what is the probability of a student scoring 910 or lower on the SAT

what is the probability of a student scoring 910 or lower on the SAT

PART1-Due Thursday PLEASE PICK OPTION A OR B. DO NOT DO BOTH!!!

Respond to one of the following in a minimum of 175 words:

Option A·      Pick a topic you would like to research. Explain one variable you could measure in each of the four scales of measurement. These may all be demographic information if that fits your topic.

Option B·      Explain what significance tells you in a way a that anyone could understand. What are the levels of significance most commonly seen? What is the most common level of significance used in psychological research?

PART2– ASSIGNMENT AND ITS RESOUREC ARE ATTACHED BELOW

PART3-

  1.    The Statistics Project assignment has been broken down into multiple parts and you will complete these parts throughout the course. This week’s assignment allows you to become familiar with opening data and viewing it in Microsoft® Excel® and using the Analysis Toolpak. In research, the individual data points are entered into databases, but for the purpose of this course, the data is provided in a spreadsheet for you in the Happiness and Engagement Dataset.
    Imagine you have been asked to enhance workplace happiness and engagement at your company. You have conducted a survey and gathered data on the gender, age, relationship with direct supervisor, telecommute schedule, relationship with coworkers, along with the ratings for workplace happiness and workplace engagement for 50 individuals in your department. You must determine what variables affect workplace happiness and engagement. The first step is to run descriptive statistics on each variable to learn more about the data you have collected.
    Calculate descriptive statistics for the following variables in the provided Microsoft® Excel® dataset:

    • Gender
    • Age
    • Relationship with Direct Supervisor
    • Telecommute Schedule
    • Relationship with Coworkers
    • Workplace Happiness Rating
    • Workplace Engagement Rating
    • Write a 125- to 175-word summary of your interpretation of the descriptive results for each variable. Copy and paste the Microsoft® Excel® output below your summary.
      Format your summary according to APA guidelines.
      Submit your Assignment.
      Note: Frequency tables are best used for nominal and ordinal variables. Other descriptive information (e.g., means and standard deviations) are best used with interval and ratio data.

PART4-ASSIGNMENT IS ATTACHED BELOW. I ONLY HAVE TO DO THE PART THATS HIGHLIGHTED, WHICH IS SECTION 3.

CHAPTER 4

Probability and Hypothesis Testing

Module 7: Probability

Basic Probability Concepts

The Rules of Probability

Probability and the Standard Normal Distribution

Review of Key Terms

Module Exercises

Critical Thinking Check Answers

Module 8: Hypothesis Testing and Inferential Statistics

Null and Alternative Hypotheses

Two-Tailed and One-Tailed Hypothesis Tests

Type I and Type II Errors in Hypothesis Testing

Probability, Statistical Significance, and Errors

Using Inferential Statistics

Review of Key Terms

Module Exercises

Critical Thinking Check Answers

Chapter 4 Summary and Review

In this chapter you will be introduced to the concepts of probability and hypothesis testing. Probability is the study of likelihood and uncertainty. Most decisions that we make are probabilistic in nature. Thus, probability plays a critical role in most professions and in our everyday decisions. We will discuss basic probability concepts along with how to compute probabilities and the use of the standard normal curve in making probabilistic decisions.

probability The study of likelihood and uncertainty; the number of ways a particular outcome can occur, divided by the total number of outcomes.

Hypothesis testing is the process of determining whether a hypothesis is supported by the results of a research project. Our introduction to hypothesis testing will include a discussion of the null and alternative hypotheses, Type I and Type II errors, and one- and two-tailed tests of hypotheses as well as an introduction to statistical significance and probability as they relate to inferential statistics.

hypothesis testing The process of determining whether a hypothesis is supported by the results of a research study.

MODULE 7

Probability

Learning Objectives

• Understand how probability is used in everyday life.

• Know how to compute a probability.

• Understand and be able to apply the multiplication rule.

• Understand and be able to apply the addition rule.

• Understand the relationship between the standard normal curve and probability.

In order to better understand the nature of probabilistic decisions, consider the following court case of The People v. Collins, 1968. In this case, the robbery victim was unable to identify his assailant. All that the victim could recall was that the assailant was female with a blonde pony tail. In addition, he remembered that she fled the scene in a yellow convertible that was driven by an African American male who had a full beard. The suspect in the case fit the description given by the victim, so the question was “Could the jury be sure, beyond a reasonable doubt, that the woman on trial was the robber?” The evidence against her was as follows: She was blonde and often wore her hair in a pony tail; her codefendant friend was an African American male with a moustache, beard, and a yellow convertible. The attorney for the defense stressed the fact that the victim could not identify this woman as the woman who robbed him, and that therefore there should be reasonable doubt on the part of the jury.

The prosecutor, on the other hand, called an expert in probability theory who testified to the following: The probability of all of the above conditions (being blonde and often having a pony tail and having an African American male friend and his having a full beard, and his owning a yellow convertible) co-occurring when these characteristics are independent was 1 in 12 million. The expert further testified that the combination of characteristics was so unusual that the jury could in fact be certain “beyond a reasonable doubt” that the woman was the robber. The jury returned a verdict of “guilty” (Arkes & Hammond, 1986; Halpern, 1996).

