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2.7: Sampling Experiment (Worksheet)


Student Learning Outcomes

  • The student will demonstrate the simple random, systematic, stratified, and cluster sampling techniques.
  • The student will explain the details of each procedure used.

In this lab, you will be asked to pick several random samples of restaurants. In each case, describe your procedure briefly, including how you might have used the random number generator, and then list the restaurants in the sample you obtained.

Note 1.7.1

The following section contains restaurants stratified by city into columns and grouped horizontally by entree cost (clusters).

Restaurants Stratified by City and Entree Cost

Restaurants Used in Sample
Entree CostUnder $10$10 to under $15$15 to under $20Over $20
San JoseEl Abuelo Taq, Pasta Mia, Emma’s Express, Bamboo HutEmperor’s Guard, Creekside InnAgenda, Gervais, Miro’sBlake’s, Eulipia, Hayes Mansion, Germania
Palo AltoSenor Taco, Olive Garden, Taxi’sMing’s, P.A. Joe’s, Stickney’sScott’s Seafood, Poolside Grill, Fish MarketSundance Mine, Maddalena’s, Spago’s
Los GatosMary’s Patio, Mount Everest, Sweet Pea’s, Andele TaqueriaLindsey’s, Willow StreetToll HouseCharter House, La Maison Du Cafe
Mountain ViewMaharaja, New Ma’s, Thai-Rific, Garden FreshAmber Indian, La Fiesta, Fiesta del Mar, DawitAustin’s, Shiva’s, MazehLe Petit Bistro
CupertinoHobees, Hung Fu, Samrat, Panda ExpressSanta Barb. Grill, Mand. Gourmet, Bombay Oven, Kathmandu WestFontana’s, Blue PheasantHamasushi, Helios
SunnyvaleChekijababi, Taj India, Full Throttle, Tia Juana, Lemon GrassPacific Fresh, Charley Brown’s, Cafe Cameroon, Faz, Aruba’sLion & Compass, The Palace, Beau Sejour
Santa ClaraRangoli, Armadillo Willy’s, Thai Pepper, PasandArthur’s, Katie’s Cafe, Pedro’s, La GalleriaBirk’s, Truya Sushi, Valley PlazaLakeside, Mariani’s

A Simple Random Sample

Pick a simple random sample of 15 restaurants.

  1. Describe your procedure.
  2. Complete the table with your sample.
    1. __________6. __________11. __________
    2. __________7. __________12. __________
    3. __________8. __________13. __________
    4. __________9. __________14. __________
    5. __________10. __________15. __________

A Systematic Sample

Pick a systematic sample of 15 restaurants.

  1. Describe your procedure.
  2. Complete the table with your sample.
    1. __________

A Stratified Sample

Pick a stratified sample, by city, of 20 restaurants. Use 25% of the restaurants from each stratum. Round to the nearest whole number.

  1. Describe your procedure.
  2. Complete the table with your sample.
    1. __________16. __________17. __________18. __________19. __________20. __________

A Stratified Sample

Pick a stratified sample, by entree cost, of 21 restaurants. Round to the nearest whole number.

  1. Describe your procedure.
  2. Complete the table with your sample.
    1. __________
    21. __________

A Cluster Sample

Pick a cluster sample of restaurants from two cities. The number of restaurants will vary.

  1. Describe your procedure.
  2. Complete the table with your sample.
    1. ________6. ________11. ________16. ________21. ________
    2. ________7. ________12. ________17. ________22. ________
    3. ________8. ________13. ________18. ________23. ________
    4. ________9. ________14. ________19. ________24. ________
    5. ________10. ________15. ________20. ________25. ________

Science Worksheets

Learn to classify animal (vertebrate) groups with these printable animal worksheets. Learn about mammals, reptiles, birds, fish, and amphibians.

Read about your all of your favorite animal species.

Learn about plant and animal cells with these diagrams, worksheets, and activities.

Print dinosaur reading comprehension articles, dinosaur puzzles, dino math pages, and more.

