I was reading a blog post by someone comparing an NFL quarterback from the 1960's and 1970's to one from this era, stating how the one back in the older era was overrated and worse in many ways than one from today. A basis of this, was in his opinion, that the older QB had a completion percentage of 50.1 as compared to 55.7 of today's QB and the interceptions thrown were higher as well for the older QB than the current player.

But to accurately say that the older players was worse, you really need to compare his numbers to the other QB in the league, the average and standard deviation come into play. I know for a fact that completion percentage in the past era was much worse than the current era. Players may be better today offensively with the complex schemes, but I know the defense isn't allowed as much freedom in today's game. For whatever the reason, have to compare statistically.

Suppose the league average completion percentage in the 1960's and 1970's was 50 percent, therefore the older QB was right on the average.

But today, a 55 percent completion percentage may be way below a league average of, say 60 percent. So in essence the older QB did better compared to the rest of the league even though his actual percentage was lower.

The same holds true for other statistical categories.

## Wednesday, June 25, 2014

## Thursday, June 19, 2014

### Solving Trigonometric Equations

A trigonometric equation is an equation which has a trigonometric
expression involving a variable, such as sinx, cosx, tanx, cotx, secx,
cscx. There are the six trigonometric functions I mentioned in a
previous introductory article on trigonometry. Some trigonometric
equations are true for all values of the variable, such as 1 + tan

Fox example, suppose the equation is sinx = √2/2. We know from the unit circle, or we can solving using a calculator, that 45 degrees is a solution for x because sin 45 = √2/2. This is not the only value that satisfies the equation. Sine is positive in the first and second quadrant, which can be seen on the sine curve. So the answer is 45 degrees and the 45 degree angle in the second quadrant, which is 135 degrees. If answering in radians, it's π/4 and 3π/4.

Let's try an equation involving a single trigonometric function such as 3cosx - 2 = 5cosx - 1. We solve such equations the same as solving a basic algebra equation involving a single variable. We isolate the function to one side of the equals sign and solve for the variable. In this case, we want to get cosx on the left side of the equation and everything else to the right side of the equals sign. Therefore, we can subtract 5cosx from both sides. That gives us -2cosx - 2 = -1. Notice we have -2 on the left side of the equation. But we all we want is cosx on that side, so add 2 to both sides. Doing that gives us -2cosx = 1. From here we simply divide both sides by -2 to get cosx = -1/2. From our knowledge of the unit circle we know it's a 60 degree angle, but since the value is negative, it's not in the first quadrant. Cosine is negative in the second and third quadrants. So the answers are 120 and 240 degrees. In radians, the answer is 2π/3 and 4π/3. Note that we convert degrees to radians by multiplying each degree measure by π/180.

Suppose the equation is in quadratic form, meaning in the form

Let's try the equation 2sin

You may also encounter a problem which involved multiple angles. An example of such an equation is cot3x = 1. To solve this equation, we are still looking for the values of x from 0 to 360 degrees (one rotation around the unit circle). We know cot is 1 at 45 degrees and 315 degrees, in radians this is π/4 and 7π/4. Therefore 3x =π/4 and 7π/4, we will add 2π to each to get 9π/4 and 15π/4 and again to complete three rotations around the unit circle. This gives us 17π/4 and 23π/4. If the problem was cot4x we'd need four rotations and so on. Now 3x is equal to all of those values. So in order to solve for x, we divide each angle to 3 to get π/12, 7π/12, 3π/4, 5π/3, 17π/12 and 23π/12.

There are more types of trigonometric equations, such as those using identities to solve, which I'll deal with in another article. Hope this guide will be useful for those that need to learn how to solve basic trigonometric equations.

^{2}x = sec^{2}x, 1 + cot^{2}x = csc^{2}x and sin^{2}x + cos^{2}x = 1. But many trigonometric equations are defined for only some values of the variable.Fox example, suppose the equation is sinx = √2/2. We know from the unit circle, or we can solving using a calculator, that 45 degrees is a solution for x because sin 45 = √2/2. This is not the only value that satisfies the equation. Sine is positive in the first and second quadrant, which can be seen on the sine curve. So the answer is 45 degrees and the 45 degree angle in the second quadrant, which is 135 degrees. If answering in radians, it's π/4 and 3π/4.