As can be seen in the previous example, the legal system operates on probability and recognizes that we can never be absolutely certain when deciding whether an individual is guilty. Thus, the standard of “beyond a reasonable doubt” was established and jurors base their decisions on probability, whether they realize it or not. Most decisions that we make on a daily basis are, in fact, based on probabilities. Diagnoses made by doctors, verdicts produced by juries, decisions made by business executives regarding expansion and what products to carry, decisions regarding whether individuals are admitted to colleges, and most everyday decisions all involve using probability. In addition, all games of chance (for example, cards, horse racing, the stock market) involve probability.

If you think about it, there is very little in life that is certain. Therefore, most of our decisions are probabilistic and having a better understanding of probability will help you with those decisions. In addition, because probability also plays an important role in science, that is another important reason for us to have an understanding of it. As we will see in later modules, the laws of probability are critical in the interpretation of research findings.

Basic Probability Concepts

Probability refers to the number of ways a particular outcome (event) can occur divided by the total number of outcomes (events). (Please note that the words outcome and event will be used interchangeably in this module.) Probabilities are often presented or expressed as proportions. Proportions vary between 0.0 and 1.0, where a probability of 0.0 means the event certainly will not occur and a probability of 1.0 means that the event is certain to occur. Thus, any probability between 0.0 and 1.0 represents an event with some degree of uncertainty to it. How much uncertainty depends on the exact probability with which we are dealing. For example, a probability close to 0.0 represents an event that is almost certain not to occur, and a probability close to 1.0 represents an event that is almost certain to occur. On the other hand, a probability of .50 represents maximum uncertainty. In addition, keep in mind that probabilities tell us about the likelihood of events in the long run, not the short run.

Let’s start with a simplistic example of probability. What is the probability of getting a “head” when tossing a coin? In this example, we have to consider how many ways there are to get a “head” on a coin toss (there is only one way, the coin lands heads up) and how many possible outcomes there are (there are two possible outcomes, either a “head” or a “tail”). So, the probability of a “head” in a coin toss is:

p(head)=NumberofwaystogetaheadNumberofpossibleoutcomes=12=.50p(head)=Number of ways to get a headNumber of possible outcomes=12=

This means that in the long run, we can expect a coin to land heads up 50% of the time.

Let’s consider some other examples. How likely would it be for an individual to roll a 2 in one roll of a die? Once again, let’s put this into basic probability terms. There is only one way to roll a 2, the die lands with the 2 side up. How many possible outcomes are there in a single roll of a die? There are six possible outcomes (any number between 1 and 6 could appear on the die). Hence, the probability of rolling a 2 on a single roll of a die would be 1/6, or about .17. Representing this in a formula as we did for the previous example:

p(2)=Numberofwaystogeta2Numberofpossibleoutcomes=16=.17p(2)=Number of ways to get a 2Number of possible outcomes=16=.17<math xmlns=”http://www.w3.org/1998/Math/MathML” display=”inline” alttext=”math”><mrow><mi>p</mi><mo stretchy=”false”>(</mo><mtext>2</mtext><mo stretchy=”false”>)</mo><mo>=</mo><mfrac><mrow><mtext>Number</mtext><mtext> </mtext><mtext>of</mtext><mtext> </mtext><mtext>ways</mtext><mtext> </mtext><mtext>to</mtext><mtext> </mtext><mtext>get</mtext><mtext> </mtext><mtext>a</mtext><mtext> </mtext><mtext>2</mtext></mrow><mrow><mtext>Number</mtext><mtext> </mtext><mtext>of</mtext><mtext> </mtext><mtext>possible</mtext><mtext> </mtext><mtext>outcomes</mtext></mrow></mfrac><mo>=</mo><mfrac><mn>1</mn><mn>6</mn></mfrac><mo>=</mo><mn>.17</mn></mrow></math>

Let’s make it a little more difficult. What is the probability of rolling an odd number in a single roll of a die? Well, there are three odd numbers on any single die (1, 3, and 5). Thus, there are three ways that an odd number can occur. Once again, how many possible outcomes are there in a single roll of a die? Six (any number between 1 and 6). Therefore, the probability of rolling an odd number on a single roll is 3/6, or .50. Represented as a formula this would be:

p(oddnumber)=NumberofwaystogetanoddnumberNumberofpossibleoutcomes=36=.50p(odd number)=Number of ways to get an odd numberNumber of possible outcomes=36=.50<math xmlns=”http://www.w3.org/1998/Math/MathML” display=”inline” alttext=”math”><mrow><mi>p</mi><mo stretchy=”false”>(</mo><mtext>odd</mtext><mtext> </mtext><mtext>number)=</mtext><mfrac><mrow><mtext>Number</mtext><mtext> </mtext><mtext>of</mtext><mtext> </mtext><mtext>ways</mtext><mtext> </mtext><mtext>to</mtext><mtext> </mtext><mtext>get</mtext><mtext> </mtext><mtext>an</mtext><mtext> </mtext><mtext>odd</mtext><mtext> </mtext><mtext>number</mtext></mrow><mrow><mtext>Number</mtext><mtext> </mtext><mtext>of</mtext><mtext> </mtext><mtext>possible</mtext><mtext> </mtext><mtext>outcomes</mtext></mrow></mfrac><mo>=</mo><mfrac><mn>3</mn><mn>6</mn></mfrac><mo>=</mo><mn>.50</mn></mrow></math>