Explore current electricity and circuits with these worksheets and activities.

Learn about the senses of sight, touch, hearing, smell, and taste with these printables.

Explore food chains and ecosystems with these printables.

Measure the volume of liquids in the graduated cylinders.

Discover the workings of the human body with these articles and worksheets.

Learn about the anatomy and life cycle of insects.

Our invertebrates page has a classroom scavenger hunt, word maze puzzles, reading comprehension passages, a cut-and-sort activity, and a classroom scavenger hunt.

Learn landform vocabulary words, such as plain, plateau, mesa, volcano, cliff, isthmus, mountain, and hill.

Life cycle wheel, scavenger hunt, worksheets, and a mini-book about butterflies

Scavenger hunt, worksheets, life cycle wheel, and a mini-book about frogs

Find out how a mealworm changes into an adult darkling beetle.

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Learn about magnetism with these experiments and worksheets.

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This page has worksheets, learning centers, and activities to help students learn about healthy and unhealthy food choices.

This page has everything you need for a unit on owls and for an owl pellet dissection project.

Learn all about plant life with these printables.

Try these fun and simple science activities in your classroom.

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Learn all about our planet with these worksheets of layers of the Earth, rocks, and volcanoes.

Use these worksheets and activities to teach students about weather.

A full index of all Math, ELA, Spelling, Phonics, Grammar, Science, and Social Studies worksheets found on this website.


Probability

Definition

The probability of an outcome A number that measures the likelihood of the outcome. e in a sample space S is a number p between 0 and 1 that measures the likelihood that e will occur on a single trial of the corresponding random experiment. The value p = 0 corresponds to the outcome e being impossible and the value p = 1 corresponds to the outcome e being certain.

Definition

The probability of an event A number that measures the likelihood of the event. A is the sum of the probabilities of the individual outcomes of which it is composed. It is denoted P ( A ) .

The following formula expresses the content of the definition of the probability of an event:

If an event E is E = < e 1 , e 2 , … , e k >, then

P ( E ) = P ( e 1 ) + P ( e 2 ) + · · · + P ( e k )

Figure 3.3 Sample Spaces and Probability

Since the whole sample space S is an event that is certain to occur, the sum of the probabilities of all the outcomes must be the number 1.

In ordinary language probabilities are frequently expressed as percentages. For example, we would say that there is a 70% chance of rain tomorrow, meaning that the probability of rain is 0.70. We will use this practice here, but in all the computational formulas that follow we will use the form 0.70 and not 70%.

Example 5

A coin is called “balanced” or “fair” if each side is equally likely to land up. Assign a probability to each outcome in the sample space for the experiment that consists of tossing a single fair coin.

With the outcomes labeled h for heads and t for tails, the sample space is the set S = < h , t >. Since the outcomes have the same probabilities, which must add up to 1, each outcome is assigned probability 1/2.

Example 6

A die is called “balanced” or “fair” if each side is equally likely to land on top. Assign a probability to each outcome in the sample space for the experiment that consists of tossing a single fair die. Find the probabilities of the events E: “an even number is rolled” and T: “a number greater than two is rolled.”

With outcomes labeled according to the number of dots on the top face of the die, the sample space is the set S = < 1,2,3,4,5,6 >. Since there are six equally likely outcomes, which must add up to 1, each is assigned probability 1/6.

Since E = < 2,4,6 >, P ( E ) = 1 ∕ 6 + 1 ∕ 6 + 1 ∕ 6 = 3 ∕ 6 = 1 ∕ 2 .

Since T = < 3,4,5,6 >, P ( T ) = 4 ∕ 6 = 2 ∕ 3 .

Example 7

Two fair coins are tossed. Find the probability that the coins match, i.e., either both land heads or both land tails.

In Note 3.8 "Example 3" we constructed the sample space S = < 2 h , 2 t , d >for the situation in which the coins are identical and the sample space S ′ = < h h , h t , t h , t t >for the situation in which the two coins can be told apart.

