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Chapter 5: Problem 15

Anna, a language major, took final exams in both French and Spanish and scored 83 on each. Her roommate Megan, also taking both courses, scored 77 on the French exam and 95 on the Spanish exam. Overall, student scores on the French exam had a mean of 81 and a standard deviation of \(5,\) and the Spanish scores had a mean of 74 and a standard deviation of \(15 .\) a) To qualify for language honors, a major must maintain at least an 85 average for all language courses taken. So far, which student qualifies? b) Which student's overall performance was better?

Short Answer

Expert verified
a) Megan qualifies. b) Anna's overall performance was better.

Step by step solution

01

Standardize French Scores for Anna

First, calculate Anna's z-score for the French exam. Use the formula for a z-score: \( z = \frac{x - \mu}{\sigma} \), where \( x = 83 \), \( \mu = 81 \), and \( \sigma = 5 \). Calculating gives:\[z = \frac{83 - 81}{5} = \frac{2}{5} = 0.4\]
02

Standardize Spanish Scores for Anna

Next, calculate Anna's z-score for the Spanish exam using the same formula: \( x = 83 \), \( \mu = 74 \), and \( \sigma = 15 \).Calculating gives:\[z = \frac{83 - 74}{15} = \frac{9}{15} = 0.6\]
03

Standardize French Scores for Megan

Calculate Megan's z-score for the French exam with \( x = 77 \), \( \mu = 81 \), and \( \sigma = 5 \).Calculating gives:\[z = \frac{77 - 81}{5} = \frac{-4}{5} = -0.8\]
04

Standardize Spanish Scores for Megan

Calculate Megan's z-score for the Spanish exam with \( x = 95 \), \( \mu = 74 \), and \( \sigma = 15 \).Calculating gives:\[z = \frac{95 - 74}{15} = \frac{21}{15} = 1.4\]
05

Determine Qualification for Language Honors (Part a)

Calculate the average scores to check if either student qualifies for honors, needing an 85 average.Anna's average score:\[\text{Average} = \frac{83 (French) + 83 (Spanish)}{2} = 83\]Megan's average score:\[\text{Average} = \frac{77 (French) + 95 (Spanish)}{2} = 86\]Megan qualifies as she has an 86 average.
06

Compare Standard Scores for Overall Performance (Part b)

Compare the sum of z-scores to assess overall performance.- For Anna:\[0.4 (French) + 0.6 (Spanish) = 1.0\]- For Megan:\[-0.8 (French) + 1.4 (Spanish) = 0.6\]Anna's overall performance is better because her sum of z-scores is higher.

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Key Concepts

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

Understanding Standard Deviation
Standard deviation is a statistical concept that helps us understand how spread out the scores in a data set are around the mean. It's a measure of variability or dispersion.

When we have a low standard deviation, it means that the data points tend to be very close to the mean. If it's high, the data points are more spread out. In the context of Anna and Megan's language courses, the standard deviations are:
  • French exam: 5
  • Spanish exam: 15
This means that Spanish scores varied more than the French scores, as a standard deviation of 15 indicates more dispersion than a deviation of 5.

Knowing the standard deviation is crucial when calculating z-scores. The z-score tells us exactly how many standard deviations a particular score is from the mean. This helps in comparing scores across different distributions, like Anna and Megan's exam scores.
Calculating the Mean
The mean, often known as the average, is the sum of all data points divided by the number of data points. It's a central measure of tendency. In language courses, when evaluating scores, knowing the mean allows us to calculate averages and understand performance.

For the French exam, the mean score is 81, while for the Spanish exam, it is 74. This indicates that French exam scores are generally higher compared to Spanish.
  • To calculate the mean, add together all scores.
  • Then, divide by the number of scores.
In the exercise, Anna's and Megan's individual averages were calculated to determine their qualification for language honors. Megan's higher average of 86 qualified her for honors versus Anna's 83.

Understanding the mean allows students to gauge how they perform relative to the rest of the class.
Comparing Language Courses
When comparing performance in language courses, it's important to look beyond just the raw scores. For Anna and Megan, the z-score helps us understand their scores in relation to the class.

- **Anna achieved:** 83 in both French and Spanish. - **Megan achieved:** 77 in French and 95 in Spanish. To determine overall performance regardless of the course:
  • Z-scores are calculated to standardize scores.
  • The sum of z-scores across courses reveals overall performance.
Though Megan qualifies for honors due to her higher average, Anna's z-score sum of 1.0 outweighs Megan's 0.6. This shows Anna performed consistently well across both exams when standardized, signaling a better relative performance overall.

This method highlights that z-scores offer a more comprehensive view when comparing across different courses with varying score distributions.

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Most popular questions from this chapter

Environmental Protection Agency (EPA) fuel economy estimates for automobile models tested recently predicted a mean of \(24.8 \mathrm{mpg}\) and a standard deviation of 6.2 mpg for highway driving. Assume that a Normal model can be applied. a) Draw the model for auto fuel economy. Clearly label it, showing what the \(68-95-99.7\) Rule predicts. b) In what interval would you expect the central \(68 \%\) of autos to be found? c) About what percent of autos should get more than \(31 \mathrm{mpg} ?\) d) About what percent of cars should get between 31 and \(37.2 \mathrm{mpg} ?\) e) Describe the gas mileage of the worst \(2.5 \%\) of all cars.

In the Normal model \(N(100,16),\) what cutoff value bounds a) the highest \(5 \%\) of all IQs? b) the lowest \(30 \%\) of the IQs? c) the middle \(80 \%\) of the IQs?

Assume the cholesterol levels of adult American women can be described by a Normal model with a mean of \(188 \mathrm{mg} / \mathrm{dL}\) and a standard deviation of 24 a) Draw and label the Normal model. b) What percent of adult women do you expect to have cholesterol levels over \(200 \mathrm{mg} / \mathrm{dL} ?\) c) What percent of adult women do you expect to have cholesterol levels between 150 and \(170 \mathrm{mg} / \mathrm{dL} ?\) d) Estimate the IQR of the cholesterol levels. e) Above what value are the highest \(15 \%\) of women's cholesterol levels?

Based on the Normal model \(N(100,16)\) describing IQ scores, what percent of people's IQs would you expect to be a) over \(80 ?\) b) under \(90 ?\) c) between 112 and \(132 ?\)

For the car speed data of Exercise 18 recall that the mean speed recorded was \(23.84 \mathrm{mph}\) with a standard deviation of \(3.56 \mathrm{mph}\). To see how many cars are speeding, John subtracts \(20 \mathrm{mph}\) from all speeds. a) What is the mean speed now? What is the new standard deviation? b) His friend in Berlin wants to study the speeds, so John converts all the original miles-per-hour readings to kilometers per hour by multiplying all speeds by 1.609 (km per mile). What is the mean now? What is the new standard deviation?
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