What if I asked you what the probability of rolling a single-digit number is in a single roll of a die? A die has six numbers on it, and each is a single-digit number. Thus, there are six ways to get a single-digit number. How many possible outcomes are there in a single roll of a die? Once again, six. Hence, the probability of rolling a single-digit number is 6/6, or 1.0. If someone asked you to place a bet on this occurring, you could not lose on this bet! Once again, as a formula this would be:

p(single-digitnumber)=Numberofwaystogetasingle-digitnumberNumberofpossibleoutcomse=66=1.0p(single-digit number)=Number of ways to get a single-digit numberNumber of possible outcomse=66=1.0<math xmlns=”http://www.w3.org/1998/Math/MathML” display=”inline” alttext=”math”><mrow><mtable columnalign=”left”><mtr columnalign=”right”><mtd columnalign=”right”><mrow><mi>p</mi><mo stretchy=”false”>(</mo><mtext>single-digit</mtext><mtext> </mtext><mtext>number</mtext><mo stretchy=”false”>)</mo></mrow></mtd><mtd columnalign=”left”><mo>=</mo></mtd><mtd columnalign=”left”><mrow><mfrac><mrow><mtext>Number</mtext><mtext> </mtext><mtext>of</mtext><mtext> </mtext><mtext>ways</mtext><mtext> </mtext><mtext>to</mtext><mtext> </mtext><mtext>get</mtext><mtext> </mtext><mtext>a</mtext><mtext> </mtext><mtext>single-digit</mtext><mtext> </mtext><mtext>number</mtext></mrow><mrow><mtext>Number</mtext><mtext> </mtext><mtext>of</mtext><mtext> </mtext><mtext>possible</mtext><mtext> </mtext><mtext>outcomse</mtext></mrow></mfrac></mrow></mtd></mtr><mtr columnalign=”right”><mtd columnalign=”right”><mrow /></mtd><mtd columnalign=”left”><mo>=</mo></mtd><mtd columnalign=”left”><mrow><mfrac><mn>6</mn><mn>6</mn></mfrac><mo>=</mo><mn>1.0</mn></mrow></mtd></mtr></mtable></mrow></math>

Now that you have a basic idea of where probabilities come from, let’s talk a little bit more about how we use probabilities. Keep in mind that probabilities tell us something about what will happen in the long run. Therefore, when we think about using some of the probabilities that we just calculated, we have to think about using them in the long run. For example, we determined that the probability of rolling a 2 on a single roll of a die was .17. This means that over many rolls of the die, it will land with the 2 side up about 17% of the time. We cannot predict what will happen on any single roll of the die, but over many rolls of the die, we will roll a 2 with a probability of .17. This means that with a very large number of trials, we can predict with great accuracy what proportion of the rolls will end up as 2s. However, we cannot predict which particular rolls will yield a 2. So when we think about using probabilities, we need to think about using them for predictions in the long run, not the short run.

1. What is the probability of pulling a king from a standard (52-card) deck of playing cards?

2. What is the probability of pulling a spade from a standard deck of playing cards?

3. What is the probability of rolling an even number on a single roll of a die?

4. Imagine that you have a bag that contains 4 black poker chips and 7 red poker chips. What is the probability of pulling a black poker chip from the bag?

The Rules of Probability

Often we are concerned with the probability of two or more events occurring and not just the probability of a single event occurring. For example, what is the probability of rolling at least one 4 in two rolls of a die, or what is the probability of getting two tails in two flips of a coin?

Let’s use the coin-toss example to determine the probability of two tails occurring in two flips of a coin. Based on what we discussed in the previous section, we know that the probability of a tail on one flip of a coin is 1/2, or 50. The same is true for the second toss, the probability of a tail is 1/2, or .50. However, let’s think about the possible outcomes for two tosses of a coin. One outcome is a head on the first toss and a head on the second toss (HH). The other outcomes would be a head followed by a tail (HT), a tail followed by a head (TH), and a tail followed by a tail (TT). These four possible outcomes are illustrated in the tree diagram in Figure 7.1.

Notice that the probability of two tails or any one of the other three possible outcomes is 1/4, or .25. But how are these probabilities calculated? The general rule that we apply here is known as the multiplication rule, or the and rule. When the events are independent and we want to know the probability of one event “and” another event, we use this rule.

