So here we can say a back-to-knee length of 26.5 inches is not significantly high because it is inside the range of values that are not considered significant.  Good job!

And that's how we do it at Aspire Mountain Academy.  Be sure to leave your comments below and let us know how good a job we did or how we can improve.  And if your stats teacher is boring or just doesn't care to help you learn stats, go to aspiremountainacademy.com, where you can find out more about accessing our lecture videos or provide feedback on what you'd like to see.  Thanks for watching!  We'll see you in the next video

Howdy!  I'm Professor Curtis of Aspire Mountain Academy here with more statistics homework help.  Today we're going to learn how to find a z-score for a standard normal distribution using StatCrunch.  Here's our problem statement: Find the indicated z-score.  The graph depicts the standard normal distribution with mean 0 and standard deviation 1.

OK, here we have a graphical depiction of a standard normal distribution curve.  Notice the indicated z-score lies to the left of 0.  That means the z-score we're looking for is negative.  We're also given the area underneath the curve that's bounded by that z-score that we're looking for.  So the simplest way to do this in StatCrunch is to go to Stat –> Calculators –> Normal.  This pulls up the Normal calculator.  Notice that the default settings for mean and standard deviation are the ones for the standard normal distribution.  These are the ones that we were instructed to use in the problem statement.

This makes solving this problem extremely easy.  All I need to do now is put this area into this field in StatCrunch.  We want the area to the left, so I need to make sure that this inequality sign drop down field is set appropriately.  This is what we want — less than or equal to — because this is like an arrow pointing in the direction that we want, and that's the direction that we want, the area to the left.  So we leave that alone.

You press Compute!, and out comes our z-score.  We are asked to round to two decimal places, so that gives me -0.62.  Well done!

And that's how we do it at Aspire Mountain Academy.  Be sure to leave your comments below and let us know how good a job we did or how we can improve.  And if your stats teacher is boring or just doesn't care to help you learn stats, go to aspiremountainacademy.com, where you can find out more about accessing our lecture videos or provide feedback on what you'd like to see.  Thanks for watching!  We'll see you in the next video.

Howdy!  I'm Professor Curtis of Aspire Mountain Academy here with more statistics homework help.  Today we're going to learn how to find a binomial distribution probability for a combined test sample.  Here's our problem statement: The probability of a randomly selected adult in one country being infected with a certain virus is 0.005.  In tests for the virus, blood samples from 13 people are combined.  What is the probability that the combined sample tests positive for the virus?  Is it unlikely for such a combined sample to test positive?  Note that the combined sample tests positive if at least one person has the virus.

To solve this problem I'm going to use StatCrunch.  Inside StatCrunch, I'm going to pull up the binomial calculator.  So to do that, I go to Stat –> Calculator and then select Binomial.  Here's my binomial distribution calculator.  Notice there are default values that are put in here.  We're going to change those values because we have different numbers for our particular problem.

n is the number of people in our sample, which is 13 because we're taking 13 samples and combining them into one big sample.  So this is the number that we use for n.  p is the probability of success.  Here we're defining success as being infected with a virus.  That probability is given to us in the problem — 0.005.  So we put that number in for p.  Now, what do we put here for our random variable?  Note that the combined sample tests positive if at least one person has the virus.  This is statement from our problem statement.  So we want the probability that at least one of those 13 are going to be affected.  This means we want to say that x is going to be greater than or equal to one, because this means at least one person has the virus of the thirteen that we're testing.

I press Compute!, and out comes my probability that the combined sample will test positive.  I'm asked to round to three decimal places.  So here that would be 0.063.  Put that answer in my answer field.  Well done!

And now the second part of the problem asks, “Is it unlikely for such a combined sample to test positive?”  Well, if we look at our answer options here, we can note the differences between them to select the proper answer.  The beginning of each of the answer options says, “It is unlikely for such a combined sample to test positive” or “It is not unlikely for such a combined sample to test positive.”  So it's either likely or unlikely.  How do we judge that?

Well, look at the last part of our answer options.  We're assessing whether it's greater than 0.05 or less than or equal to 0.05.  So we're using 5% as a threshold value to test whether it's likely or unlikely the combined sample is going to test positive.  A threshold value of 5% is a commonly used standard in many applications.

