Statistics: what is a z-score?

Tutoring some high school math courses, statistics is prominent.  The tutor introduces an important definition.

 
Statistics focuses on the mean and the standard deviation. But why is that? The reason is that while most phenomena follow a normal distribution, very few follow the standard normal distribution. Yet, the standard normal distribution is what we have tables for.

The z-score of a measurement “standardizes” it; i.e., it tells where that measurement would fall in the standard normal distribution, based on its placement in its own distribution.

The z-score of measurement x is given by

z=(x-μ)/σ

where

μ = mean of population x is from
σ = standard deviation of the population x is from

Example:

Find the z score of the exam mark 72 if the mean mark is 61 and the standard deviation is 12.

Solution:

z = (x-μ)/σ = (72-61)/12 = 11/12 = 0.9167

Looking at a z-score table or else the Sharp EL-520W (see here), we get the answer 0.82035: 82% of the students got that mark or below. Therefore, only 18% got above it.

z-scores can be seen as leading to the “bell curve” people refer to in connection with exam marks.

HTH:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Statistics: normal distribution calculations with the Sharp EL-520W

The tutor continues to find handy features on the Sharp EL-520W.  Today: probability associated with the normal distribution.

The Sharp EL-520W just keeps on giving.  Sometimes I wonder, “Will it do that?” Usually, it turns out that it does.

Relating to statistics, I already wrote an article (see here) about how to find standard deviation on the Sharp EL-520W. Today I’ll discuss one of its handy functions for finding normal probabilites.

Example 1: Assume z has a standard normal distribution. Find probability(z<-0.44). Solution: Step 1: Go into STAT mode, by keying MODE, then 1. You will be transferred to another menu: choose SD (which is 0). The screen should now say Stat 0.

Step 2: Press MATH. You will arrive in a menu. To calculate p(z<-0.44), choose P (which is choice 1). The screen should now say P(_. It’s waiting for you to enter your value.

Step 3: Key in -0.44, then close the bracket. Next, press equals. Hopefully, you see the answer 0.329968

Example 2: Find prob(z<1.1). Solution: following the steps above, entering 1.1 instead of -0.44, you hopefully arrive at the answer 0.864334

There are even more functions to discuss, but that’ll suffice for this post.

The values the calculator is giving match the values in the standard normal table (aka, z-score table) found in Appendix B of my source listed below. That’s nice confirmation:)

Source:

Zimmer, Cooke et al. Mathematics of Data Management. Toronto: Nelson, 2003.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Consumer Education: the Consumer Price Index

When the tutor was in high school, a course called Consumer Education was mandatory.

The beautiful, energetic high school students of today will, sooner or later, be workers. They will gain one kind of freedom at the expense of another.

Even at age 18, I never paid attention to prices.  By age 22, I did.  A university student sharing a 1BR apartment, I was very poor.  I finished my degree at age 25, always working a minimum wage job on the side.  Being young was fun in ways, but making a living was sobering.

Statistics Canada has a measure called the Consumer Price Index or CPI .  For many, its immediate use is tracking the cost of living.

The scale works as follows:  the base cost of 100 represents the year 2002.  Each year (or month) thence is given a number that compares its cost of living to that of 2002.  For BC, the November 2014 CPI was 118.8 (statcan.gc.ca), meaning the cost of living had increased by 18.8% since 2002.

Of course, the cost of living is only half the story.  To begin to know how much harder, or easier, it is to make a living, we need to compare the increase in cost of living with the increase in wages.

Among the people most affected by changing prices are those earning minimum wage. In 2002, the BC minimum wage was $8/hr (hrvoice.org); now, it’s $10.25 (globalnews.ca).  The percent increase is (10.25-8)/8 = 0.28 or 28%.

Although living’s not easy on minumum wage, it seems the wage itself has gone up by 28% since 2002, while the cost of living has increased by 18.8% over that period. That’s for BC.