The multiplication rule says that the probability of a series of outcomes occurring on successive trials is the product of their individual probabilities, when the events are independent (do not impact one another). Thus, when using the multiplication rule, we multiply the probability of the first event by the probability of the second event. Therefore, for the present problem, the probability of a tail in the first toss is 1/2, or .50, and the probability of a tail in the second toss is 1/2, or .50. When we multiply these two probabilities, we have .50 × .50 = .25. This should make some sense to you because the probability of both events occurring should be less than that of either event alone. We can represent the problem as follows:

multiplication rule A probability rule stating that the probability of a series of outcomes occurring on successive trials is the product of their individual probabilities, when the sequence of outcomes is independent.

p(tailonfirsttossandtailonsecondtoss)=p(tailonfirsttoss)×p(tailonsecondtoss)p(tail on first toss and tail on second toss)=p(tail on first toss) ×p(tail on second toss)<math xmlns=”http://www.w3.org/1998/Math/MathML” display=”inline” alttext=”math”><mrow><mtable columnalign=”left”><mtr columnalign=”right”><mtd columnalign=”right”><mrow><mi>p</mi><mo stretchy=”false”>(</mo><mtext>tail</mtext><mtext> </mtext><mtext>on</mtext><mtext> </mtext><mtext>first</mtext><mtext> </mtext><mtext>toss</mtext><mtext> </mtext><mtext>and</mtext><mtext> </mtext><mtext>tail</mtext><mtext> </mtext><mtext>on</mtext><mtext> </mtext><mtext>second</mtext><mtext> </mtext><mtext>toss</mtext><mo stretchy=”false”>)</mo></mrow></mtd><mtd columnalign=”left”><mo>=</mo></mtd><mtd columnalign=”left”><mrow><mi>p</mi><mtext>(tail</mtext><mtext> </mtext><mtext>on</mtext><mtext> </mtext><mtext>first</mtext><mtext> </mtext><mtext>toss)</mtext></mrow></mtd></mtr><mtr columnalign=”right”><mtd columnalign=”right”><mrow /></mtd><mtd columnalign=”left”><mrow /></mtd><mtd columnalign=”left”><mrow><mtext> </mtext><mo>×</mo><mi>p</mi><mo stretchy=”false”>(</mo><mtext>tail</mtext><mtext> </mtext><mtext>on</mtext><mtext> </mtext><mtext>second</mtext><mtext> </mtext><mtext>toss)</mtext></mrow></mtd></mtr></mtable></mrow></math>

Let’s try another example. Assuming that the probabilities of having a girl and having a boy are both .50 for single-child births, what is the probability that a couple planning a family of three children would have the children in the following order: girl, girl, boy?

You can see in the tree diagram in Figure 7.2 that the probability of girl, girl, boy is .125. Let’s use the and rule to double-check this probability. The probability of a girl as the first child is 1/2, or .50. The same is true for the probability of a girl as the second child (.50) and the probability of a boy as the third child (.50). In order to determine the probability of this sequence of births, we multiply: .50 × .50 × .50 = .125.

In addition to being able to calculate probabilities based on a series of independent events (as in the preceding examples), we can also calculate the probability of one event or another event occurring on a single trial when the events are mutually exclusive. Mutually exclusive means that only one of the events can occur on a single trial. For example, a coin toss can either be heads or tails on a given trial, but not both. When dealing with mutually exclusive events, we apply what is known as the addition rule, which states that the probability of one outcome or the other outcome occurring on a particular trial is the sum of their individual probabilities. In other words, we are adding the two probabilities together. Thus, the probability of having either a girl or a boy when giving birth would be:

addition rule A probability rule stating that the probability of one outcome or another outcome occurring on a particular trial is the sum of their individual probabilities, when the outcomes are mutually exclusive.

p(girlorboy)=p(girl)+p(boy)=.50+.50=1.00p(girl or boy) = p(girl) + p(boy) = .50 + .50=1.00<math xmlns=”http://www.w3.org/1998/Math/MathML” display=”inline” alttext=”math”><mrow><mi>p</mi><mo stretchy=”false”>(</mo><mtext>girl</mtext><mtext> </mtext><mi>o</mi><mi>r</mi><mtext> </mtext><mtext>boy)</mtext><mtext> </mtext><mtext>=</mtext><mtext> </mtext><mi>p</mi><mo stretchy=”false”>(</mo><mtext>girl)</mtext><mtext> </mtext><mtext>+</mtext><mtext> </mtext><mi>p</mi><mo stretchy=”false”>(</mo><mtext>boy</mtext><mo stretchy=”false”>)</mo><mtext> </mtext><mo>=</mo><mtext> </mtext><mn>.50</mn><mtext> </mtext><mo>+</mo><mtext> </mtext><mn>.50</mn><mo>=</mo><mn>1.00</mn></mrow></math>

This is sometimes referred to as the or rule because we are determining the probability of one event or the other event.