So let's look back at our probability.  0.063 is 6.3%.  This is greater than 5%.  Therefore, we're not going to select answer options B or D because these answers are options ending with “less than or equal than 0.05.”  This means the correct answer must be either option A or C.  So is it likely that the sample will test positive or unlikely that the sample will test positive?  Well, this sample testing positive has a probability of 6.3%.   It's greater than 5%.  Therefore, it's going to be likely that we'll test positive.  If the probability were less than 5%, then it would be unlikely.  But our probability here is greater than 5%.   Therefore, it is likely (or not unlikely).  So the answer that we're going to select is answer option C.  Fantastic!

And that's how we do it at Aspire Mountain Academy.  Be sure to leave your comments below and let us know how good a job we did or how we can improve.  And if your stats teacher is boring or just doesn't care to help you learn stats, go to aspiremountainacademy.com, where you can find out more about accessing our lecture videos or provide feedback on what you'd like to see.  Thanks for watching!  We'll see you in the next video!

Howdy!  I'm Professor Curtis of Aspire Mountain Academy here with more statistics homework help.  Today we're going to learn how to find the mean and standard deviation for a probability distribution using StatCrunch. Here's our problem statement: Five males with an x-linked genetic disorder have one child each.  The random variable x is the number of children among the five who inherit the x-linked genetic disorder.  Determine whether a probability distribution is given.  If a probability distribution is given, find its mean and standard deviation.  If a probability distribution is not given, identify the requirements that are not satisfied.

Here we're given a table listing the values and our probability distribution, so the random variable x is 0, 1, 2, 3, 4, 5, and then corresponding probabilities are in the other column of the table.

Our first question reads, “Does the table show a probability distribution?  Select all that apply.”  To check the values in our probability table to see if we have a probability distribution, we must ascertain that three requirements are met.  The first requirement is that each of the probability values in the probability table must be a probability value, which means they must be between 0 & 1 inclusive.  The second requirement is that each of the values for random variables must have a probability associated with it.  And the third requirement is that the probabilities in the table must add up to equal 1.  So let's check these requirements off.

First, the probabilities that are listed in our table have to be between 0 & 1 inclusive.  And we can see here that each one of these values in the probability column is between 0 & 1 inclusive.  So there's our first requirement.

The second requirement is that a probability value must be assigned to each random variable.  And we can see that each random variable has an associated probability with it.  Notice that in this case for random variables x=2 and x=3, we have the same probability value.  That's OK.  The associations don't need to be unique; you could have the same probability value for more than one random variable.  There just needs to be a probability value assigned to each random variable.  And we have that here, so there's the second requirement.

The third requirement is that each of the probabilities added together must equal 1.  You could punch these out in your calculator, or you could load it up into StatCrunch and get the sum.  Here I actually find it easier to get the sum by loading this into Excel.  So if I dump this table into Excel — and we have to wait a little while for the program to open up because someone is sleeping inside my computer — excellent!

OK, here we go.  So now the table is here in my — hold on, hold still, hold still — so now we have the table here in Excel.  And the easy way to get through this is we simply select the values in the probability column.  And notice how some statistics are calculated down here at the bottom of my window in Excel.  One of those statistics is the sum.  So here we see the sum equals 1, and it's exactly 1.  This means that we have a probability distribution, because the third of the three requirements has now been met.

You have to be careful with this.  It has to be 1 exactly.  So if the sum reads 1.000 something, or 0.999 something, it's not 1 exactly, and therefore it's not a probability distribution.  The numbers have to add up to equal 1 exactly.  So here we have a probability distribution.

If I wanted to calculate the sum in StatCrunch, that's easy enough to do.  Let's just load the table into StatCrunch.  We’ll move this window over a little bit.  So here in StatCrunch, if I go up to Stat –> Summary Stats –> Columns, select my probability, and then down here under Statistics I'm looking for the sum.  And here we see the sum is 1 exactly.

I think StatCrunch is a little bit more clunky; there's a little bit more clicking to do with the mouse.  It's much easier to just load it into Excel, select the cells that you want to add up together, and then it automatically does it for you there at the bottom.  It's a little quicker in Excel.  But if you prefer StatCrunch, there I've showed you how to do it.

Either way, we get the same answer: We do have a probability distribution.  So I'm going to select that from my answer options.  Good job!

Now the second part of our problem says, “Find the mean of the random variable x.  Select the correct choice below and, if necessary, fill in the answer box to complete your choice.”  You can do this in either Excel or Statcrunch.  For this particular part of the problem, it's easier to work in StatCrunch, so that's what I'm going to do.