There is endless material for discussion in the world of home economics. Look for more posts about it coming soon:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Physics: the speed of sound

While giving a couple of numerical examples, the tutor reminisces about sound.

When I was a kid I played baseball.  In that town, there wasn’t much else to do except play it or watch it.  The men played softball.

I recall lying in the summer heat, watching from the grass far beyond the left field fence. The pitcher threw underhand, the ball arced through the air, and the batter swung.  The ball started up and away.  Then, I heard the metallic “conk” of the aluminum bat striking the ball.  I remember being surprised how late the sound reached me after I’d seen the ball take off.

Looking back, I might have been about 180m from home plate.  The temperature was likely around 28C.  The speed of sound, s, in metres per second, is given by

s=(332 + 0.6T),   where T is in degrees C.

The speed of sound that day would likely have been 332 + 0.6(28) = 348.8 ≈ 349m/s.

Given my distance of 180m, and the fact that

time=distance/speed

it would have taken about 180/349 = 0.516 ≈ 0.52s, or about half a second, for the sound to reach me.

One more recollection: Maybe the same year as the event above, I saw a documentary about the first nuclear test in the American southwest. One of the scientists had watched the explosion from an observation post that, I believe, was 10 miles away. He commented that the light flash from the explosion was instantaneous. “Then, a long time later…the sound.”

Let’s imagine the temp in that situation was also 28C, in which case the speed of sound, as calculated above, would have been 349m/s. If I recall, there is roughly 1600m to a mile. The scientist’s observation post was 10(1600m) = 16000m from the blast site. How long did the sound take to reach him?

time=distance/speed=16000/349 = 45.8s ≈ 49s

The scientist might have waited about 49 seconds between the light flash and the sound from the atomic test.

Well, that’s all for today. I hope you enjoyed the memories:)

Source:

Heath, Robert et al. Fundamentals of Physics. Heath Canada, 1981.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Perl: the time vector

The tutor continues about Perl’s handling of time.

 
In yesterday’s post I began about Perl’s time utilities.

Besides localtime, there is simply time. It returns the number of seconds since January 1, 1970 on the Windows system I use, as well as on most others (tutorialspoint.com).

These few lines of code

#!/usr/bin/perl
$tim0=time;
print “The time is $tim0\n”;

will hopefully give output like

The time is 1421093858

There is even more to Perl’s handling of time than we’ve seen so far. Perl offers a veritable vector of time information to those desiring it. Consider the following little program:

#!/usr/bin/perl
@timarray=localtime;
$length=@timarray; #gives length of the array
for($i=0; $i<$length; $i++){
print “Element $i of Perl’s time array is $timarray[$i]\n”;
}

Running the lines above, you may be greeted with output like this:

Element 0 of Perl’s time array is 2 //seconds
Element 1 of Perl’s time array is 44 //minutes
Element 2 of Perl’s time array is 12 //hours
Element 3 of Perl’s time array is 12 //the day this month
Element 4 of Perl’s time array is 0 //the month: Jan=0
Element 5 of Perl’s time array is 115 //the year: 1900=0
Element 6 of Perl’s time array is 1 //day of the week: Sun=0
Element 7 of Perl’s time array is 11 //the day this year:Jan 1=0
Element 8 of Perl’s time array is 0 //daylight savings: no=0, yes=1

For now, this wraps up my coverage of Perl time functions. HIH:)

Source: McGrath, Mike. Perl in easy steps. Southam: Computer Step, 2004.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Perl: the time: localtime

Tutoring computer science, you need to discuss the built-in functions of the language(s) you use.  The tutor opens a discussion about Perl’s handling of time.

Perl, apparently a language meant to process data in large organizations, has a few built-in ways to present the time.

One of Perl’s time functions is localtime. On my systems, these few lines

#!/usr/bin/perl
$tim=localtime;
print “The time is $tim”;

give output like

The time is Sun Jan 11 15:43:32 2015

The above code works for me on both Windows and Linux. However, it uses only one aspect of Perl’s time handling.

