Howdy!  I’m Professor Curtis of Aspire Mountain Academy here with more statistics homework help.  Today were going to learn how to find a binomial distribution probability of “no more than.”  Here’s our problem statement: A survey showed that 82% of adults need correction (eyeglasses, contacts, surgery, etc.) for their eyesight.  If 19 adults are randomly selected, find the probability that no more than one of them need correction for their eyesight.  Is 1 a significantly low number of adults requiring eyesight correction?

OK, the first part of this problem asks us to find the probability that no more than one of the 19 adults require eyesight correction.   To do this, envelope stack French doors is no icon here in the problem that lets you load stack crunch automatically

So I was like to keep a separate copy open in another window so that, in case I ever need it, it's there for me when I'm solving the problems.  Once I'm in StatCrunch, I want to find the binomial calculator because what we need is the binomial distribution.  We know it's a binomial distribution because you need correction for your eyesight or you don't, so it's one or the other.  That's the binomial distribution.  To get there, I go to Stat –> Calculators –> Binomial.

So here is my binomial calculator.  I will adjust my sample size to match that of the problem is to be 19 in the probability of success really consider needing eyesight success so as to be 82% in the rest of find the probability that no more than one so as to be less than or equal to its no more than Celeste that stays the same as change this number here one computer and there's my answer a really small number this he represents the number 10 in the -13's exponent and replacing with the tenses attended the -13th so that means this decimal point is 13 places to the left of zero before you actually get to the first actual nonzero number here.  So this is a very, very small number — practically zero.  So I put that here for my probability.   Fantastic!

And now the second part of this problem asks us, “Is 1 a significantly low number of adults requiring eyesight correction?  Note that a small probability is one that is less than 5%.”  Well, zero is definitely less than 5%, so we have a small probability that this is going to occur.  And therefore, we would say that 1 is a significantly low number.  So I'm going to select the answer option that corresponds with that.  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 want to help you learn stats, go to aspiremountainacademy.com, where you can learn 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 use the Range Rule of Thumb to find significant values.  Here's our problem statement: The accompanying table describes results from groups of 10 births from 10 different sets of parents.  The random variable x represents the number of girls among 10 children.  Use the Range Rule of Thumb to determine whether one girl in 10 births is a significantly low number of girls.

OK, so the first part of this problem is asking us to determine the maximum value using that Range Rule of Thumb.  To do that, let's actually come down here and click on this icon to view the table of our data.  And we do that and find that here we have a probability distribution.  We can easily get the mean and standard deviation to use for our Range Rule of Thumb by dumping this data into StatCrunch.  So I’m going to click this icon right here and dump my data into StatCrunch.  Great.  Now we’re going to resize his window here so we can see more of what's going on.  Excellent!

OK, so we need to get the mean and standard deviation for this distribution.  Now if I go directly into Stat –> Summary Stats and then Grouped/Binned Data, I’m going to find that I'm going to get frustrated because this option requires us to have integer values; it wants frequency counts here.  What we have are percentages because it's a probability distribution.  So what I need to do here is provide integers.  And the easiest way to do that is to round each of these probabilities which are really percentages — multiply by 100 and then round to the nearest integer.

So if I do that here for this first column, I'm gonna find I get 0.4 because I move that decimal point to places to the right.  That gives me 0.4, which rounds down to zero.  The next value I get — move it over two places — 1.1, so that’s going to round to 1.  And I just keep doing the same thing for each of the numbers here in my probability distribution.

OK, now that I actually have counts here, I'm going to go into Stat –> Summary Stats –> Grouped/Binned Data.  The bins are going to be your x-values; that’s your random variable.  The counts are going to be what we just calculated in the new column there.  I want the mean and the standard deviation.  There is my mean and standard deviation.

So now all I need to do is apply the Range Rule of Thumb, come back over here, and get my maximum value.  So to do that, I whip out my calculator.  I’m going to take the mean value, which in this case is 5.14, and add it to twice the standard deviation.  I don’t need to type the whole number in because I'm only asked to round to one decimal place, so this four decimal places should be more than sufficient.  And there’s my maximum value.  I round to one decimal place.  Fantastic!

Now, the second part asks us for the minimum value.  I’m going to do the same thing — use the mean and standard deviation, make the same calculation, only this time I’m going to subtract everything out.  So there's the mean value, 5.14, subtracting out twice the standard deviation.  Round to one decimal place.  Good job!