We have the data already loaded here into StatCrunch, so now all we need to do is get the distribution into a calculator.  To do that, we click Stat –> Calculators and then go down here to Custom.  The distribution that we have is not identified as being a particular distribution.  It's just a general probability distribution, and so we're going to select this Custom option at the bottom of the list.

Here in our options window, the values will be the random variable (which is the x), the weights will be our probabilities.  Press Compute!, and here comes my distribution calculator.

And look here!  The mean and the standard deviation are already calculated for me.  So I'm asked for the mean here.  I'll just put that number into the answer field.  Good job!

Finally we're asked, “Find the standard deviation of a random variable x.  Select the correct choice below and, if necessary, fill in the answer box to complete your choice.”  We have a standard deviation already calculated here.  We're asked to round to one decimal place, so that would be here 1.1.  Put that in my answer field.  Good job!

And now a final word of caution about the Custom calculator in StatCrunch.  StatCrunch will not check to see if you have a probability distribution before it calculates mean and standard deviation inside the Custom calculator.  So when you use the Custom calculator, it will just take whatever data it gives you and calculate mean and standard deviation for you.  It won't check to see if you have a probability distribution.

So you have to run the check first to see if you have a probability distribution before you use the Custom calculator.  If you just rush off and use the Custom calculator, then these answer options here for Parts 2 and 3 where the table does not show a probability distribution may in fact be the right answer.  You don't want to go selecting the wrong answer, so make sure you check to see that you have a probability distribution before you start taking mean and standard deviation values from your data.

And that's how we do it at Aspire Mountain Academy.  Be sure to leave your comments below and let us know how good a job we did or how we can improve.  And if your stats teacher is boring or just doesn't care to help you learn stats go to aspiremountainacademy.com, where you can find out more about accessing our lecture videos or provide feedback on what you'd like to see.  Thanks for watching!  We'll see you in the next video.

Howdy I'm Professor Curtis of Aspire Mountain Academy here with more statistics homework help.  Today we're going to learn how to find the number of possible selections from more than enough candidates.  Here's our problem statement: A clinical test on humans of a new drug is normally done in three phases.  Phase I is conducted with a relatively small number of healthy volunteers.  For example, a Phase I test of a specific drug involved only seven subjects.  Assume that we want to treat seven healthy humans with this new drug, and we have eleven suitable volunteers available.  Complete Parts A through C below.

Part A says, “If the subjects are selected and treated in sequence so that the trial is discontinued if anyone displays adverse effects, how many different sequential arrangements are possible if seven people are selected from the eleven that aren't available?  Choose the correct answer below.”  So here we're trying to find the number of possibilities.  We have eleven people to select from, and we only need to select seven.  So we're selecting from eleven seven at a time.

This is a problem involving permutations and combinations.  To help us calculate the permutations and combinations, I have here a calculator from the online website calculator.net.  I'm not getting any kickback money or advertising revenue from calculator.net.  This is totally a freebie.  I'm trying to use something online that I can use in my video.  Normally when I calculate permutations and combinations, I use my calculator to calculate it.  Many calculators have combination and permutation functions built into them, so you'll have to learn how to use the particular model that you have because they're all a little different.  But I'm going to use this online calculator to illustrate how this is actually done.

The big question when working these types of problems is “Do we want to calculate a permutation, or do you want to calculate a combination?” The key determinate in answering that question is the answer to this question: Does the order matter?  If the order in which we arrange the elements matters, then we want to calculate permutations.  If it doesn't matter, then we want to calculate combinations.

Let's look back at our problem statement. Notice here that we're saying we're treating people in sequence.  So because there's a sequence here, that means there's an order that we want to treat people in.  So the order is going to matter.  Therefore, we want to calculate permutations.  At this point I go over to my online calculator.  I'm going to put in the total number of the set, which is 11 because there are 11 people that we want to treat.  The total that we're selecting from is 7, so we're taking 11 people selected 7 at a time.  I hit Calculate, and this is the number of permutations that I'm looking for.  So I go over here to my answer field, select that as our answer option.  Excellent!

Now Part B says, “If 7 subjects are selected from the 11 that are available, and the 7 selected subjects are all treated at the same time, how many different treatment groups are possible?”  Well, if we're treating all the people at the same time, obviously the order doesn't matter.  So here we want combinations.  So here we're going to calculate combinations.  This is the number we want here from the online calculator; we've already calculated it.  So I'm just going to put that number, 330, into my answer field.  Fantastic!