Let's try an equation involving a single trigonometric function such as 3cosx - 2 = 5cosx - 1. We solve such equations the same as solving a basic algebra equation involving a single variable. We isolate the function to one side of the equals sign and solve for the variable. In this case, we want to get cosx on the left side of the equation and everything else to the right side of the equals sign. Therefore, we can subtract 5cosx from both sides. That gives us -2cosx - 2 = -1. Notice we have -2 on the left side of the equation. But we all we want is cosx on that side, so add 2 to both sides. Doing that gives us -2cosx = 1. From here we simply divide both sides by -2 to get cosx = -1/2. From our knowledge of the unit circle we know it's a 60 degree angle, but since the value is negative, it's not in the first quadrant. Cosine is negative in the second and third quadrants. So the answers are 120 and 240 degrees. In radians, the answer is 2π/3 and 4π/3. Note that we convert degrees to radians by multiplying each degree measure by π/180.

Suppose the equation is in quadratic form, meaning in the form

*au*^{2}+*bu*+*c*= 0. where*u*is a trigonometric function and*a, b*and*c*are real numbers and*a*cannot equal zero. We generally can solve this kind of equation by factoring, which I explained in a previous article.Let's try the equation 2sin

^{2}x + sinx - 1 = 0. We'll attempt to solve this by factoring. Remember we use two sets of parentheses, each with two terms separated by an + or - sign. The first term in each must multiply to 2sin2x, the last terms must multiply to -1 and when multiplied out must give you sinx in the middle. Putting it together we get (2sinx - 1)(sinx + 1) = 0. We know this is correct because multiplying this out we get 2sin^{2}x + 2sinx - sinx - 1, which is 2sinx^{2}x + sinx - 1, the original expression. We set each factor equal to zero and solve. Therefore we have 2sinx - 1 = 0 and sinx + 1 = 0. Solving for x in the same manner we did in the previous example, we get x = 30 and 150 degrees and x = 270 degrees. In radians, the answers are π/6, 5π/6 and 3π/2.You may also encounter a problem which involved multiple angles. An example of such an equation is cot3x = 1. To solve this equation, we are still looking for the values of x from 0 to 360 degrees (one rotation around the unit circle). We know cot is 1 at 45 degrees and 315 degrees, in radians this is π/4 and 7π/4. Therefore 3x =π/4 and 7π/4, we will add 2π to each to get 9π/4 and 15π/4 and again to complete three rotations around the unit circle. This gives us 17π/4 and 23π/4. If the problem was cot4x we'd need four rotations and so on. Now 3x is equal to all of those values. So in order to solve for x, we divide each angle to 3 to get π/12, 7π/12, 3π/4, 5π/3, 17π/12 and 23π/12.

There are more types of trigonometric equations, such as those using identities to solve, which I'll deal with in another article. Hope this guide will be useful for those that need to learn how to solve basic trigonometric equations.

## Thursday, June 12, 2014

### Discontinuity

In some cases, a function will have discontinuity, which is a value for x where the function is undefined. It could be removable discontinuity, in which case there is a "hole" in the graph, or non-removable discontinuity, which in many cases will be an asymptote.

Take for example the problem f(x) = (x^2 - 1)/(x - 1). This is undefined for x = 1.

Factored this becomes (x - 1)(x + 1)/(x - 1). When simplifying we are in essence "removing" the x -1 from the numerator and the denominator and are left with x + 1.

The graph of x + 1 is a line with slope 1 and y-intercept of 1. But since we know x is undefined at 1, we have a hole at the coordinate (1,2) represented by an open circle.

Here's another example which has some discontinuity.

f(x) = (x^2 + 6x + 8)/(x^2 -2x - 8)

When factoring we get (x + 4)(x + 2)/[(x -4 )(x + 2)]

This function is undefined at x = 4 and x = -2. Notice the removable discontinuity at x = -2. The other discontinuity at x = 4 is non-removable.

Take for example the problem f(x) = (x^2 - 1)/(x - 1). This is undefined for x = 1.

Factored this becomes (x - 1)(x + 1)/(x - 1). When simplifying we are in essence "removing" the x -1 from the numerator and the denominator and are left with x + 1.

The graph of x + 1 is a line with slope 1 and y-intercept of 1. But since we know x is undefined at 1, we have a hole at the coordinate (1,2) represented by an open circle.

Here's another example which has some discontinuity.

f(x) = (x^2 + 6x + 8)/(x^2 -2x - 8)

When factoring we get (x + 4)(x + 2)/[(x -4 )(x + 2)]

This function is undefined at x = 4 and x = -2. Notice the removable discontinuity at x = -2. The other discontinuity at x = 4 is non-removable.

## Monday, June 9, 2014

### Which math should I take in college?

If you are a math major or statistics major, the majority of your courses will be laid out for you with the exception of major electives, in which you'll have a variety of math courses to choose from. Take all the statistics and probability courses you can if you are a statistics major.

If you are going into a science based field of study or any type of engineering major, take calculus your freshman year. Make sure you take a course in complex variables, which has engineering applications.

If you are not going into a math, science, statistics, or engineering based field, your college may offer college algebra. If not, take an introductory statistics course. You may only be required to take a year's worth of math.