Let’s try another problem using the or rule. What is the probability of drawing either a club or a heart when drawing one card from a deck of cards? The probability of drawing a club is 13/52, or .25. The same holds for drawing a heart (13/52 = .25). Thus, the probability of drawing either a club or a heart card on a single draw would be .25 + .25 = .50.

p(cluborheart)=p(club)+p(heart)=.25+.25=.50p(club or heart) = p(club) + p(heart) = .25 + .25 = .50<math xmlns=”http://www.w3.org/1998/Math/MathML” display=”inline” alttext=”math”><mrow><mi>p</mi><mo stretchy=”false”>(</mo><mtext>club</mtext><mtext> </mtext><mi>o</mi><mi>r</mi><mtext> </mtext><mtext>heart)</mtext><mtext> </mtext><mtext>=</mtext><mtext> </mtext><mi>p</mi><mo stretchy=”false”>(</mo><mtext>club</mtext><mo stretchy=”false”>)</mo><mtext> </mtext><mo>+</mo><mtext> </mtext><mi>p</mi><mtext>(heart</mtext><mo stretchy=”false”>)</mo><mtext> </mtext><mo>=</mo><mtext> </mtext><mn>.25</mn><mtext> </mtext><mo>+</mo><mtext> </mtext><mn>.25</mn><mtext> </mtext><mo>=</mo><mtext> </mtext><mn>.50</mn></mrow></math>

THE RULES OF PROBABILITY

RuleExplanationExampleThe Multiplication RuleThe probability of a series of independent outcomes occurring on successive trials is the product of their individual probabilities. This is also known as the and rule because we want to know the probability of one event and another event.In order to determine the probability of one coin toss of a head followed by (and) another coin toss of a head, we multiply the probability of each individual event: 50 × .50 = .25The Addition RuleThe probability of one outcome or another outcome occurring on a particular trial is the sum of their individual probabilities when the two outcomes are mutually exclusive. This is also known as the or rule because we want to know the probability of one event or another event.In order to determine the probability of tossing a head or a tail on a single coin toss, we sum the probability of each individual event: 50 +.50 = 1.0

1. Which rule, the multiplication rule or the addition rule, would be applied in each of the following situations?

a. What is the probability of a couple having a girl as their first child followed by a boy as their second child?

b. What is the probability of pulling a spade or a diamond from a standard deck of cards on a single trial?

c. What is the probability of pulling a spade (and then putting it back in the deck) followed by pulling a diamond from a standard deck of cards?

d. What is the probability of pulling a jack or a queen from a standard deck of cards on a single trial?

2. Determine the probability for each of the examples in exercise 1.

Probability and the Standard Normal distribution

As you might remember from Chapter 3z scores can be used to determine proportions under the standard normal curve. In that chapter, we used z scores and the area under the standard normal curve to determine percentile ranks. We will now use z scores and the area under the standard normal curve (Table A.1) to determine probabilities. As you might remember, the standard normal curve has a mean of 0 and a standard deviation of 1. In addition, as discussed in Chapter 3, the standard normal curve is symmetrical and bell-shaped and the mean, median, and mode are all the same. Take a look at Figure 7.3, which represents the area under the standard normal curve in terms of standard deviations. We looked at this figure in the previous module (Figure 6.1), and based on this figure, we see that approximately 68% of the observations in the distribution fall between −1.0 and + 1.0 standard deviations from the mean. This approximate percentage holds for all data that are normally distributed. Notice also that approximately 13.5% of the observations fall between −1.0 and −2.0 and another 13.5% between +1.0 and +2.0, and that approximately 2% of the observations fall between − 2.0 and − 3.0 and another 2% between + 2.0 and + 3.0. Only .13% of the scores are beyond a z score of either ±3.0. If you sum the percentages in Figure 7.3, you will have 100%—all of the area under the curve, representing everybody in the distribution. If you sum half of the curve, you will have 50%—half of the distribution.

We can use the areas under the standard normal curve to determine the probability that an observation falls within a certain area under the curve. Let’s use a distribution that is normal to illustrate what we mean. Intelligence test scores are normally distributed with a mean of 100 and a standard deviation of 15. We could use the standard normal curve to determine the probability of randomly selecting someone from the population who had an intelligence score as high or higher than a certain amount. For example, if a school psychologist wanted to know the probability of selecting a student from the general population who had an intelligence test score of 119 or higher, we could use the area under the standard normal curve to determine this. First we have to convert the intelligence test score to a z score. As you might remember, the formula for a z score is:

z=X−μσz=X−μσ<math xmlns=”http://www.w3.org/1998/Math/MathML” display=”inline” alttext=”math”><mrow><mi>z</mi><mo>=</mo><mfrac><mrow><mi>X</mi><mo>−</mo><mi>μ</mi></mrow><mi>σ</mi></mfrac></mrow></math>

where X represents the individual’s score on the intelligence test, Âµ represents the population mean, and σ represents the population standard deviation. Using this formula, we can calculate the individual’s z score as follows:

z=X−μσ=119−10015=1915=+1.27z=X−μσ=119−10015=1915=+1.27<math xmlns=”http://www.w3.org/1998/Math/MathML” display=”inline” alttext=”math”><mrow><mi>z</mi><mo>=</mo><mfrac><mrow><mi>X</mi><mo>−</mo><mi>μ</mi></mrow><mi>σ</mi></mfrac><mo>=</mo><mfrac><mrow><mn>119</mn><mo>−</mo><mn>100</mn></mrow><mrow><mn>15</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>19</mn></mrow><mrow><mn>15</mn></mrow></mfrac><mo>=</mo><mo>+</mo><mn>1.27</mn></mrow></math>

Thus, we know that this individual’s z score falls +1.27 standard deviations above the mean. As in Chapter 3, it is helpful to represent this on a figure where the z score of + 1.27 is indicated. This is illustrated in Figure 7.4.