Now, the last part asks, “Based on the result, is one girl in 10 births a significantly low number of girls?  Explain.”  Well, if its’ going to be significantly low, that's means it’s outside the range of usual values on the low side.  So what we’re really looking for is something outside this range on the low side, which would be less than the minimum value.  1 is less than our minimum value of 1.7, so we conclude, yes, this is a significantly low number girls.  So I want yes, it's significantly low because it's less than the minimum value that's not significant.  Excellent!

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 want to help you learn stats, go to aspiremountainacademy.com, where you can learn more about accessing our lecture videos or provide feedback on what you 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 probabilities using permutations and combinations.  Here’s our problem statement: A corporation must appoint a president, chief executive officer, chief operating officer, and chief financial officer.  It must also appoint a planning committee with three different members.  There are 15 qualified candidates, and officers can also serve on the committee.  Complete Parts A through C below.

OK, Part A asks, “How many different ways can the officers be appointed?”  Well, to get this value, we need to calculate either a permutation or a combination.  And since the scientific calculator in my computer is not equipped to calculate permutations and combinations, I’ve gone out to a website called Calculators.net because it has this handy permutations and combinations calculator.  So we put in the total number for our set — we’re choosing among 15 qualified candidates, so I’ll put 15 in here.  And then the amount of each subset — that's how many people we have to select for.  So how many positions do we have here?  A president, CEO, COO, and CFO — so that's 1, 2, 3, 4 positions.  We’ve got four positions, so I calculate that.

And here are my values, but do I take the permutation, or do I take the combination?  Well, the question we need to ask ourselves is “Do we have any order that matters?  Is order important to what we’re trying to calculate?”  We’re calculating this for the officers that are going to be appointed.  The officers have a hierarchy, which is a type of order.  Therefore, order is important and so we’re going to select the permutation.  So there are, in this case, 32,760 different ways to appoint the officers.  Excellent!

Now, Part B asks, “How many different ways can the committee be appointed?”  The committee has 15 candidates that we’re choosing from, and we have three different members for the committee.  So we want three of 15 total.  And hear the order doesn't matter because were just choosing people for committee; once you’re on the committee, you're on the committee.  There's no indication that there's any order here with the committee.  So therefore that means we want to calculate a combination, which in this case you see here is 455.  Nice work!

And finally, Part C asks, “What is the probability of randomly selecting the committee members and getting the three youngest of the qualified candidates?”  Well, the probability is going to be the part over the whole, so we’ll put a fraction in here.  The part goes on top.  How many of the ways that you can pick qualified candidates represent the three youngest?  Well, there's only one way in which the three youngest members get on the committee.  How many ways are there total?  Well, for the committee we already calculated that there’s 455 different ways, so that’s going to be the whole.  I check my answer.  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 want to help you learn stats, go to aspiremountainacademy.com, where you can learn more about accessing our lecture videos or provide feedback on what you 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 were going to learn how to find the probability of at least one.  Here’s our problem statement: Subjects for the next presidential election poll are contacted using telephone numbers in which the last four digits are randomly selected with replacement.  Find the probability that for one such phone number the last four digits include at least one zero.

OK, so what we’re looking at here is the probability of at least one of those four digits the end of the phone number being zero.  And to do that, we actually remember that, whenever looking for the probability of at least one, that's one minus the probability of none.  So the probability that none of those last four numbers are zero subtracted from one is going to give us a probability of at least one.

So what’s the probability that none of them are zero?  Well, it’s the probability that the first number is not zero and the probability that the second number is not zero and the probability that the third number is not zero and the probability that the fourth number is not zero.

OK, so from this point what do we do?  Well, what’s the probability that the first one is not zero?  Well, we go through and substitute, we find that it's a part over the whole.  So there are nine numbers that are not zero, and there's 10 numbers total (0 through 9), so the part over the whole is 9/10.  This is the probability that the first number is not zero.  And of course is just statistics speak for multiplication, so that's what we have here.  And then we can continue on and find that it's the same thing for the rest of the numbers.  So the probability that the second one is not zero — you got 10 numbers to choose from, one of them is not zero, so there’s nine to choose from out of the 10 total.  And it just goes on like that, so we can just continue on like this.

Now, at this point in the expression, we can simplify this.  We got nine tenths multiplied by itself.  There’s four of them, so I can actually just simplify that by saying nine tenths to the fourth power, which, when I punch that out on my calculator, I get 0.6561.  Subtract that from 1, and I get 0.3439.  I’m asked to round to three decimal places.  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 want to help you learn stats, go to aspiremountainacademy.com, where you can learn more about accessing our lecture videos or provide feedback on what you 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 probability of accepting a batch in manufacturing.  Here's our problem statement: Among 7897 cases of heart pacemaker malfunctions, 340 were found to be caused by firmware, which is software programmed into the device.  If the firmware is tested in three different pacemakers randomly selected from this batch of 7897, and the entire batch is accepted if there are no failures, what is the probability that the firmware in the entire batch will be accepted?  Is this procedure likely to result in the entire batch being accepted?