If you are in a business related field, take math that relates to finance or accounting principles.

If you are going into a science based field of study or any type of engineering major, take calculus your freshman year. Make sure you take a course in complex variables, which has engineering applications.

If you are not going into a math, science, statistics, or engineering based field, your college may offer college algebra. If not, take an introductory statistics course. You may only be required to take a year's worth of math.

If you are in a business related field, take math that relates to finance or accounting principles.

## Wednesday, June 4, 2014

Those
who have studied a course in statistics are familiar with a histogram
or frequency distribution. In statistical process control or SPC, the
Pareto chart is a frequency distribution or histogram of attribute data
arranged by categories. For example, suppose we consider defects on an
submarine used in a naval application. We can plot the total frequency
of occurrence of each type of defect on the y-axis against various types
of defects on the x-axis. With this type of chart a user can quickly
see which types of defects occur most frequently and less frequently.
The cause of the defects that occur most frequently should probably be
identified and addressed first.

One must notice that the Pareto
chart doesn't list the defect in order of importance, rather just in
order of frequency. For example, a certain defect may only occur three
percent of the time, but if it's one that could end up in scrapping the
submarine. So, in some cases, the defect occurring most frequently may
not be the one to take care of first. How do we determine what to take
care of first? There are two methods, first you could use a weighted
average to modify the frequency counts, meaning that we emphasize the
defects that are more significant. You could also have a second chart to
consult that lists that has a large cost exposure.

There are variations to the basic Pareto chart I discussed above. Suppose we have components on a circuit board labeled by part number and assume that we know the percentage of defective components. We could have a chart that lists the percentage of defects on the vertical axis and the component number on the horizontal axis.

Suppose we know the components come from three suppliers, A, B and C. We could have a chart with the number of defective components on the vertical axis and the component number listed by supplier on the horizontal axis. For example, suppose for component number 1 that there are 30 overall defects, 20 from supplier A, 8 from supplier B and only 2 from supplier C. The vertical bar above component 1 would extend to 20 (labeled A), from 20 to 28 (labeled B) and from 29 to 30 (labeled C). A chart of this sort quickly enables a used to see which supplier provides a disproportionally large amount of defects.

Another variation that is extremely useful is one that has two scales on the y-axis. On the left side is the number of defects and the right side is total percentage of defects. Suppose there are defects of types A, B, C and D. There are 100 defects total, 35 of which are type A, 30 are type B, 25 are type C and 10 are type D. We have a bar that extends vertically above A up to 35, a bar vertically over B up to 30 and so on. Now we place dots which match the cumulative percent on the right side. For example, 35 is 35 percent of the total defects, so place a dot above defect A which lines up with 35 on the right side. Now B is 30, but since we are doing cumulative, we add it to 35 to get 65 and place a dot that lines up with 65 above B. With C, the dot will go at 90 since (65 + 25 = 90) and the final dot goes above D at 100. Connect the dots to complete the chart.

Generally speaking, the Pareto chart is a very useful SPC problem solving tool. There are five others tools which I may discuss in future articles on statistical process control.

There are variations to the basic Pareto chart I discussed above. Suppose we have components on a circuit board labeled by part number and assume that we know the percentage of defective components. We could have a chart that lists the percentage of defects on the vertical axis and the component number on the horizontal axis.

Suppose we know the components come from three suppliers, A, B and C. We could have a chart with the number of defective components on the vertical axis and the component number listed by supplier on the horizontal axis. For example, suppose for component number 1 that there are 30 overall defects, 20 from supplier A, 8 from supplier B and only 2 from supplier C. The vertical bar above component 1 would extend to 20 (labeled A), from 20 to 28 (labeled B) and from 29 to 30 (labeled C). A chart of this sort quickly enables a used to see which supplier provides a disproportionally large amount of defects.

Another variation that is extremely useful is one that has two scales on the y-axis. On the left side is the number of defects and the right side is total percentage of defects. Suppose there are defects of types A, B, C and D. There are 100 defects total, 35 of which are type A, 30 are type B, 25 are type C and 10 are type D. We have a bar that extends vertically above A up to 35, a bar vertically over B up to 30 and so on. Now we place dots which match the cumulative percent on the right side. For example, 35 is 35 percent of the total defects, so place a dot above defect A which lines up with 35 on the right side. Now B is 30, but since we are doing cumulative, we add it to 35 to get 65 and place a dot that lines up with 65 above B. With C, the dot will go at 90 since (65 + 25 = 90) and the final dot goes above D at 100. Connect the dots to complete the chart.

Generally speaking, the Pareto chart is a very useful SPC problem solving tool. There are five others tools which I may discuss in future articles on statistical process control.

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