The theory of probability does not tell us how to assign probabilities to the outcomes, only what to do with them once they are assigned. Specifically, using sample space S, matching coins is the event M = < 2 h , 2 t >, which has probability P ( 2 h ) + P ( 2 t ) . Using sample space S ′ , matching coins is the event M ′ = < h h , t t >, which has probability P ( h h ) + P ( t t ) . In the physical world it should make no difference whether the coins are identical or not, and so we would like to assign probabilities to the outcomes so that the numbers P ( M ) and P ( M ′ ) are the same and best match what we observe when actual physical experiments are performed with coins that seem to be fair. Actual experience suggests that the outcomes in S ′ are equally likely, so we assign to each probability 1∕4, and then

P ( M ′ ) = P ( h h ) + P ( t t ) = 1 4 + 1 4 = 1 2

Similarly, from experience appropriate choices for the outcomes in S are:

P ( 2 h ) = 1 4 P ( 2 t ) = 1 4 P ( d ) = 1 2

which give the same final answer

P ( M ) = P ( 2 h ) + P ( 2 t ) = 1 4 + 1 4 = 1 2

The previous three examples illustrate how probabilities can be computed simply by counting when the sample space consists of a finite number of equally likely outcomes. In some situations the individual outcomes of any sample space that represents the experiment are unavoidably unequally likely, in which case probabilities cannot be computed merely by counting, but the computational formula given in the definition of the probability of an event must be used.

Example 8

The breakdown of the student body in a local high school according to race and ethnicity is 51% white, 27% black, 11% Hispanic, 6% Asian, and 5% for all others. A student is randomly selected from this high school. (To select “randomly” means that every student has the same chance of being selected.) Find the probabilities of the following events:

  1. B: the student is black,
  2. M: the student is minority (that is, not white),
  3. N: the student is not black.

The experiment is the action of randomly selecting a student from the student population of the high school. An obvious sample space is S = < w , b , h , a , o >. Since 51% of the students are white and all students have the same chance of being selected, P ( w ) = 0.51 , and similarly for the other outcomes. This information is summarized in the following table:

  1. Since B = < b >, P ( B ) = P ( b ) = 0.27 .
  2. Since M = < b , h , a , o >, P ( M ) = P ( b ) + P ( h ) + P ( a ) + P ( o ) = 0.27 + 0.11 + 0.06 + 0.05 = 0.49
  3. Since N = < w , h , a , o >, P ( N ) = P ( w ) + P ( h ) + P ( a ) + P ( o ) = 0.51 + 0.11 + 0.06 + 0.05 = 0.73

Example 9

The student body in the high school considered in Note 3.18 "Example 8" may be broken down into ten categories as follows: 25% white male, 26% white female, 12% black male, 15% black female, 6% Hispanic male, 5% Hispanic female, 3% Asian male, 3% Asian female, 1% male of other minorities combined, and 4% female of other minorities combined. A student is randomly selected from this high school. Find the probabilities of the following events:

  1. B: the student is black,
  2. M F : the student is minority female,
  3. F N : the student is female and is not black.

Now the sample space is S = < w m , b m , h m , a m , o m , w f , b f , h f , a f , o f >. The information given in the example can be summarized in the following table, called a two-way contingency table:

Gender Race / Ethnicity
White Black Hispanic Asian Others
Male 0.25 0.12 0.06 0.03 0.01
Female 0.26 0.15 0.05 0.03 0.04
  1. Since B = < b m , b f >, P ( B ) = P ( b m ) + P ( b f ) = 0.12 + 0.15 = 0.27 .
  2. Since M F = < b f , h f , a f , o f >, P ( M ) = P ( b f ) + P ( h f ) + P ( a f ) + P ( o f ) = 0.15 + 0.05 + 0.03 + 0.04 = 0.27
  3. Since F N = < w f , h f , a f , o f >, P ( F N ) = P ( w f ) + P ( h f ) + P ( a f ) + P ( o f ) = 0.26 + 0.05 + 0.03 + 0.04 = 0.38

Key Takeaways

  • The sample space of a random experiment is the collection of all possible outcomes.
  • An event associated with a random experiment is a subset of the sample space.
  • The probability of any outcome is a number between 0 and 1. The probabilities of all the outcomes add up to 1.
  • The probability of any event A is the sum of the probabilities of the outcomes in A.