Speaking of time: today, I won’t take much of yours:) I will be discussing more about Perl’s built-in time features in future posts.

Source:

McGrath, Mike. Perl in easy steps. Southam: Computer Step, 2004.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Math: solving a system by substitution: an investment problem

Heading towards semester exams, some students face questions like this one. The tutor works this example.

Consider the following problem:

Sue has $2200 to invest for one year.  She has a choice two investments:  investment A pays 4%, while investment B pays 5.4%.  She wants to earn $100 interest for the year. How much should she invest in each choice?

Solution:

Let x=amount of money she puts in A (at 4%)

Let y=amount of money she puts in B (at 5.4%)

Now, we know that

x + y = 2200 (equation 1)

To get the percent of a number, you change the percent into a decimal (by dividing by 100), then multiply.  For instance, 29% of $246=0.29($246)=$71.34.

Therefore, 4% of x is 0.04x and 5.4% of y is 0.054y.  We have

0.04x + 0.054y = 100 (equation 2)

We can solve this system of two equations using substitution:  notice that equation 1 implies

y = 2200 – x

We can substitute 2200 – x, instead of y, in equation 2:

0.04x + 0.054(2200 – x) = 100

Now we simplify and solve:

0.04x + 118.8 – 0.054x = 100

Subtract 118.8 from both sides:

0.04x – 0.054x = -18.8

Simplify the left:

-0.014x = -18.8

Divide both sides by -0.014:

x = -18.8/-0.014 = 1342.86

Now, looking back at equation 1

x + y = 2200

we see that

1342.86 + y = 2200

Subtracting 1342.86 from both sides gives

y = 857.14

Looking back at our let statements, we see that

x = amount invested at 4%

y = amount invested at 5.4%

We now know that she should invest $1342.86 at 4% and $857.14 at 5.4%.

Check:  1342.86(0.04) + 857.14(0.054) = 53.71 + 46.29 = 100

From a high school math point of view, this question is settled.

Questions arise about the investment aspect of this problem.  However, they belong to a different discipline.  We might explore some of those questions in future posts:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Physics: force of gravity on other planets

Tutoring physics, this is a matter of interest.  The tutor discusses a topic that may also be of interest to space-philes….

On Earth, gravity is -9.8N/kg. The negative just means downward.

Occasionally, as a kid, I’d see a sci-fi show in which an enemy was from a “high-gravity” planet. That person was likely, of course, much stronger than the protagonist. There might be a scene in which the enemy chased the protagonist about, throwing furniture that a normal person could barely lift. I think such a scene played in an episode of Buck Rogers in the 25th Century.

Although I only saw that few minutes of the episode, it raised my awareness that other planets may not have the same gravitional force as Earth. Indeed, the planets in our solar system generally have different gravitational pulls. Let’s see if we can predict the gravitational force on a given planet.

Theoretically, the force Fp due to gravity on a given planet can be calculated by

Fp=Fe*mp_rel_e/d^2p_rel_e in which

Fe = force of gravity on Earth (once again: -9.8N/kg)

m p_rel_e= mass of the given planet, relative to earth’s

dp_rel_e= diameter of the given planet, relative to earth’s

Let’s see if our formula works. Looking at the very convenient planetary fact sheet provided online by NASA, we see that Jupiter has the following statistics:

mass: 317.8 times that of Earth

diameter: 11.21 times that of Earth

Our formula suggests that the force of gravity on Jupiter should be given by

Fj=-9.8*(317.8/11.21^2)=-24.8N/kg

The same planetary fact sheet reports Jupiter’s gravity to be 2.36 times that of Earth, which would be 2.36(-9.8N/kg)=-23.1N/kg.

