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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 for honors. b) Anna's overall performance was better.

Step by step solution

01

Calculate Average Scores

First, calculate each student's average score for the language exams. Anna's scores are 83 in French and 83 in Spanish. The average is \( \frac{83 + 83}{2} = 83 \). Megan's scores are 77 in French and 95 in Spanish. The average is \( \frac{77 + 95}{2} = 86 \).
02

Evaluate Language Honors Qualification

To determine who qualifies for language honors, compare each student's average to the requirement of at least 85. Anna's average is 83, which is below 85, so she does not qualify. Megan's average is 86, which is above 85, so she qualifies for language honors.
03

Standardize Scores (Calculate Z-scores)

Calculate the Z-score for each exam to compare the students' performance relative to their peers. The formula for a Z-score is \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the score, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.- For Anna in French: \( Z = \frac{83 - 81}{5} = 0.4 \)- For Anna in Spanish: \( Z = \frac{83 - 74}{15} = 0.6 \)- For Megan in French: \( Z = \frac{77 - 81}{5} = -0.8 \)- For Megan in Spanish: \( Z = \frac{95 - 74}{15} = 1.4 \)
04

Compare Overall Performance Using Z-scores

Add each student's Z-scores to compare overall performance. For Anna, the sum is \(0.4 + 0.6 = 1.0\). For Megan, the sum is \(-0.8 + 1.4 = 0.6\). A higher total Z-score indicates better relative performance among peers.

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Key Concepts

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

Understanding Z-Scores
Z-scores help us understand how a specific data point compares with the average of a dataset. It's like finding where you stand in a group. The Z-score tells you how many standard deviations away from the mean your score is. Here's how to calculate it:
  • Identify the score you want to analyze, called the `X`.
  • Find the mean (`\mu`) of the dataset.
  • Determine the standard deviation (`\sigma`).
  • Apply the formula: \( Z = \frac{X - \mu}{\sigma} \)
Higher Z-scores mean you scored better compared to the group. If your Z-score is positive, you're above average. A negative Z-score means you're below average. In the problem, Megan achieves a higher Z-score in Spanish, while Anna does better in French. This helps us understand who excelled more based on standardized scores.
Grasping Standard Deviation
Standard deviation is a measure that tells us how spread out the numbers in a data set are. Imagine it as the distance or variation in scores. When you have a low standard deviation, it means most numbers are close to the mean (i.e., the average). A high standard deviation indicates a wide spread of scores.
  • Consider a class of students. If everyone's test scores are very similar, the standard deviation is low.
  • If there's a wide range of scores, the standard deviation is higher.
In the exercise, French scores have a standard deviation of 5, indicating most students scored close to the mean of 81. Spanish scores have a standard deviation of 15, showing more variation in scores around the mean of 74. Understanding standard deviation helps us put Z-scores into context.
Mean: The Average Score
The mean is what we commonly call the "average". It's calculated by summing up all the values and then dividing by the number of values. Here's why it's important:
  • The mean provides a central value for data, which is useful in comparing relative performances.
  • It can be skewed by very high or very low scores, but it often gives a good snapshot.
In the exercise, the mean scores for French and Spanish exams are 81 and 74, respectively. Calculating the mean helps us to understand the baseline or reference point for comparing individual scores using Z-scores.
Language Honors Qualification
Language honors qualification refers to meeting certain academic criteria to receive an award or recognition. In this exercise, a student must maintain an average score of at least 85 across all language courses to qualify.
  • Anna's average score is calculated as \( \frac{83 + 83}{2} = 83 \), which doesn't meet the criteria.
  • Megan's average is \( \frac{77 + 95}{2} = 86 \), so she qualifies for the honors.
To determine who qualifies, simply calculate the average and compare it to the required score. This system incentivizes students to perform consistently well across different language courses.

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Most popular questions from this chapter

A popular band on tour played a series of concerts in large venues. They always drew a large crowd, averaging 21,359 fans. While the band did not announce (and probably never calculated) the standard deviation, which of these values do you think is most likely to be correct: \(20,200,2000\), or 20,000 fans? Explain your choice.

A town's January high temperatures average \(36^{\circ} \mathrm{F}\) with a standard deviation of \(10^{\circ}\), while in July the mean high temperature is \(74^{\circ}\) and the standard deviation is \(8^{\circ} .\) In which month is it more unusual to have a day with a high temperature of \(55^{\circ} ?\) Explain.

A high school senior uses the Internet to get information on February temperatures in the town where hell be going to college. He finds a Web site with some statistics, but they are given in degrees Celsius. The conversion formula is \({ }^{\circ} \mathrm{F}=9 / 5^{\circ} \mathrm{C}+32\). Determine the Fahrenheit equivalents for the summary information below. Maximum temperature \(=11^{\circ} \mathrm{C} \quad\) Range \(=33^{\circ}\) Mean \(=1^{\circ} \quad\) Standard deviation \(=7^{\circ}\) Median \(=2^{\circ} \quad\) IQR \(=16^{\circ}\)

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?

The mean of the 100 car speeds in Exercise 20 was \(23.84 \mathrm{mph}\), with a standard deviation of \(3.56 \mathrm{mph}\). a) Using a Normal model, what values should border the middle \(95 \%\) of all car speeds? b) Here are some summary statistics. $$ \begin{array}{lll} \hline \text { Percentile } & & \text { Speed } \\ \hline 100 \% & \text { Max } & 34.060 \\ 97.5 \% & & 30.976 \\ 90.0 \% & & 28.978 \\ 75.0 \% & \text { Q3 } & 25.785 \\ 50.0 \% & \text { Median } & 23.525 \\ 25.0 \% & \text { Q1 } & 21.547 \\ 10.0 \% & & 19.163 \\ 2.5 \% & & 16.638 \\ 0.0 \% & \text { Min } & 16.270 \\ \hline \end{array} $$

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