Statistics Assignment — Handout 4210

Statistics Notes — How to Use that Calculator to Find Binomial Probability

The results of a recent survey indicate that 58% of households in the United States own a gas grill. If you randomly select 100 households, what is the probability that exactly 65 households will own a gas grill?

P(65) = ₁₀₀C₆₅ (.58)⁶⁵ (.42)³⁵ = 0.0299216472

Let me show you how to use that nice calculator of yours.

Press 2ND DISTR -> :binompdf( -> ENTER

Enter the values for n, p, and r separated by commas

binompdf(100,.58,65) -> ENTER

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What is the probability that at least 65 households will own a gas grill?

P(at least 65) = P(65) + P(66) + P(67) + P(68) + P(69) + P(70) + P(71) +… + P(100)

Press 2ND DISTR -> :binomcdf( -> ENTER

binomcdf(100,.58,64) -> ENTER

CDF means cumulative density function and it allows for the computation for the probability of “r and fewer” successes. It gives you a cumulative total of probabilities. The reason I entered 64 instead of 65 is I wanted to find the probability of 0 successes through 64 successes. I want to subtract the probabilities from 0 through 64. This is the results that is on the screen – this number represents the total cumulative probabilities from 0 to 64.

Now in order to find the probability of at least 65, which is the complement of the event 64 and fewer, I will subtract this result from 1 and the solution is given.

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Example: The results of a recent survey indicate that 71% of people in the United States use more than one topping on their hotdogs. If you randomly select 250 people in the United States, what is the probability that exactly 178 of them will use more than one topping?

binompdf(250,.71,178) = 0.0555119581

What is the probability that at least 178 of them will use more than one topping?

1 – binomcdf(250,.71,177) = 0.5039008221

Statistics Notes — Variance of a Probability Distribution — A Shortcut Formula

A shortcut formula for the variance of a probability distribution is

You still have to do an abbreviated table. Plus I think it is easy to cross up the columns you are using. Use it at your own risk. I do think it is quicker though.

The Rosetta Stone and The Rosetta Project

The ‘Rosetta Stone’ is an Ancient Egyptian artifact which was instrumental in advancing modern understanding of hieroglyphic writing. The stone is a Ptolemaic era stele with carved text made up of three translations of a single passage: two in Egyptian language scripts (hieroglyphic and Demotic) and one in classical Greek. It was created in 196 BC, discovered by the French in 1799 at Rashid (a harbor on the Mediterranean coast in Egypt which the French referred to as Rosetta during Napoleon Bonaparte’s campaign in Egypt), and contributed greatly to the decipherment of the principles of hieroglyph writing in 1822 by the British scientist Thomas Young and the French scholar Jean-François Champollion. Comparative translation of the stone assisted in understanding many previously undecipherable examples of hieroglyphic writing. The text of the Rosetta Stone is a decree from Ptolemy V, describing the repealing of various taxes and instructions to erect statues in temples.

The Stone is 45 inches high at its highest point, 28.5 inches wide, and 11 inches thick. Weighing approximately 1,676 lb, it was originally thought to be granite or basalt but is currently described as granodiorite of a dark pinkish-grey color. The stone has been on public display at The British Museum since 1802.

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The Rosetta Project is a global collaboration of language specialists and native speakers working to build a publicly accessible digital library of human languages. Since becoming a National Science Digital Library collection in 2004, the Rosetta Archive has more than doubled its collection size, now serving nearly 100,000 pages of material documenting over 2,500 languages—the largest resource of its kind on the Net.

A major concern of the project is the drastic and accelerated loss of the world’s languages. Just as globalization threatens human cultural diversity, the languages of small, unique, localized human societies are at serious risk. In fact, linguists predict that as much as 90% of the world’s linguistic diversity will be lost within the next century. Language is both an embodiment of human culture, as well as the primary means of its maintenance and transmission. When languages are lost, the transmission of traditional culture is often abruptly severed meaning the loss of cultural diversity is tightly connected to loss of linguistic diversity. To stem the tide and help reverse this trend, the Rosetta Project is working to promote human cultural and linguistic diversity, as well as to make sure that no language vanishes without a trace.

“Rosetta Stone” is also used as a metaphor to refer to anything that is a critical key to a process of decryption, translation, or a difficult problem.

Statistics Notes — Expected Value

Expected value

At a carnival, you pay $2.00 to play a coin-flipping game with three fair coins. On each coin one side has the number 0 and the other side has the number 1. You flip the three coins at one time and you win $1.00 for every 1 that appears on top. Are your expected earnings equal to the cost to play? We’ll answer this question in several steps.

a. In this game, the random variable of interest counts the number of 1s that show. What is the sample space for the values of this random variable?  The sample space is (0, 1, 2, 3), since any of these numbers of 1s can appear.

b. There are eight equally likely outcomes for throwing three coins. They are 000, 001, 010, 011, 100, 101, 110, and 111.