And finally Part C — “If 7 subjects are randomly selected and treated at the same time, what is the probability of selecting the 7 youngest subjects?”  Well, remember that probability is simply calculated by finding the part and dividing by the whole.  So we're gonna have a fraction here, so I'm gonna put my fraction element here in my answer field.

The part goes on top.  What's the part?  The part is the seven youngest subjects.  Now out of all the possible outcomes that we could have in our sample space, how many of them have the seven youngest subjects?  There's only going to be one of those outcomes in our sample space that has the seven youngest subjects.  So that is the part that we put at the top of our fraction.

In the bottom of our fraction, we put the whole.  The whole is going to be either the permutations or the combinations because we want the total number that's in our sample space.  But which is it?  Is that the permutations or the combinations?  Well, we're treating people here at the same time.  This means that the order is not important, and therefore we want combinations.  So down here in the whole part of my fraction, I'm going to put the number of combinations that I have.  Well done!

And that's how we do it at Aspire Mountain Academy.  Be sure to leave your comments below and let us know how good a job we did or how we can improve.  And if your stats teacher is boring or just doesn't care to help you learn stats, go to aspiremountainacademy.com, where you can find out more about accessing our lecture videos or provide feedback on what you'd like to see.  Thanks for watching!  We'll see you in the next video.

Howdy!  I'm Professor Curtis of Aspire Mountain Academy here with more statistics homework help.  Today we're going to learn how to find a product reliability given the failure rate.  Here's our problem statement: Assume that there is an 11 percent rate of disk drive failure in a year.  Part A: If all your computer data is stored on a hard disk drive with a copy stored on a second hard disk drive what is the probability that during a year you can avoid catastrophe with at least one working drive?  Part B: If copies of all your computer data are stored on three independent hard disk drives what is the probability that during a year you can avoid catastrophe with at least one working drive?

OK, so here we have given a failure rate for a disk drive.  We want to know what's the reliability of our system if we have two of these disk drives in place.  So we have one disk drive and then a copy of all the data on a second disk drive.  What's the reliability of our system?  Well, the probability that we're looking for is that at least one of these two are going to be working, because if one of them fails, that's okay; we still have everything on the other drive.

So we want the probability that at least one of these two is going to work.  That's going to be the same as 1 minus the probability that none of them works.  Remember that when you have the probability of “at least one” the way we calculate that is by taking 1 minus the complement of “at least one.”  And the complement of at least one is “none.”  So what then is the probability that none of them work?  Well, the probability that none of them work is going to be the probability that the first one fails and the probability that the second one fails.  If none of them work, both of them have to fail.

Well, we can calculate this probability.  We know the probability that the first one's going to fail. That's given to us in the problem statement — “The disk drive as a failure rate of 11%.”  So the probably that none of them work is equal to 0.11 (that's the probability that the first one will fail) and translates into multiplication (when we're dealing with probability problems and you see the word and, you know you need to use multiplication) and the probability the second one will fail is the same as the probability of the first to fail because the disk drives are identical.  So now we have 0.11 times 0.11.  We multiply that out, and we get 0.121.  This then is the probability that none of them work.

But remember we were looking for the probability that at least one of the two work, so we have to take the complement of this in order to get the probability that we're actually looking for.  And so we go ahead and do that.  The probability that at least one of the two work is 1 minus the probability that none of them works.  We know that the probability that none of them works is 0.0121,  so we substitute that in here.  And we get, when we punch that out in our calculator, 0.9879.  So this then is the probability that catastrophe can be avoided.  That's the reliability of our system.  We're asked to round to four decimal places, and that's what we have here.  So I'm just going to put that number in my answer field.  Nice work!

Now Part B says with three hard disk drives what is the probability that catastrophe can be avoided is . . . .  So now we have three instead of two.  Well, the problem is going to work pretty much the same way.  The probability that at least one of the three work is 1 minus the probability that none of them work.  But the probability that none of them work as we’ve seen from the previous problem statement is simply the failure rate of one, the failure rate of the second, and the failure rate of the third all multiplied together.

We can simplify this by saying 0.11 to the third power instead of 0.11 times 0.11 times 0.11.  We see these types of problems where you have different parts of a system and they're all independent of each other.  Typically, the way that we calculate the probability that none of them work is we just take the probability of one and raise it to the power of however many of these similar parts that we have in redundancy.  So here we've got three of them together, so we're going to raise this to the third power.