The difference between the two figures for Jupiter’s gravity is -1.7N/kg, or 7%. Close enough? You be the judge:)

Source:

Giancoli, Douglas C. Physics, 5th Ed. New Jersey: Prentice Hall, 1998.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Calories burned during running, continued….

The tutor continues an exploration of calorie burn during running.

In my previous post I calculated, using chemistry methods, the number of calories I burned during a 45-min, 8km (5mile) run.  While the chemistry was sound, the process hinged on my estimated breathing rate:  1 litre exhaled every 2 seconds.

How much air a person really inhales or exhales is very tricky to estimate just by feel. I think my estimate was the best I could do, but it may well be significantly different from reality.

Well, I looked at two websites today to get other opinions on the calories I might have burned during yesterday’s run. For a 180lb runner, Nutristrategy.com gives an hourly rate for a 6.7mph pace (5 miles in 45 min) of 899 calories. 45 minutes is 0.75 of that, implying 0.75(899)=674 calories.

Healthstatus.com gives 704 calories for the same activity at a 7mph pace. To pro-rate it down to 6.7mph, I do 704(6.7/7)=674 calories.

Impressively, both websites give the exact same calorie count for my 45 minute, 5-mile run. My figure of 288 calories suggests my breathing estimate was quite far off.

In experimental science, a committed estimate is often required – especially at the beginning. The estimate’s accuracy will be determined in due time by repetition and cross-referencing.

Perhaps I did better during yesterday’s run than I’d thought:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Back to it, part II: post-holiday exercise

A tutor, just like anyone else, is likely conscious of society’s love of fitness (and thinness).

Pretty much on my 40th birthday, I could, all of a sudden, gain weight just by eating.  I’d always been a bone rack; in my 30s I could gain weight by exercise, but my body wouldn’t put on fat.  My, oh my:  what a different world I live in now, at age 4_.

Over the holidays, like so many people, I put on about 10 lbs.  As of this morning, five of those 10 lbs were already gone.  Then, I went for a 45 min run (which, by the way, turned out to be much cooler and drier than I’d expected).

Here’s the topical question:  how many calories was that run worth?

There are many ways to calculate the answer to that question.  Today we’ll do it chemically.

To start with, according to the bbc, inhaled air has 0.04% CO2, while exhaled air has 4% CO2. We’ll take that to mean you breath in virtually no CO2, while a breath out is 4% CO2. We are going to track the consumption of calories via the production of CO2.

I’d estimate, during that run, I exhaled a litre of air every 2 seconds. Therefore, I exhaled 4% of that, or 0.04L, of CO2 every 2 seconds. In one minute, that’s 30(0.04)=1.2L; in 45 min, it’s 45(1.2L)=54L.

At STP (standard temperature and pressure: 0°C, 1atm pressure), there is 22.4L of gas in a mole (for what a mole is, see my post here). The conditions of my run were virtually STP (it’s January). 54L of CO2 is 54/22.4 = 2.4 moles CO2. The molar mass (see my post here for more about the molar mass) of CO2 is 12 +16+16=44g. Therefore, on the run, I produced 2.4x44g=105.6g CO2.

The chemical equation for burning glucose is

C6H12O6 + 6O2 → 6CO2 + 6H2O

Translating to molar masses, we could write the equation as

180g glucose + 192g oxygen → 264g carbon dioxide + 108g water

The equation indicates a consumption ratio of 180g glucose for every 264g CO2 produced. Since I produced 105.6g CO2, my glucose consumption would have been 105.6(180/264)=72g. Apparently, I consumed 72g glucose on that run.

Calorie King tells me that glucose is 4 calories per gram. Therefore, the 72g glucose consumed on that run is worth 72(4)=288 calories.

Apparently, I burned 288 calories during my 45 minute run. We’ll look into how else one could calculate the calorie burn, how similar those findings are, and much more about this topic in future posts:)

Source:

Mortimer, Charles E. Chemistry, 6th Ed. Belmont: Wadsworth, 1986.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.