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d. The expected value is μ = ΣxP(x) = 0 + 0.375 + 0.750 + 0.375 = $1.50

The expected value is $1.50. It cost $2.00 to play the game; the expected value is less than the cost. The carnival is making money. In the long run, the carnival can expect to make an average of about 50 cents per player.

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Another way is to consider the gain of the game. You pay $2.00.
If you get zero 1s, your gain is -2, actually its a loss, not a gain.
If you get one 1, your gain is -1, again its a loss, but play along.
If you get two 1s, your gain is 0.
If you get three 1s, your gain is 1.

The expected value of your gain is μ = ΣxP(x) = -0.250 – 0.375 + 0 + 0.125 = -$0.50

The expected value of your gain is -50 cents. It cost $2.00 to play the game, you can expect to lose 50 cents each time you play. The carnival is making money. Overall, all players should expect to lose 50 cents each time they play, money the carnival will gladly take from them.

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In gambling, an expected value of 0 implies that a game is a fair game — an unlikely occurrence.

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At a raffle, 1000 tickets are sold at $5 each for four prizes of $1000, $200, $100, and $50. You buy one ticket. What is the expected value of each ticket? What is the expected value of your gain?

The expected value is μ = ΣxP(x) = 1 + 0.2 + 0.1 + 0.05 + 0 = $1.35

The expected value is $1.50. It cost $5.00 to play the game; the expected value is less than the cost. The fundraiser is making money. The fundraisers can expect to make an average of $3.65 per player.

The expected value of your gain is μ = ΣxP(x) = 0.995 + 0.195 + 0.095 + 0.045 – 4.98 = -$3.65

The expected value of your gain is -$3.65. It cost $5.00 to buy a raffle ticket, you can expect to lose $3.65 on each ticket you buy. The fundraiser will make money. Overall, all buyers should expect to lose $3.65 each time they buy a raffle ticket, money to buy wonderful equipment for wonderful kids. Thanks, we appreciate you.

Statistics Notes — Mean, Variance, and Standard Deviation of a Probability Distribution

A probability distribution can be thought of as a relative frequency distribution based on a very large n. Thus, it has a mean and a standard deviation.

The mean of a probability distribution is often called the expected value of the distribution. This terminology reflects the idea that the mean represents a “central point” or “cluster point” for the entire distribution. Of course, the mean or expected value is an average value, and it is usually not a point of the sample space.

The mean, the variance, and the standard deviation of a discrete population probability distribution are found by using these formulas:

where x is the value of a random variable, P(x) is the probability of that variable.

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Example. Are we influenced to buy a product by an ad we saw on TV?
National Infomercial Marketing Association determined the number of times buyers of a product watched a TV infomercial before purchasing the product. The results are shown here:

We can treat the information shown as an estimate of the probability distribution because the events are mutually exclusive and the sum of the percentages is 100%. Compute the mean and standard deviation of the distribution.

Statistic Assignment — Book Chapter 4

Statistics — Chapter 4 Syllabus

Statistics

Chapter 4

1. Book 4.1.1 page 169: 7 – 26, 33 – 36

2. Book 4.1.2 page 169: 1 – 6, 27 – 32, 37 – 44

3. Handout 4110

4. Handout 4120

5. Book 4.2.1 page 183: 5 – 14

6. Book 4.2.2 page 183: 15 – 20

7. Handout 4210

8. Handout 4220

9. Handout 4310

10. Book 4.3 page 192: 1 – 14

11. Book Uses and Abuses page 197

12. Book Chapter 4 Review page 198

13. Book Chapter 4 Quiz page 203

TEST

Statistics Notes — Discrete Probability Distribution

A random variable has a probability distribution whether it is discrete or continuous. The probability distribution is simply an assignment of probabilities to the specific values of the random variable or to a range of values of the random variable.

A discrete probability distribution lists every possible value the random variable can assume, together with its probability.

1. The probability distribution of a discrete random variable has a probability assigned to each value of the random variable, a value between 0 and 1, inclusive.
2. The sum of these probabilities must be 1.

Guidelines for constructing a discrete probability distribution
1. Make a frequency distribution for the possible outcomes, x.
2. Find the sum of the frequencies.
3. Find the probability of each possible outcomes, P(x), by dividing its frequency by the sum of the frequencies.
4. Check that each probability is between 0 and 1 and that the sum is 1.