Exercises

Basic

A box contains 10 white and 10 black marbles. Construct a sample space for the experiment of randomly drawing out, with replacement, two marbles in succession and noting the color each time. (To draw “with replacement” means that the first marble is put back before the second marble is drawn.)

A box contains 16 white and 16 black marbles. Construct a sample space for the experiment of randomly drawing out, with replacement, three marbles in succession and noting the color each time. (To draw “with replacement” means that each marble is put back before the next marble is drawn.)

A box contains 8 red, 8 yellow, and 8 green marbles. Construct a sample space for the experiment of randomly drawing out, with replacement, two marbles in succession and noting the color each time.

A box contains 6 red, 6 yellow, and 6 green marbles. Construct a sample space for the experiment of randomly drawing out, with replacement, three marbles in succession and noting the color each time.

In the situation of Exercise 1, list the outcomes that comprise each of the following events.

In the situation of Exercise 2, list the outcomes that comprise each of the following events.

  1. At least one marble of each color is drawn.
  2. No white marble is drawn.
  3. More black than white marbles are drawn.

In the situation of Exercise 3, list the outcomes that comprise each of the following events.

  1. No yellow marble is drawn.
  2. The two marbles drawn have the same color.
  3. At least one marble of each color is drawn.

In the situation of Exercise 4, list the outcomes that comprise each of the following events.

  1. No yellow marble is drawn.
  2. The three marbles drawn have the same color.
  3. At least one marble of each color is drawn.

Assuming that each outcome is equally likely, find the probability of each event in Exercise 5.

Assuming that each outcome is equally likely, find the probability of each event in Exercise 6.

Assuming that each outcome is equally likely, find the probability of each event in Exercise 7.

Assuming that each outcome is equally likely, find the probability of each event in Exercise 8.

A sample space is S = < a , b , c , d , e >. Identify two events as U = < a , b , d >and V = < b , c , d >. Suppose P ( a ) and P ( b ) are each 0.2 and P ( c ) and P ( d ) are each 0.1.

A sample space is S = < u , v , w , x >. Identify two events as A = < v , w >and B = < u , w , x >. Suppose P ( u ) = 0.22 , P ( w ) = 0.36 , and P ( x ) = 0.27 .

A sample space is S = < m , n , q , r , s >. Identify two events as U = < m , q , s >and V = < n , q , r >. The probabilities of some of the outcomes are given by the following table:

A sample space is S = < d , e , f , g , h >. Identify two events as M = < e , f , g , h >and N = < d , g >. The probabilities of some of the outcomes are given by the following table:

Applications

The sample space that describes all three-child families according to the genders of the children with respect to birth order was constructed in Note 3.9 "Example 4". Identify the outcomes that comprise each of the following events in the experiment of selecting a three-child family at random.

  1. At least one child is a girl.
  2. At most one child is a girl.
  3. All of the children are girls.
  4. Exactly two of the children are girls.
  5. The first born is a girl.

The sample space that describes three tosses of a coin is the same as the one constructed in Note 3.9 "Example 4" with “boy” replaced by “heads” and “girl” replaced by “tails.” Identify the outcomes that comprise each of the following events in the experiment of tossing a coin three times.

  1. The coin lands heads more often than tails.
  2. The coin lands heads the same number of times as it lands tails.
  3. The coin lands heads at least twice.
  4. The coin lands heads on the last toss.

Assuming that the outcomes are equally likely, find the probability of each event in Exercise 17.

Assuming that the outcomes are equally likely, find the probability of each event in Exercise 18.