You punch this out on our calculator — 0.11 to the third power is 0.001331.  Subtract that from 1, and we get 0.99869.  We're asked to round to six decimal places, and that's exactly how many we have here.  So I'm just going to put that number here in my answer field.  Excellent!

And that's how we do it at Aspire Mountain Academy.  Be sure to leave your comments below.  Let us know how good a job we did or how we can improve.  And if your stats teacher is boring or just doesn't care to help you learn stats, go to aspiremountainacademy.com, where you can find out more about accessing our lecture videos or provide feedback on what you'd like to see.  Thanks for watching!  We'll see you in the next video.

Howdy!  I'm Professor Curtis of Aspire Mountain Academy here with more statistics homework help.  Today we're going to learn how to find the reliability for redundant systems.  Here's our problem statement: The principle of redundancy is used when system reliability is improved through redundant or backup components.  Assume that a student's alarm clock has a 12.8% daily failure rate.  Complete Parts A through D below.

Part A says, “What is the probability that the student's alarm clock will not work on the morning of an important final exam?”  So what we're looking for is the probability that one alarm clock will not work.  This is the probability that the alarm clock will fail, and we're told in the problem statement “Assume that a student's alarm clock has a 12.8% daily failure rate.”  So guess what the probability of one clock not working is going to be.  If you guessed 12.8%, you win a prize!  I don't know what the prize is, but you win something.  So that's the answer that we're going to put here in our answer field.  Good job!

Part B says, “If the student has two such alarm clocks, what is the probability that they both fail on the morning of an important final exam?”  So what we're looking for now is the probability that two alarm clocks do not work.  This is going to be the probability that the first alarm clock doesn't work multiplied by the probability that the second alarm clock doesn't work, because what we're looking for is the probability the first one doesn't work and the probability that the second one doesn't work.  That word and indicates we need to use multiplication, so we're going to multiply the probability that each alarm clock doesn't work together.  This is the same as taking the probability that one alarm clock doesn't work and squaring it — squaring it because we have two alarm clocks.   Punch that out in your calculator, and you get 0.016384.  However, the instructions tell us to round to five decimal places, so when we do that, we get 0.01638.  Excellent!

Part C says, “What is the probability of not being awakened if the student uses three independent alarm clocks?”  Well, again we are looking for the probability that three alarm clocks acting separately do not work.  I hope you can see a pattern here.  If we have one alarm clock, we just take the failure rate for one alarm clock.  If we have two alarm clocks, we take the failure rate for two alarm clocks and multiply them together.  So it stands to reason for three alarm clocks we're going to take the probability that each of three alarm clocks doesn't work and multiply them together.   That gives us 0.002097152.  Again, we are asked to round to five decimal places, so that brings us 0.00210.  Good job!

Part D says, “Do the second and third alarm clocks result in greatly improved reliability?”  Well, if we were to actually make a table of the results that we've just obtained and then convert those to reliability, we could then chart a graph to see what the effect is of adding alarm clocks to our clock system.

To get reliability out here, we have the failure rates that we calculated from the previous three parts of the problem.  Reliability is simply the complement of failure.  So something that you can rely on is something that works, and the probability something works is the complement of something not working, or something failing.  So we simply take 1 and subtract the probability of failure to get the reliability.  Notice the probability of failure is expressed in decimal form, but reliability is expressed in percent form.

We could then take these reliability numbers and plot them on a graph.  Notice there is a pretty significant jump by adding one extra alarm clock to the system.  So if we have one alarm clock and we add just one extra clock — so now we have two — that's a pretty sizable jump.  Adding one additional alarm clock after the second one to get three alarm clocks doesn't provide us with the same sort of jump or increase in reliability.  So having two clocks is much better than one, but having three clocks isn't all that much better than two.

Let's go back here to our problem and look at our answer options.  Answer Option A says, “Yes, because you can always be certain that at least one alarm clock will work.”  That may be a viable answer, but let's check the other answer options first.  Obviously Answer Options B and C aren't going to be correct because the graph clearly shows having an extra alarm clock is going to be helpful.  Answer Option D says, “Yes, because total malfunction would not be impossible but it would be unlikely.”

This is a much better answer than Answer Option A.  Answer Option A says, “because you can always be certain that at least one alarm clock will work.”  Technically, that's not true.  As this graph shows you can get increased reliability and get closer and closer to 100% reliability, which is certainty that you're going to have work what you want to have work.