Let’s look at a discrete probability distribution and its graph.

Example: Discrete probability distribution

Dr. Smith and Dr. Johnson developed a test to measure boredom tolerance. They administered it to a group of 200 statistics students between the ages of 25 and 35. The possible scores were 0, 1, 2, 3, 4, 5, and 6, with 6 indicating the highest tolerance for boredom. The test results for this group are shown in the table.

Test Scores for 200 Subjects

a. If a subject is chosen at random from this group, the probability that they have a score of 1 is 26/200, or 0.13. In a similar way, we can use the relative frequency to compute the probabilities for the other scores. These probability assignments make up the probability distribution. Notice that the scores are mutually exclusive. No one subject has two scores and the sum of the probabilities of all the scores is 1.

Probability Distribution of Scores on Boredom Tolerance Test

b. The graph of this distribution is simply a relative-frequency histogram in which the height of the bar over a score represents the probability of that score. Since each bar is one unit wide, the area of the bar over a score equals the height and thus represents the probability of that score. Since the sum of the probabilities is 1, the area under the graph is also 1.

It is important for you to establish the correspondence between the area under regions of the probability histogram and the probability that the random variable takes on a particular value.

c. The Spike Factory needs to hire someone with a score on the boredom tolerance test of 5 or 6 to operate the metal press machine. Since the scores 5 and 6 are mutually exclusive, the probability that someone in the group who took the boredom tolerance test made either a 5 or a 6 is the sum

P(5 or 6) = P(5) + P(6) = 0.08 + 0.02 = 0.10

Notice that to find P(5 or 6), we could have simply added the areas of the bars over 5 and over 6. One out of 10 of the group who took the boredom tolerance test would qualify for the position at The Spike Factory.

Statistics Notes — Random Variables, Discrete and Continuous Random Variables

A statistical experiment is any process by which measurements are obtained. Examples are
1. Counting the number of eggs in a robin’s nest
2. Counting the number of defective light bulbs in a case of bulbs
3. Measuring daily rainfall
4. Measuring the weight of a polar bear cub

Let x represent a quantitative variable that is measured or observed in an experiment. We are interested in the numerical values that x can take on. So x = number of eggs in a robin’s nest and x = weight in kilograms of a polar bear cub would be examples of such quantitative variables. Furthermore, we say that the quantitative variable x is a random variable because the value that x takes on in a given experiment is a chance or random outcome. We will study two types of random variables: discrete random variables and continuous random variables.

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Discrete random variable: When the observations of a quantitative random variable can take on only a finite number of values or a countable number of values, we say that the variable is a discrete random variable.

What is a countable number of values? As an example, let the random variable x be the number of wells that are drilled before the first productive well is found. Then x could be any of the values 1, 2, 3 …. In theory, we have an infinite number of possibilities for the values of x. The set of values of x corresponds to the set of counting numbers. Therefore, this type of infinity is called countable, and we say x has a countable number of values.

In most of the cases we will consider, a discrete random variable will be the result of a count. For instance, the number of students in a certain section of a statistics course this term is a discrete random variable. The value must be a counting number such as 14, or 20, or 34, and so forth. The values 25.34 or 25 1/2 are not possible.

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Continuous random variable: When the observations of a quantitative random variable can take on any of the countless number of values in a line interval, we say that the variable is a continuous random variable.

We will see most continuous random variables occurring as the result of a measurement. For example, the air pressure in an automobile tire represents a continuous random variable. The air pressure could in theory take on any value from 0 lb/in² or psi to the bursting pressure of the tire. Values such as 28.10 psi, 28.12 psi, and so forth are possible. Another example is the height of students in your statistics class. The heights could take on any value from a low of 3 feet to a high of 7.5 feet.

In general, measurements of quantities such as length, weight, volume, temperature, or time yield continuous random variables. If the temperature changes from 72°F to 73°F, for example, it must take on all the temperature values between 72 and 73. Temperatures cannot just jump from one reading to the next. Discrete random variables often come from counts, such as the number of passing scores on an exam or the number of weeds in a garden.

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Examples: Which of the following random variables are discrete and which are continuous?
a. The time it takes a student selected at random to register for the fall term. Time can take on any value, so this is a continuous random variable.
b. The number of bad checks drawn on a bank on a day selected at random. The number of bad checks can be only a whole number such as 0, 1, 2, 3, etc. This is a discrete variable.
c. The amount of gasoline needed to drive your car 200 miles. We are measuring volume, which can assume any value, so this is a continuous random variable.
d. The number of voters pick in a random sample of 50 registered voters in a district in order to find the number who voted in the last county election. This is a count, so the variable is discrete.