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SAT math scores follow a normal distribution with an approximate \(\mu=500\) and \(\sigma=100\). Also ACT math scores follow a normal distrubution with an approximate \(\mu=21\) and \(\sigma=4.7\). You are an admissions officer at a university and have room to admit one more student for the upcoming year. Joe scored 600 on the SAT math exam, and Kate scored 25 on the ACT math exam. If you were going to base your decision solely on their performances on the exams, which student should you admit? Explain.

Short Answer

Expert verified
Joe should be admitted because his z-score is higher (1 vs approximately 0.85).

Step by step solution

01

Understand the Question

We need to compare Joe's performance in the SAT math exam to Kate's performance in the ACT math exam by determining their respective positions within their respective normal distributions. To do this, we will calculate the standardized scores (z-scores) for each student's exam score and base our decision on which student performed better relative to their test population.
02

Calculate Joe's SAT Z-Score

Joe scored 600 on the SAT. The z-score formula is given by: \[ z = \frac{X - \mu}{\sigma} \]Substitute the given SAT values into the formula: \( X = 600, \mu = 500, \sigma = 100 \)\[ z_{SAT} = \frac{600 - 500}{100} = \frac{100}{100} = 1 \]Joe's z-score for the SAT exam is 1.
03

Calculate Kate's ACT Z-Score

Kate scored 25 on the ACT. Using the same z-score formula:\[ z = \frac{X - \mu}{\sigma} \]Substitute the given ACT values into the formula: \( X = 25, \mu = 21, \sigma = 4.7 \)\[ z_{ACT} = \frac{25 - 21}{4.7} = \frac{4}{4.7} \approx 0.85 \]Kate's z-score for the ACT exam is approximately 0.85.
04

Compare Z-Scores

Joe's z-score is 1, and Kate's z-score is approximately 0.85. Since a higher z-score indicates a better performance relative to their respective distributions, Joe performed better compared to the average SAT math score than Kate did compared to the average ACT math score.
05

Make the Decision

Based solely on their exam performances relative to their respective populations, Joe should be admitted since his z-score is higher, indicating he performed better on the SAT compared to the average in his group.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Distribution
In statistics, the normal distribution is a fundamental concept often referred to as the bell curve due to its shape. This distribution describes how data points are distributed around the mean (average). It's characterized by two key parameters: the mean (\(\mu\)) and the standard deviation (\(\sigma\)). The mean is the central value where the peak of the curve lies, while the standard deviation controls how spread out the data is around the mean.

The normal distribution is crucial in educational assessment since it's used to interpret test scores and understand how a student's performance compares to a larger group. It allows educators to determine where most students' scores fall (close to the mean) and identify outliers or exceptional performers. When test scores are normally distributed, most students will score around the average, with fewer students scoring extremely high or low.
  • Mean (\(\mu\)): Central value of the distribution.
  • Standard deviation (\(\sigma\)): Measures variability from the mean.
  • Bell curve shape represents the distribution of scores.
Z-scores
Z-scores are standardized scores used to interpret individual performance relative to a normal distribution. By converting raw scores to z-scores, it becomes easier to compare scores from different tests or populations. The formula for calculating a z-score is \[ z = \frac{X - \mu}{\sigma} \] where \(X\) is the raw score, \(\mu\) is the mean, and \(\sigma\) is the standard deviation of the distribution.

In educational settings, z-scores help determine how well a student did compared to others. A positive z-score indicates a score above the average, while a negative z-score reflects a below-average performance. A z-score of zero means the score is exactly average. By comparing z-scores, admissions officers can see how students stack up against their peers, regardless of the test taken.
  • Used for comparing different test scores.
  • Positive z-score: Above average performance.
  • Negative z-score: Below average performance.
Standardized Testing
Standardized testing is a method of evaluation designed to assess students uniformly across different locations and schools. Tests are designed to have consistent questions, administration, and scoring processes so that everyone is compared on the same scale. This creates a level playing field, where students' abilities are judged based on the same criteria.

Common standardized tests, like the SAT and ACT, follow a normal distribution, which allows for scores to be compared across different test versions and populations. By using standardized tests, educators and admissions officers can make objective decisions based on data that reflect students' relative performance. This standardized method is helpful for college admissions to ensure fairness and consistency in evaluating candidates from diverse educational backgrounds.
  • Ensures fairness in assessment.
  • Consistent criteria and scoring.
  • Used widely in college admissions.
Educational Assessment
Educational assessment is the process of measuring students’ understanding, skills, and knowledge. Through various forms of testing and evaluation, educators gain insight into a student's academic prowess. The purpose of educational assessments can vary from aiding learning and improving teaching practices to making informed decisions about student placement and advancement.