Now, in order to determine the probability of selecting a student with an intelligence test score of 119 or higher, we need to turn to Table A.1 in Appendix A. We begin by looking up a z score of 1.27 and find that for this score, a proportion of .39797 of the scores fall between the score and the mean of the distribution and a proportion of .10203 of the scores fall beyond the score. Referring to Figure 7.4, we see that the proportion of the curve in which we are interested is the area beyond the score, or .10203. This means that the probability of randomly selecting a student with an intelligence test score of 119 or higher is .10203, or just slightly higher than 10%. We can represent this problem in standard probability format as follows:

p(X≥119)=.10203p(X≥119)=.10203<math xmlns=”http://www.w3.org/1998/Math/MathML” display=”inline” alttext=”math”><mrow><mi>p</mi><mo stretchy=”false”>(</mo><mi>X</mi><mo>≥</mo><mn>119</mn><mo stretchy=”false”>)</mo><mo>=</mo><mn>.10203</mn></mrow></math>

Let’s try a couple more probability problems using the intelligence test score distribution. First, what is the probability of the school psychologist randomly selecting a student with an intelligence test score of 85 or higher? Secondly, what is the probability of the school psychologist selecting a student with an intelligence test score of 70 or lower?

Let’s begin with the first problem. We need to convert the intelligence test score to a z score and then consult Table A.1.

z=X−μσ=85−10015=−1515=−1.0z=X−μσ=85−10015=−1515=−1.0<math xmlns=”http://www.w3.org/1998/Math/MathML” display=”inline” alttext=”math”><mrow><mi>z</mi><mo>=</mo><mfrac><mrow><mi>X</mi><mo>−</mo><mi>μ</mi></mrow><mi>σ</mi></mfrac><mo>=</mo><mfrac><mrow><mn>85</mn><mo>−</mo><mn>100</mn></mrow><mrow><mn>15</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mo>−</mo><mn>15</mn></mrow><mrow><mn>15</mn></mrow></mfrac><mo>=</mo><mo>−</mo><mn>1.0</mn></mrow></math>

When we consult Table A.1, we find that for a z score of −1.0, 15866 of the scores fall below this score and .34134 of the scores fall between this score and the mean of the distribution. This z score is illustrated in Figure 7.5 along with the area in which we are interested—the probability of a student with an intelligence test score of 85 or higher being selected.

In order to determine the probability of selecting a student with an intelligence test score this high or higher, we take the area between the mean and the z score (.34134) and add the .50 from the other half of the distribution to it. Hence, the probability of selecting a student with an intelligence test score of 85 or higher is .84134, or approximately 84%. You should see that the probability of this happening is fairly high because when we look at Figure 7.5 we are talking about a large proportion of people who fit this description. This can be represented as follows:

P(X≥85)=.84134P(X≥85)=.84134<math xmlns=”http://www.w3.org/1998/Math/MathML” display=”inline” alttext=”math”><mrow><mi>P</mi><mo stretchy=”false”>(</mo><mi>X</mi><mo>≥</mo><mn>85</mn><mo stretchy=”false”>)</mo><mo>=</mo><mn>.84134</mn></mrow></math>

Let’s work the second problem, the probability of selecting a student with an intelligence test score of 70 or lower. Once again we begin by converting this score into a z score.

z=X−μσ=70−10015=−3015=−2.0z=X−μσ=70−10015=−3015=−2.0<math xmlns=”http://www.w3.org/1998/Math/MathML” display=”inline” alttext=”math”><mrow><mi>z</mi><mo>=</mo><mfrac><mrow><mi>X</mi><mo>−</mo><mi>μ</mi></mrow><mi>σ</mi></mfrac><mo>=</mo><mfrac><mrow><mn>70</mn><mo>−</mo><mn>100</mn></mrow><mrow><mn>15</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mo>−</mo><mn>30</mn></mrow><mrow><mn>15</mn></mrow></mfrac><mo>=</mo><mo>−</mo><mn>2.0</mn></mrow></math>

Next, we represent this on a figure with the z score indicated along with the area in which we are interested (anyone with this score or a lower score). This is illustrated in Figure 7.6.