However you'll never actually reach 100% as we saw from calculating these failure rates up above in the problem statement.  The numbers that we get by adding on additional clocks get smaller and smaller and smaller, but the only way you can get zero failure rate is if you multiply by zero.  And since you're always multiplying by a number that's greater than zero, this number will continue to get smaller and smaller but never actually reach zero.

Therefore, the reliability will get closer and closer to 100% but never actually reach there.  Certainty means 100%, so you can't always be certain; you'll always have some amount of probability of failure.  So the answer that we're going to select is Answer Option D.  Nice work!

And that's how we do it at Aspire Mountain Academy.  Be sure to leave your comments below and let us know how good a job we did or how we can improve.  And if your stats teacher is boring or just doesn't care to help you learn stats, go to aspiremountainacademy.com where you can find out more about accessing our lecture videos or provide feedback on what you'd like to see.  Thanks for watching!  We'll see you in the next video.

Howdy!  I'm Professor Curtis of Aspire Mountain Academy here with more statistics homework help.  Today we're going to learn how to find the reliability for series and parallel configurations.  Here's our problem statement: Refer to the figure below in which search protectors p and q are used to protect an expensive high definition television.  If there is a surge in the voltage, the surge protector reduces it to a safe level.  Assume that each surge protector has a 98% probability of working correctly when a voltage surge occurs.  Complete parts (a) through (c) below.

We see in the figure they're showing us the difference between a series configuration and a parallel configuration.  In a series configuration, both of the surge protectors are placed one after the other.  This means that in order for a surge to come through and fry our TV, both of these surge protectors have to fail.  If one of them works, we’re protected.  Contrast that with the parallel configuration.  In a parallel configuration, only one of these surge protectors needs to fail in order for a surge to come through and fry our TV.

Now let's read Part A.  Part A says, “If the two surge protectors are arranged in series, what is the probability that a voltage surge will not damage the television?”  Well, what we're looking for here is the series configuration, and we're trying to ascertain the probability that at least one of these works.  The probability that a voltage surge will not damage the television is the probability that at least one of these works, because in the series configuration we only need one of them to work to be protected.

We know that the converse — or rather the complement — of “at least one” is none, so the probability of at least one is going to be one minus the probability that none of them work.  This comes from the principle of complements.  The complement of at least one is none, so in order to solve this probability, we simply find the complement.

What is the probability that none of them work?  Well, here we use the fundamental counting rule to find that probability.  The probability that none of them work is the probability that one of them doesn't work multiplied by the probability that the other one doesn't work.  We have a 98% probability of working correctly.  That means we have a 2% probability of failure, because the most any probability can be is 100%, or 1.  So if there's a 98% probability of working right, there's a 2% probability of not working right, or a 2% probability of failure.  So the probability that it doesn't work is 2%.

We raise that to the second power because there are two surge protectors in our series configuration.  If we had three search protectors in series, then we would be raising this number to the third power.  But we only have two, so we're going to raise it to the second power.  This then is easy to punch out on your calculator.  When we do that, we get 0.9996, almost perfect reliability.  That's outstanding!  So this is the number I'm going to put in here.  Notice the instructions say, “Do not round.”  Excellent!

Part B says, “If the two search protectors are arranged in parallel, what is the probability that a voltage surge will not damage the television?”  Well, here we have the parallel configuration.  Notice in the parallel configuration we only need one of these to fail in order for the surge to come through and fry our TV.  So the probability that both of these are working is what we're looking for here, because both of these have to work in the parallel configuration in order for our TV to be protected.

Well, the probability that both of them are going to work is simply the probability that one of them works multiplied by the probability that the other one works.  We have a 98% probability of working correctly, so therefore the probability that both of them work is simply 0.98 raised to the second power.  Again, if we had three surge protectors in parallel, then we would raise this number to the third power.  But we only have two surge protectors, so we raise it to the second power.  This is also easy to punch out on a calculator.  When we do, we get 0.9604, so that's the number I'm going to put here.  Well done!

And now the last part, Part C, says, “Which arrangement should be used for better protection?”  Well, if we come over here and look at our reliabilities, the one with the higher reliability is the one that's going to offer us better protection.  And clearly, that's the series configuration because, while 96% reliability is pretty good, 99+% reliability is even better.  So we want the series configuration because it gives us a higher probability of protection.  And looking at the answer options, that's going to be this one here.  Nice work!

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