Additional Exercises

The following two-way contingency table gives the breakdown of the population in a particular locale according to age and tobacco usage:

A person is selected at random. Find the probability of each of the following events.

  1. The person is a smoker.
  2. The person is under 30.
  3. The person is a smoker who is under 30.

The following two-way contingency table gives the breakdown of the population in a particular locale according to party affiliation (A, B, C, or None) and opinion on a bond issue:

Affiliation Opinion
Favors Opposes Undecided
A 0.12 0.09 0.07
B 0.16 0.12 0.14
C 0.04 0.03 0.06
None 0.08 0.06 0.03

A person is selected at random. Find the probability of each of the following events.

  1. The person is affiliated with party B.
  2. The person is affiliated with some party.
  3. The person is in favor of the bond issue.
  4. The person has no party affiliation and is undecided about the bond issue.

The following two-way contingency table gives the breakdown of the population of married or previously married women beyond child-bearing age in a particular locale according to age at first marriage and number of children:

Age Number of Children
0 1 or 2 3 or More
Under 20 0.02 0.14 0.08
20–29 0.07 0.37 0.11
30 and above 0.10 0.10 0.01

A woman is selected at random. Find the probability of each of the following events.

  1. The woman was in her twenties at her first marriage.
  2. The woman was 20 or older at her first marriage.
  3. The woman had no children.
  4. The woman was in her twenties at her first marriage and had at least three children.

The following two-way contingency table gives the breakdown of the population of adults in a particular locale according to highest level of education and whether or not the individual regularly takes dietary supplements:

Education Use of Supplements
Takes Does Not Take
No High School Diploma 0.04 0.06
High School Diploma 0.06 0.44
Undergraduate Degree 0.09 0.28
Graduate Degree 0.01 0.02

An adult is selected at random. Find the probability of each of the following events.

  1. The person has a high school diploma and takes dietary supplements regularly.
  2. The person has an undergraduate degree and takes dietary supplements regularly.
  3. The person takes dietary supplements regularly.
  4. The person does not take dietary supplements regularly.

Large Data Set Exercises

Note: These data sets are missing, but the questions are provided here for reference.

Large Data Sets 4 and 4A record the results of 500 tosses of a coin. Find the relative frequency of each outcome 1, 2, 3, 4, 5, and 6. Does the coin appear to be “balanced” or “fair”?


A heterogeneous mixture is a mixture in which the composition is not uniform throughout the mixture. Vegetable soup is a heterogeneous mixture. Any given spoonful of soup will contain varying amounts of the different vegetables and other components of the soup.

A phase is any part of a sample that has a uniform composition and properties. By definition, a pure substance or a homogeneous mixture consists of a single phase. A heterogeneous mixture consists of two or more phases. When oil and water are combined, they do not mix evenly, but instead form two separate layers. Each of the layers is called a phase.

Oil and water do not mix, instead forming two distinct layers called phases. The oil phase is less dense than the water phase, and so the oil floats on top of the water.

In the case of vegetable soup, one phase would be the liquid soup itself. This phase has vitamins, minerals, and other components dissolved in the water. This phase would be homogeneous. The carrots, peas, corn, or other vegetables represent other phases of the soup.The various vegetables are not mixed evenly in the soup, but are spread around at random.

There are a large number of heterogeneous mixtures around us. Soil is composed of a variety of substances, and is often of different composition depending on the sample taken. One shovelful may come up with dirt and grass, while the next shovelful could contain an earthworm.

Smog is another example of a heterogeneous mixture. This murky collection of pollutants can be a mixture of water and contaminants from burning gasoline or plastics, mixed with nitric oxide derivatives and ozone. You can see that the smog distribution in the air illustrated below is not evenly spread out, but varies from one part of the atmosphere to another.


Sample Space In Probability

In these lessons, we will learn simple probability, experiments, outcomes, sample space and probability of an event.

The following diagram shows how the sample space for an experiment can be represented by a list, a table, and a tree diagram. Scroll down the page for examples and solutions.