OK, we have two parts.  The first part is asking for the probability.  To get the probability of the firmware actually passing the inspection, the probability that the lot is going to be accepted is going to be determined by all three of those randomly selected pacemakers passing inspection.  So the first one has to pass, and the second one has to pass, and the third one has to pass.  Well, the probability of the first one passing is simply one minus the probability of that first one failing, because there's only two options — I mean, the part either passes or it fails.  So there is our failure rate for the first one; 1 minus the failure is going to give us the pass.  We can produce the same thing for each of the other two pieces that we’re going to test.

So now here's a probability of acceptance.  And the probability that the first one is going to fail is going to be equal to 340/7897 because you can see from our problem statement that 340 of those failures were caused by the firmware, and that's what we’re testing for with our acceptance test.  So 340 is the part over the whole 7897 — that's the probability that the first one fails.  So that's what we put in here.  And is simply probability talk for multiplication.  So when you see and, that simply translates into multiplication.  The probability that the second one is going to fail is going to be just the same as the probability of the first; it's the same failure rate for all the different pieces, so I can just write the same thing here for the second piece.  And again translates into multiplication.  And then we find that the failure of the third piece is going to have the same probability as the other two pieces.  So now you got this written.

Now what we have to do is simplify this and solve.  So notice how you have the same expression multiplied by itself three times, so I can just write this as that expression raised to the third power.  And then I can go and solve for that expression.  So when I do that, 340 divided by 7097 gives me 0.0431.  I subtract that from 1 and get 0.9569.  I raise that to third power, and I get 0.8763.  This is my probability that the lot will be accepted.  I’m asked to round to three decimal places.

And then here on the second part, it’s a drop-down menu.  So “this procedure is <blank> to result in the entire batch being excepted.”  There’s three options.  Well, it’s not going to be certain because certain means the probability is 100%, and that’s not what we have.  It’s not going to be unlikely, because unlikely means that the probability is less than 50%; we have more than 50% here.  So we’re going to choose likely, because the probability is greater than 50%.  I check my answer.  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 want to help you learn stats, go to aspiremountainacademy.com, where you can learn more about accessing our lecture videos or provide feedback on what you 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 probability of winning at roulette.  Here's our problem statement: A modified roulette wheel has 36 slots.  One slot is 0, another is 00, and the others are numbered 1 through 34, respectively.  You are placing a bet that the outcome is an odd number in roulette (0 and 00 are neither odd nor even).

OK, the first part of this problem asks for our probability of winning.  And to do that, we're going to calculate probability the way we always calculate probability, which is the part over the whole.  Well, what's the part?  The part is the number of slots that are actually odd, because that's the outcome that we're looking for — how many odd numbers do we have here?

Well, 0 and 00 are not odd, and so we have to look through 1 through 34.  Half of those will be odd, and half will be even, so if I take my calculator here and divide 34 by 2, I get 17.  So this is the number of odd slots that we have.  So here in my answer field, I will set up a fraction.  In the top part of the fraction, I put the part; it's the number of odd slots on the wheel.

And then on the bottom of my fraction, I put the total number of slots; this is the whole part, the whole set.  So that's gonna be 36.  Well done!

Now the second part asks for the actual odds against winning.  OK, here we have the odds against winning, which means we're going to calculate this as a fraction, although here it's listed in ratio form.  And this first number, which would be the number on top of our fraction, will be the part that's against winning, the part that corresponds with losing.  And then on the bottom of the fraction, which corresponds with the second number, that's going to be the part for the winning.

Well, we know what that is; that's going to be the 17.  What's the part for the losing?  Well, I'm just going to subtract the 17 from the total 36, and that gives me the part that's losing, because you either lose or you're win.  So this first part here is going to be 19, and the second part is going to be 17.  Well done!

Now, the third part asks for how much profit we make if the payoff odds are one to one and we bet $10 and win.  Well, if we win, we're gonna get the $10 back that we bet plus an additional payoff according to the payoff odds, which here is one to one.  That means for every $1 we bet, we get $1 of payoff.  So if you bet $10, your payoff is gonna be $10.  Well done!

