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Chapter 5: Problem 15

Scores on the common final exam given in a large enrollment multiple section course were normally distributed with mean 69.35 and standard deviation \(12.93 .\) The department has the rule that in order to receive an A in the course his score must be in the top \(10 \%\) of all exam scores. Find the minimum exam score that meets this requirement.

Short Answer

Expert verified
The minimum exam score for the top 10% is 86.

Step by step solution

01

Understanding Normal Distribution

Exam scores are normally distributed with a mean \( \mu = 69.35 \) and a standard deviation \( \sigma = 12.93 \). To find the minimum exam score for the top 10% of the distribution, we need to identify the z-score corresponding to the 90th percentile.
02

Finding the Z-score for the 90th Percentile

The z-score is a value from the standard normal distribution that corresponds to a cumulative probability of 0.90 (since we're looking for the top 10% cutoff). We can find this value using a standard normal distribution table or a calculator. The z-score for the 90th percentile is approximately 1.28.
03

Converting the Z-score to an Exam Score

To find the corresponding exam score, we use the z-score formula: \[ z = \frac{X - \mu}{\sigma} \]where \( z = 1.28 \), \( \mu = 69.35 \), and \( \sigma = 12.93 \). We need to solve for \( X \), the score. Rearranging the formula gives:\[ X = z \cdot \sigma + \mu \]Substituting the values, we get:\[ X = 1.28 \times 12.93 + 69.35 \]
04

Calculating the Minimum Score

Perform the calculation:\[ X = 1.28 \times 12.93 + 69.35 = 16.5504 + 69.35 = 85.9004 \]Rounding this value to the nearest whole number, the minimum exam score is 86.

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Key Concepts

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

Mean and Standard Deviation
Understanding mean and standard deviation is crucial in analyzing exam scores. The mean, often referred to as the average, is a measure indicating the central point of a data set. In the context of exam scores, it represents the typical performance. For our exam, it is given as 69.35.

The standard deviation provides insight into how much the exam scores spread out from the mean. A smaller standard deviation indicates that the scores are closely clustered around the mean, while a larger standard deviation suggests more variability in scores. The standard deviation for this exam is 12.93, signaling how far scores typically deviate. This helps in understanding how exceptional a particular score is compared to the average.
Z-Score
A z-score is a statistical measurement that describes a value's relationship to the mean, using the standard deviation as a unit of measure. Calculating the z-score helps determine how far a specific score is from the mean.
  • A positive z-score indicates the value is above the mean.
  • A negative z-score means it is below the mean.
In our problem, we are interested in the top 10% of scores, corresponding to a cumulative probability of 0.90. To find this, we use a z-score table, which tells us that a z-score of about 1.28 correlates to the 90th percentile.
Percentiles
Percentiles are a way to express how a particular score compares to others by indicating the percentage of scores it exceeds. For example, if a score is in the 90th percentile, it means it is higher than 90% of the distribution.

In this exercise, we were tasked with finding which exam score lies at the 90th percentile, since the top 10% of scores (100% - 10% = 90th percentile) receive an A. By identifying the z-score associated with this percentile, we can convert it back into the original exam scale to find the required score.
Exam Scores Analysis
Exam scores analysis enables us to make informed decisions about grading and performance. In this scenario, by applying the properties of a normal distribution, we determined the minimum score needed to be in the top 10% and thus receive an A.

By converting the z-score to an exam score using the formula:
  • \[ X = z \cdot \sigma + \mu \]
we calculated that the minimum score required is approximately 86, ensuring students understand their standing and motivating them to achieve target percentiles for better performance.

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Most popular questions from this chapter

A machine for filling 2 -liter bottles of soft drink delivers an amount to each bottle that varies from bottle to bottle according to a normal distribution with standard deviation 0.002 liter and mean whatever amount the machine is set to deliver. a. If the machine is set to deliver 2 liters (so the mean amount delivered is 2 liters) what proportion of the bottles will contain at least 2 liters of soft drink? b. Find the minimum setting of the mean amount delivered by the machine so that at least \(99 \%\) of all bottles will contain at least 2 liters.

Heights \(X\) of adult women are normally distributed with mean 63.7 inches and standard deviation 2.71 inches. Romeo, who is 69.25 inches tall, wishes to date only women who are shorter than he but within 4 inches of his height. Find the probability that the next woman he meets will have such a height.

Dogberry's alarm clock is battery operated. The battery could fail with equal probability at any time of the day or night. Every day Dogberry sets his alarm for 6: 30 a.m. and goes to bed at 10: 00 p.m. Find the probability that when the clock battery finally dies, it will do so at the most inconvenient time, between 10:00 p.m. and 6: 30 a.m.

All students in a large enrollment multiple section course take common in- class exams and a common final, and submit common homework assignments. Course grades are assigned based on students' final overall scores, which are approximately normally distributed. The department assigns a C to students whose scores constitute the middle \(2 / 3\) of all scores. If scores this semester had mean 72.5 and standard deviation \(6.14,\) find the interval of scores that will be assigned a \(\mathrm{C}\)

The amount \(X\) of beverage in a can labeled 12 ounces is normally distributed with mean 12.1 ounces and standard deviation 0.05 ounce. A can is selected at random. a. Find the probability that the can contains at least 12 ounces. b. Find the probability that the can contains between 11.9 and 12.1 ounces.
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