Standardized tests, such as the SAT and ACT, are a substantial component of educational assessment. They provide data that help educators understand how a student performs relative to other students, both regionally and nationally. This information can guide curriculum development, highlight areas for improvement, and support policy decisions within educational institutions. Importantly, these assessments highlight the need for continuous development of fair and effective evaluation methods.
  • Measures students' abilities and knowledge.
  • Informs curriculum and teaching improvements.
  • Guides student placement and advancement decisions.

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Most popular questions from this chapter

An energy study in Gainesville, Florida, found that in March 2006 , household use of electricity had a mean of 673 and a standard deviation of 556 kilowatthours. (Source: Data from Todd Kamhoot, Gainesville Regional Utilities.) a. Suppose the distribution of energy use was normal. Using a table, calculator, or software that can give normal probabilities, find the proportion of households with electricity use greater than 1000 kilowatt- hours. b. Based on the mean and standard deviation given, do you think that the distribution of energy use actually is normal? Why or why not?

The state of Ohio has several statewide lottery options. One is the Pick 3 game in which you pick one of the 1000 three-digit numbers between 000 and 999\. The lottery selects a three-digit number at random. With a bet of \(\$ 1,\) you win \(\$ 500\) if your number is selected and nothing (\$0) otherwise. (Many states have a very similar type of lottery.) (Source: Background information from www.ohiolottery.com.) a. With a single \(\$ 1\) bet, what is the probability that you win \(\$ 500 ?\) b. Let \(X\) denote your winnings for a \(\$ 1\) bet, so \(x=\$ 0\) or \(x=\$ 500\). Construct the probability distribution for \(X\). c. Show that the mean of the distribution equals 0.50 , corresponding to an expected return of 50 cents for the dollar paid to play. Interpret the mean. d. In Ohio's Pick 4 lottery, you pick one of the 10,000 four-digit numbers between 0000 and 9999 and (with a \(\$ 1\) bet ) win \(\$ 5000\) if you get it correct. In terms of your expected winnings, with which game are you better off - playing Pick 4 , or playing Pick 3 in which you win \(\$ 500\) for a correct choice of a three-digit number? Justify your answer.

An exit poll is taken of 3000 voters in a statewide election. Let \(X\) denote the number who voted in favor of a special proposition designed to lower property taxes and raise the sales tax. Suppose that in the population, exactly \(50 \%\) voted for it. a. Explain why this scenario would seem to satisfy the three conditions needed to use the binomial distribution. Identify \(n\) and \(p\) for the binomial. b. Find the mean and standard deviation of the probability distribution of \(X\). c. Using the normal distribution approximation, give an interval in which you would expect \(X\) almost certainly to fall, if truly \(p=0.50\). (Hint: You can follow the reasoning of Example 14 on racial profiling.) d. Now, suppose that the exit poll had \(x=1706 .\) What would this suggest to you about the actual value of \(p ?\)

For a normal distribution, use Table A, software, or a calculator to find the probability that an observation is a. at least 1 standard deviation above the mean. b. at least 1 standard deviation below the mean. c. In each case, sketch a curve and show the tail probability.

A roulette wheel consists of 38 numbers, 0 through 36 and \(00 .\) Of these, 18 numbers are red, 18 are black, and 2 are green ( 0 and 00 ). You are given \(\$ 10\) and told that you must pick one of two wagers, for an outcome based on a spin of the wheel: (1) Bet \(\$ 10\) on number 23\. If the spin results in \(23,\) you win \(\$ 350\) and also get back your \(\$ 10\) bet. If any other number comes up, you lose your \(\$ 10,\) or (2) Bet \(\$ 10\) on black. If the spin results in any one of the black numbers, you win \(\$ 10\) and also get back your \(\$ 10\) bet. If any other color comes up, you lose your \(\$ 10\). a. Without doing any calculation, which wager would you prefer? Explain why. (There is no correct answer. Peoples' choices are based on their individual preferences and risk tolerances.) b. Find the expected outcome for each wager. Which wager is better in this sense?

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