Now the last part of the problem asks us the same question but with the caveat that we've somehow convinced the casino to change the payoff odds so that their the same as the actual odds against winning.  Well, the odds against winning are right here — 19 to 17.  So what we do is we actually convert this to a fraction.  We can actually then multiply that by the amount of our bet, and that gives us the amount of our payoff.

That's essentially what we were doing up here in this third part.  We took the bet, which is 10, we multiplied it by 1 over 1 — and 1 over 1 is 1, and so anything times 1 is itself — and so the numbers came out the same.  Here we're gonna do the same thing, only we're gonna use a different number than 1.  So I've got 19 divided by 17.  So now I take that and multiply it by my bet, and there's my payoff, which I round to the nearest cent.  Good job!

And that's how we do it in Aspire Mountain Academy.  Be sure to leave your comments below let us know how good the job we did or how we can improve.  And if your stats teacher is boring or just doesn't want to help you learn stats, go to aspiremountainacademy.com, where you can learn 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 convert z-scores into real-world values with StatCrunch.  Here's our problem statement: Consider a value to be significantly low if its z-score is less than or equal to -2, or consider a value to be significantly high if its z-score is greater than or equal to 2.  A test is used to assess readiness for college.  In a recent year, the mean test score was 22.4 and the standard deviation was 5.1.  Identify the test scores that are significantly low or significantly high.

OK, this first part is asking us for the test scores that are significantly low.  Now, if I wanted to, I could go “old school” and use the formula that I talked about in the lecture to calculate — with my calculator, punch the numbers out on my calculator and use that formula to convert from a z-score to a real-world value.  But I love the 21st century because it allows us to use technology like StatCrunch.  So I'm gonna use StatCrunch, and I'm gonna show you how easy this is in StatCrunch.

First, since we're dealing with z-scores, we need to get our normal distribution calculator out.  To do that, we're going to go to Stat –> Calculators –> Normal.  Here's my normal calculator.  You'll note that the default values here are for the standard normal distribution.  So what we want to do first is, since we're dealing with scores that are significantly low and scores that are significantly high, I'm gonna take this Between option up here at the top of the calculator.  That way I get the scores here that are significantly low here in the left tail of my distribution and the ones that are significantly high will be in the right tail of my distribution.

The bounds for these tails you can see here in the problem statement.  It says anything less than -2 and anything more than 2 is going to be low or high, respectively.  So I'm just gonna put those values in here.  I hit Compute!, and this is the number that I want to select.  And I want to copy it, the reason being is that now what I'm going to do is take the mean value and the standard deviation value that was given to me in the problem statement, and I'm just going to put that here in my distribution so StatCrunch can do all of the calculation for me.  I'm going to clear out these values in the probability fields, and then here I'm gonna paste that value that we had before.  Now when I hit Compute!, I've got my distribution.

And you can see the bounds here.  So anything less than this left bound is going to be significantly low, and anything more than this right bound is going to be significantly high.  And those numbers are given to me right here in the probability field.  So what test scores are significantly low?  Well, test scores are going to be less than, in this case, 12.2.  Good job!

And what test scores are significantly high?  Well, we've already done the work here.  So all I got to do is just take the number and put it in — anything greater than 32.6.  Nice work!

And that's how we do it in 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 want to help you learn stats, go to aspiremountainacademy.com, where you can learn 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 use Chebyshev's Theorem to derive proportions.  Here's our problem statement: Using the accompanying table of data, a blood platelet counts of women have a bell-shaped distribution with a mean of 255.0 and a standard deviation of 65.3.  (All units are thousand cells per microliter.)  Using Chebyshev's Theorem, what is known about the percentage of women with platelet counts that are within three standard deviations of the mean?  What are the minimum and maximum possible platelet counts that are within three standard deviations of the mean?

OK, the first part asks for Chebyshev's Theorem, and we want to know, you know, what percentage of the data are within three standard deviations of the mean.  Well, Chebyshev's Theorem — it's just basically tabulated data.  It comes from an equation that he derived, but you don't have to use the equation.  You just have to just take it straight off the table.  So you want the value for within three standard deviations.  Well, just take it straight off the table.  We're supposed to round to the nearest integer.  Well done!

Now, the second part asks, “What are the minimum and maximum possible platelet counts that are within three standard deviations of the mean?”  Well, they give us values for the mean and the standard deviation here in the problem statement.  So all I need to do now is whip out my calculator and punch that out.  So the minimum is going to be the mean minus three standard deviations.  So I'm going to go ahead and calculate that here.  And then to get the maximum, I'm going to do the same thing, only instead of subtracting, now I have to add.  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 want to help you learn stats, go to aspiremountainacademy.com, where you can learn 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.