Sample Space

In the study of probability, an experiment is a process or investigation from which results are observed or recorded.

An outcome is a possible result of an experiment.

A sample space is the set of all possible outcomes in the experiment. It is usually denoted by the letter S. Sample space can be written using the set notation, < >.

Experiment 1: Tossing a coin
Possible outcomes are head or tail.
Sample space, S =

Experiment 2: Tossing a die
Possible outcomes are the numbers 1, 2, 3, 4, 5, and 6
Sample space, S =

Experiment 3: Picking a card
In an experiment, a card is picked from a stack of six cards, which spell the word PASCAL.
Possible outcomes are P, A 1, S, C, A 2 and L.
Sample space, S = 1, S, C, A 2 L>. There are 2 cards with the letter ‘A’

Experiment 4: Picking 2 marbles, one at a time, from a bag that contains many blue and red marbles.
Possible outcomes are: (Blue, Blue), (Blue, Red), (Red, Blue) and (Red, Red).
Sample space, S = <(B,B), (B,R), (R,B), (R,R)>.

A simple explanation of Sample Spaces for Probability

Sample Space Of An Event

Sample space is all the possible outcomes of an event. Sometimes the sample space is easy to determine. For example, if you roll a dice, 6 things could happen. You could roll a 1, 2, 3, 4, 5, or 6.

Sometimes sample space is more difficult to determine, so you can make a tree diagram or a list to help you figure out all the possible outcomes.

Example 1:
You are ordering pizza. You can choose a small, medium or large pizza and you can choose cheese or pepperoni. What are the possible ways that you could could order a pizza? How many combinations could you have?

Example 2:
Daisy has 3 pairs of shorts, 2 pairs of shoes and 5 t-shirts. How many outfits can she make?

This lesson is on finding simple probabilities and sample spaces.

Example:
When you roll a die,

Example:
Use the spinner below to answer the following questions:

  1. What is the sample space?
  2. P(Blue)
  3. P(Orange or Green)
  4. P(Not Red)
  5. P(Purple)

The following video explains simple probability, experiments, outcomes, sample space and probability of an event. It also gives an example of a simple probability problem.

Example:
A jar contains five balls that are numbered 1 to 5. Also, two of the balls are yellow and the others are red. They are numbered and colored as shown below.

  1. Find the probability of randomly selecting a red ball.
  2. Find the probability of randomly selecting an even number ball.

Lists and Sample Spaces - Probability

Example:
Entrees - Ribs, Chicken
Sides - Mac and Cheese, Veggies, Mashed Potatoes
Drinks - Water, Coffee, Milk
What are the different possibilities for the menu?

Explains three methods for listing the sample space of an event and introduces conditional probability: List, Table, Tree Diagram.

Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.


The sample data on this page is sales data for an imaginary office supply company, and each row represents one order. Each row shows:

  • OrderDate: when the order was placed,
  • Region: geographical area in which the sale was made
  • Rep: sales representative's name
  • Item: name of the item sold
  • Units: number of units sold
  • UnitCost: cost of one unit
  • Total: total cost of the order - Units x UnitCost

Separating Mixtures Through Physical Changes

Homogeneous mixtures (solutions) can be separated into their component substances by physical processes that rely on differences in some physical property, such as differences in their boiling points. Two of these separation methods are distillation and crystallization. Distillation makes use of differences in volatility, a measure of how easily a substance is converted to a gas at a given temperature. A simple distillation apparatus for separating a mixture of substances, at least one of which is a liquid. The most volatile component boils first and is condensed back to a liquid in the water-cooled condenser, from which it flows into the receiving flask. If a solution of salt and water is distilled, for example, the more volatile component, pure water, collects in the receiving flask, while the salt remains in the distillation flask.

Figure (PageIndex<3>): The Distillation of a Solution of Table Salt in Water. The solution of salt in water is heated in the distilling flask until it boils. The resulting vapor is enriched in the more volatile component (water), which condenses to a liquid in the cold condenser and is then collected in the receiving flask.