Consulting Table A.1, we find that for a z score of −2.0, .02275 of the scores are below the score (beyond it) and .47725 of the scores are between the score and the mean of the distribution. We are interested in the probability of selecting a student with an intelligence test score of 70 or lower. Can you figure out what that would be? If you answered .02275, you are correct. Therefore, there is slightly more than a 2% chance of selecting a student with an intelligence test score this low or lower—a fairly low probability event. This can be represented as follows:

p(X≤70)=.02275p(X≤70)=.02275<math xmlns=”http://www.w3.org/1998/Math/MathML” display=”inline” alttext=”math”><mrow><mi>p</mi><mo stretchy=”false”>(</mo><mi>X</mi><mo>≤</mo><mn>70</mn><mo stretchy=”false”>)</mo><mo>=</mo><mn>.02275</mn></mrow></math>

Let’s apply what we have learned about using the standard normal curve to calculate probabilities together with the addition rule from earlier in the module to determine the probability of selecting a child whose intelligence test score is 70 or lower or 119 or higher. We have already determined the z scores for each of these intelligence test scores in our previous problems. The intelligence test score of 70 converts to a z score of −2.0 and the intelligence test score of 119 converts to a z score of + 1.27. Moreover, we have already determined that the probability of selecting a student with a score of 70 or lower is .02275 and that the probability of selecting a student with an intelligence test score of 119 or higher is .10203. Thus, applying the addition rule, the probability of selecting a student with a score that is 70 or lower or 119 or higher would be the sum of these two probabilities, or .02275 + .10203. These two probabilities sum to .12478, or just about 12.5%. This can be represented as follows:

=p(X≤70orX≥119)=p(X≤70)+p(X≥119)=p(X≤70orX≥119)=(.02275)+(.10203)=p(X≤70orX≥119)=(.02275)+(.10203)=p(X≤70 or X≥119)= p(X≤70)+p(X≥119)=p(X≤70 or X≥119)= (.02275)+(.10203)=p(X≤70 or X≥119)=(.02275)+(.10203)<math xmlns=”http://www.w3.org/1998/Math/MathML” display=”inline” alttext=”math”><mtable columnalign=”left”><mtr><mtd><mo>=</mo><mi>p</mi><mo stretchy=”false”>(</mo><mi>X</mi><mo>≤</mo><mn>70</mn><mtext> </mtext><mi>o</mi><mi>r</mi><mtext> </mtext><mi>X</mi><mo>≥</mo><mn>119</mn><mo stretchy=”false”>)</mo><mo>=</mo><mtext> </mtext><mi>p</mi><mo stretchy=”false”>(</mo><mi>X</mi><mo>≤</mo><mn>70</mn><mo stretchy=”false”>)</mo><mo>+</mo><mi>p</mi><mo stretchy=”false”>(</mo><mi>X</mi><mo>≥</mo><mn>119</mn><mo stretchy=”false”>)</mo></mtd></mtr><mtr><mtd><mo>=</mo><mi>p</mi><mo stretchy=”false”>(</mo><mi>X</mi><mo>≤</mo><mn>70</mn><mtext> </mtext><mi>o</mi><mi>r</mi><mtext> </mtext><mi>X</mi><mo>≥</mo><mn>119</mn><mo stretchy=”false”>)</mo><mo>=</mo><mtext> </mtext><mo stretchy=”false”>(</mo><mn>.02275</mn><mo stretchy=”false”>)</mo><mo>+</mo><mo stretchy=”false”>(</mo><mn>.10203</mn><mo stretchy=”false”>)</mo></mtd></mtr><mtr><mtd><mo>=</mo><mi>p</mi><mo stretchy=”false”>(</mo><mi>X</mi><mo>≤</mo><mn>70</mn><mtext> </mtext><mi>o</mi><mi>r</mi><mtext> </mtext><mi>X</mi><mo>≥</mo><mn>119</mn><mo stretchy=”false”>)</mo><mo>=</mo><mo stretchy=”false”>(</mo><mn>.02275</mn><mo stretchy=”false”>)</mo><mo>+</mo><mo stretchy=”false”>(</mo><mn>.10203</mn><mo stretchy=”false”>)</mo></mtd></mtr></mtable></math>

Now let’s turn to using the multiplication rule, discussed earlier in the module, with the area under the standard normal curve (Table A.1). In this case, we want to determine the probability of selecting two students who fit different descriptions. For example, what is the probability of selecting one student with an intelligence test score equal to or below 80, followed by another student with an intelligence test score equal to or above 125? Once again, we begin by converting the scores to z scores.

z=X−μσ=80−10015=−2015=−1.33z=X−μσ=80−10015=−2015=−1.33<math xmlns=”http://www.w3.org/1998/Math/MathML” display=”inline” alttext=”math”><mrow><mi>z</mi><mo>=</mo><mfrac><mrow><mi>X</mi><mo>−</mo><mi>μ</mi></mrow><mi>σ</mi></mfrac><mo>=</mo><mfrac><mrow><mn>80</mn><mo>−</mo><mn>100</mn></mrow><mrow><mn>15</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mo>−</mo><mn>20</mn></mrow><mrow><mn>15</mn></mrow></mfrac><mo>=</mo><mo>−</mo><mn>1.33</mn></mrow></math>z=X−μσ=125−10015=2515=+1.67z=X−μσ=125−10015=2515=+1.67<math xmlns=”http://www.w3.org/1998/Math/MathML” display=”inline” alttext=”math”><mrow><mi>z</mi><mo>=</mo><mfrac><mrow><mi>X</mi><mo>−</mo><mi>μ</mi></mrow><mi>σ</mi></mfrac><mo>=</mo><mfrac><mrow><mn>125</mn><mo>−</mo><mn>100</mn></mrow><mrow><mn>15</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>25</mn></mrow><mrow><mn>15</mn></mrow></mfrac><mo>=</mo><mo>+</mo><mn>1.</mn><mn>6</mn><mn>7</mn></mrow></math>

Consequently, the intelligence test scores convert to z scores of −1.33 and +1.67, respectively. Next we use Table A.1 to determine the probability of each of these events. Consulting Table A.1, we find that the probability of selecting a student with a score of 80 or lower is .09175 and the probability of selecting a student with a score of 125 or higher is .04745. These z scores and proportions are represented in Figure 7.7.