Mixtures of two or more liquids with different boiling points can be separated with a more complex distillation apparatus. One example is the refining of crude petroleum into a range of useful products: aviation fuel, gasoline, kerosene, diesel fuel, and lubricating oil (in the approximate order of decreasing volatility). Another example is the distillation of alcoholic spirits such as brandy or whiskey. This relatively simple procedure caused more than a few headaches for federal authorities in the 1920s during the era of Prohibition, when illegal stills proliferated in remote regions of the United States.

Another example for using physical properties to separate mixtures is filtration (Figure (PageIndex<4>)). Filtration is any of various mechanical, physical or biological operations that separate solids from fluids (liquids or gases) by adding a medium through which only the fluid can pass. The fluid that passes through is called the filtrate. There are many different methods of filtration all aim to attain the separation of substances. Separation is achieved by some form of interaction between the substance or objects to be removed and the filter. The substance that is to pass through the filter must be a fluid, i.e. a liquid or gas. Methods of filtration vary depending on the location of the targeted material, i.e. whether it is dissolved in the fluid phase or suspended as a solid.

Figure (PageIndex<4>): Filtration for the separation of solids from a hot solution. (CC BY-SA 4.0 Suman6395).

Summary

  • Chemists make a distinction between two different types of changes that they study - physical changes and chemical changes.
  • Physical changes are changes that do not alter the identity of a substance.
  • Chemical changes are changes that occur when one substance is turned into another substance.
  • Chemical changes are frequently harder to reverse than physical changes. Observations that indicate a chemical change occurred include color change, temperature change, light given off, formation of bubbles, formation of a precipitate, etc.

Here is a simple variation of the Brain Bag/Box of Science project. Get 5 to 7 different socks. Put different small objects into each sock. Have other people try to guess what is inside of each sock by touching and feeling the object on the outside of the sock. If they can't guess what the object is, have them put their hand into the sock and feel it. By actually touching the object, you can get more information about the characteristics of the object. For example, its roughness and texture. Try numbering each sock. Then prepare a worksheet identifying the object in each numbered sock.


Survey Methods

This interactive module explores methods used to survey large animal populations, and what they have revealed about the current state of the African elephant population.

African elephants provide a powerful case study for how science can inform conservation. Knowing how many elephants are left and where they live is important in devising strategies to protect them. In this Click & Learn, students consider the advantages and disadvantages of different methods and approaches that scientists use to measure and monitor elephant populations.

The accompanying worksheet guides students’ exploration.

The “Resource Google Folder” link directs to a Google Drive folder of resource documents in the Google Docs format. Not all downloadable documents for the resource may be available in this format. The Google Drive folder is set as “View Only” to save a copy of a document in this folder to your Google Drive, open that document, then select File → “Make a copy.” These documents can be copied, modified, and distributed online following the Terms of Use listed in the “Details” section below, including crediting BioInteractive.


Half-life is defined as the amount of time it takes a given quantity to decrease to half of its initial value. The term is most commonly used in relation to atoms undergoing radioactive decay, but can be used to describe other types of decay, whether exponential or not. One of the most well-known applications of half-life is carbon-14 dating. The half-life of carbon-14 is approximately 5,730 years, and it can be reliably used to measure dates up to around 50,000 years ago. The process of carbon-14 dating was developed by William Libby, and is based on the fact that carbon-14 is constantly being made in the atmosphere. It is incorporated into plants through photosynthesis, and then into animals when they consume plants. The carbon-14 undergoes radioactive decay once the plant or animal dies, and measuring the amount of carbon-14 in a sample conveys information about when the plant or animal died.

Below are shown three equivalent formulas describing exponential decay:

    where
    N0 is the initial quantity
    Nt is the remaining quantity after time, t
    t1/2 is the half-life
    τ is the mean lifetime
    λ is the decay constant

If an archaeologist found a fossil sample that contained 25% carbon-14 in comparison to a living sample, the time of the fossil sample's death could be determined by rearranging equation 1, since Nt, N0, and t1/2 are known.