We now apply the multiplication rule to determine the probability of selecting the first person followed by the second person. Thus, we multiply the first probability by the second probability, or .09175 × .04745 = .00435. This can be represented as follows:

=p(X≤80andX≥125)=p(X≤80)+p(X≥125)=p(X≤80andX≥125)=(.09175)+(.04745)=p(X≤80andX≥119)=.00435=p(X≤80 and X≥125)= p(X≤80)+p(X≥125)=p(X≤80 and X≥125)= (.09175)+(.04745)=p(X≤80 and X≥119)= .00435<math xmlns=”http://www.w3.org/1998/Math/MathML” display=”inline” alttext=”math”><mtable columnalign=”left”><mtr><mtd><mo>=</mo><mi>p</mi><mo stretchy=”false”>(</mo><mi>X</mi><mo>≤</mo><mn>80</mn><mtext> </mtext><mi>a</mi><mi>n</mi><mi>d</mi><mtext> </mtext><mi>X</mi><mo>≥</mo><mn>125</mn><mo stretchy=”false”>)</mo><mo>=</mo><mtext> </mtext><mi>p</mi><mo stretchy=”false”>(</mo><mi>X</mi><mo>≤</mo><mn>80</mn><mo stretchy=”false”>)</mo><mo>+</mo><mi>p</mi><mo stretchy=”false”>(</mo><mi>X</mi><mo>≥</mo><mn>125</mn><mo stretchy=”false”>)</mo></mtd></mtr><mtr><mtd><mo>=</mo><mi>p</mi><mo stretchy=”false”>(</mo><mi>X</mi><mo>≤</mo><mn>80</mn><mtext> </mtext><mi>a</mi><mi>n</mi><mi>d</mi><mtext> </mtext><mi>X</mi><mo>≥</mo><mn>125</mn><mo stretchy=”false”>)</mo><mo>=</mo><mtext> </mtext><mo stretchy=”false”>(</mo><mn>.09175</mn><mo stretchy=”false”>)</mo><mo>+</mo><mo stretchy=”false”>(</mo><mn>.04745</mn><mo stretchy=”false”>)</mo></mtd></mtr><mtr><mtd><mo>=</mo><mi>p</mi><mo stretchy=”false”>(</mo><mi>X</mi><mo>≤</mo><mn>80</mn><mtext> </mtext><mi>a</mi><mi>n</mi><mi>d</mi><mtext> </mtext><mi>X</mi><mo>≥</mo><mn>119</mn><mo stretchy=”false”>)</mo><mo>=</mo><mtext> </mtext><mn>.00435</mn></mtd></mtr></mtable></math>

Thus, the probability of the first event followed by the second event has a very low probability of less than 1%.

1. If SAT scores are normally distributed with a mean of 1,000 and a standard deviation of 200, what is the probability of a student scoring 1,100 or higher on the SAT?

2. For this hypothetical SAT distribution, what is the probability of a student scoring 910 or lower on the SAT?

3. For this hypothetical SAT distribution, what is the probability of a student scoring 910 or lower or 1,100 or higher?

4. For this hypothetical SAT distribution, what is the probability of selecting a student who scored 910 or lower followed by a student who scored 1,100 or higher on the SAT?

REVIEW OF KEY TERMS

addition rule (p. 117)

hypothesis testing (p. 111)

multiplication rule (p. 116)

probability (p. 111)

MODULE EXERCISES

(Answers to odd-numbered questions appear in Appendix B.)

1. Imagine that I have a jar that contains 50 blue marbles and 20 red marbles.

a. What is the probability of selecting a red marble from the jar?

b. What is the probability of selecting a blue marble from the jar?

c. What is the probability of selecting either a red or a blue marble from the jar?

d. What is the probability of selecting a red marble (with replacement) followed by a blue marble?

2. What is the probability of a couple having children in the following birth order: boy, boy, boy, boy?

3. What is the probability of selecting either a 2 or a 4 (of any suit) from a standard deck of cards?

4. If height is normally distributed with a mean of 68 inches and a standard deviation of 5 inches, what is the probability of selecting someone who is 70 inches or taller?

5. For the distribution described in exercise 4, what is the probability of selecting someone who is 64 inches or shorter?

6. For the distribution described in exercise 4, what is the probability of selecting someone who is 70 inches or taller or 64 inches or shorter?

7. For the distribution described in exercise 4, what is the probability of selecting someone who is 70 inches or taller followed by someone who is 64 inches or shorter?

CRITICAL THINKING CHECK ANSWERS

Critical Thinking Check 7.1

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