This means that the fossil is 11,460 years old.

Derivation of the Relationship Between Half-Life Constants

Using the above equations, it is also possible for a relationship to be be derived between t1/2, τ, and λ. This relationship enables the determination of all values, as long as at least one is known.


AP Lab 5 Sample 7

The human body has to have energy in order to perform the functions that allow life. This energy comes from the process of cellular respiration. Cellular respiration releases energy that the body can use in the form of ATP from carbohydrates by using oxygen. Cellular respiration is not just one singular reaction, it is a metabolic pathway made up of several reactions that are enzyme mediated. This process begins with glycolysis in the cytosol of the cell. In glycolysis, glucose is split into two three-carbon compounds called pyruvate, producing a small amount of ATP The final two steps of cellular respiration occur in the mitochondria. These final two steps are the electron transport system and the Krebs Cycle. The overall equation for cellular respiration is

C6H12O6 + 6O2 -> 6CO2 + 6H2O + 686 kilocalories of energy per mole of glucose oxidized.

There are three ways to measure the rate of cellular respiration. These three ways are by measuring the consumption of oxygen gas, by measuring the production of carbon dioxide, or by measuring the release of energy during cellular respiration. In order to measure the gases, the general gas law must be understood. The general gas law state: PV=nRT where P is the pressure of the gas, V is the volume of the gas, n is the number of molecules of gas, R is the gas constant, and T is the temperature of the gas (in K). The gas law also shows concepts about gases. If temperature and pressure are kept constant, then the volume of the gas is directly proportional to the number of molecules of the gas. If the temperature and volume remain constant, then the pressure of the gas changes in direct proportion to the number of molecules of gas present. If the number of gas molecules and the temperature remain constant, then the pressure is inversely proportional to the volume. If the temperature changes and the number of gas molecules is kept constant, then either pressure of volume will change in direct proportion to the temperature.

In this experiment, the rate of cellular respiration will be measured by measuring the oxygen gas consumption by using a respirometer in water. This experiment measures the consumption of oxygen by germinating and non-germinating at room temperature and at ice water temperature. The carbon dioxide produced in cellular respiration will be removed by potassium hydroxide (KOH). As a result of the carbon dioxide being removed, the change in the volume of gas in the respirometer will be directly related to the amount of oxygen consumed. The respirometer with glass beads alone will show any changes in volume due to atmospheric pressure changes or temperature changes.

The germinating peas will have a higher rate of respiration, than the beads and non-germinating peas.

This lab requires two thermometers, two water baths, beads, germinating and non-germinating peas, beads, six vials, twelve pipettes, 100 mL graduated cylinder, scotch tape, tap water, ice, KOH, absorbent and non-absorbent cotton, six washers, six rubber stoppers, scotch tape, and a one mL dropper.

Start the experiment by setting up two water baths, one at room temperature and the other at 10 degrees Celsius. Then, find the volume of twenty-five germinating peas. Next, put 50 mL of water in a graduated cylinder and put twenty-five non-germinating peas in it. Then, add beads until the volume is the same as twenty-five germinating peas. Next, pour our the peas and beads, refill the graduated cylinder with 50 mL of water, and add only beads until the volume is the same as the twenty-five germinating peas. Repeat these steps for another set of peas and beads. Also, put together the six respirometers by gluing a pipette to a stopper and taping another pipette to the pipette for all six respirometers. Then, put two absorbent cotton balls, several drops of KOH, and half of a piece of non-absorbent cotton into all six vials. Next, add the peas and beads to the appropriate respirometers. Place one set of respirometers into the room temperature water bath and the other set in the ice water bath. Elevate the respirometers by setting the pipettes onto masking tape and allow them to equilibrate for five minutes. Next, lower the respirometers into the water baths and take reading at 0, 5, 10, 15, and 20 minutes. Record the results in the table.


Watch the video: Cluster Sampling (November 2021).