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Chapter 6: Problem 111

Consider the variable \(x=\) time required for a college student to complete a standardized exam. Suppose that for the population of students at a particular university, the distribution of \(x\) is well approximated by a normal curve with mean 45 minutes and standard deviation 5 minutes. a. If 50 minutes is allowed for the exam, what proportion of students at this university would be unable to finish in the allotted time? b. How much time should be allowed for the exam if you wanted \(90 \%\) of the students taking the test to be able to finish in the allotted time? c. How much time is required for the fastest \(25 \%\) of all tudents to complete the exam

Short Answer

Expert verified
a. The proportion of students who would be unable to finish the exam in 50 minutes can be calculated using the Z-score and the standard normal distribution table. b. The time that should be allowed for the exam, given we want \(90\%\) of all students to be able to finish in time can be calculated after determining the corresponding Z-score for \(90\%\), and using it in the formula \(X = Z\sigma + \mu\). c. To find the time required for the fastest \(25\%\) of students to complete the exam, get the Z-score for \(25\%\), and compute using same formula \(X = Z\sigma + \mu\).

Step by step solution

01

Determine the Proportion Who Would Be Unable to Finish on Time

To find the proportion of students who would be unable to finish in 50 minutes, we need to calculate the z-score for 50 minutes using formula \(Z = (X - \mu) / \sigma \), where \(X=50\), \(\mu=45\) minutes and \(\sigma=5\). After calculating the Z-score, we can use the standard normal distribution table to find the proportion. This will give us the probability that a randomly chosen student will finish the exam in 50 minutes. However, we want to find the proportion of students who would be unable to finish in this time, therefore we need to subtract the calculated probability from 1.
02

Determine the Time That Should Be Allowed for Exam

To find out how much time should be allowed for the exam if you wanted \(90\%\) of the students taking the test to be able to finish in the allotted time, we need to find the Z-score corresponding to \(90\%\) from the standard normal distribution table. Then we can use formula \(X = Z\sigma + \mu\) to calculate the time, where Z is the Z-score found from the table, \(\mu = 45\) and \(\sigma = 5\).
03

Determine the Time Required for the Fastest Students

To find the time required for the fastest \(25\%\) of students to complete the exam, we first need to find the Z-score corresponding to \(25\%\) from the standard normal distribution table, then again using the formula \(X = Z\sigma + \mu\), we can calculate the time needed.

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Key Concepts

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

Z-score Calculation
The z-score is a crucial concept when dealing with normal distributions. It allows us to determine how far a specific point is from the mean of the distribution, measured in standard deviations. To calculate a z-score, we use the formula: \[ Z = \frac{X - \mu}{\sigma} \]where:
  • \(X\) is the score or value you are examining.
  • \(\mu\) is the mean of the distribution.
  • \(\sigma\) is the standard deviation of the distribution.
After finding the z-score, we consult a standard normal distribution table to find the probability related to that z-score.For instance, in our task: calculating the z-score for a time of 50 minutes gives us an insight into how unusual or common it is for students to complete a test in that time under the assumed normal distribution. A positive z-score indicates the value is above the mean, while a negative score shows it's below.
Standard Deviation
Standard deviation is a statistical measure that represents the spread or dispersion of a set of data points. In the context of a normal distribution, it tells us how much the values typically vary from the mean. For example, our given data shows a standard deviation of 5 minutes for test completion times, which suggests most completion times cluster around the mean (45 minutes). Understanding the size of the standard deviation helps in assessing whether a particular score is typical or unusual. In practical terms:
  • A smaller standard deviation means the data points are close to the mean.
  • A larger standard deviation indicates more spread out data.
This information is essential when calculating times for a range of students, such as determining how much additional time allows most students to finish their exams on time.
Probability Analysis
Probability analysis involves determining the likelihood of specific events occurring, based on the z-scores and the standard normal distribution.Let's break it down further:
  • To calculate the proportion of students unable to finish a 50-minute test, we find the z-score and then interpret it through the standard normal distribution table. This gives us the probability of completion within 50 minutes, and subtracting from 1 shows those unable to finish.
  • For allowing 90% of students to complete the test, we look for a z-score with that cumulative probability. The corresponding exam time is found by rearranging the standard z-formula to \(X = Z\sigma + \mu\).
  • When considering the fastest 25% of students, we again use probability analysis to find the z-score, which helps us determine the maximum time required for this subset of students to complete the exam, ensuring they are counted among the fastest finishers.
Understanding these probabilities is key to tailoring exam durations to student performance levels effectively.

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Most popular questions from this chapter

A pizza company advertises that it puts 0.5 pound of real mozzarella cheese on its medium-sized pizzas. In fact, the amount of cheese on a randomly selected medium pizza is normally distributed with a mean value of 0.5 pound and a standard deviation of 0.025 pound. a. What is the probability that the amount of cheese on a medium pizza is between 0.525 and 0.550 pound? b. What is the probability that the amount of cheese on a medium pizza exceeds the mean value by more than 2 standard deviations? c. What is the probability that three randomly selected medium pizzas each have at least 0.475 pound of cheese?

A grocery store has an express line for customers purchasing at most five items. Let \(x\) be the number of items purchased by a randomly selected customer using this line. Make two tables that represent two different possible probability distributions for \(x\) that have the same mean but different standard deviations.

Let \(z\) denote a random variable that has a standard normal distribution. Determine each of the following probabilities: a. \(P(z<2.36)\) b. \(P(z \leq 2.36)\) c. \(P(z<-1.23)\) d. \(P(1.142)\) g. \(P(z \geq-3.38)\) h. \(P(z<4.98)\)

The Los Angeles Times (December 13,1992 ) reported that what \(80 \%\) of airline passengers like to do most on long flights is rest or sleep. Suppose that the actual percentage is exactly \(80 \%,\) and consider randomly selecting six passengers. Then \(x=\) the number among the selected six who prefer to rest or sleep is a binomial random variable with \(n=6\) and \(p=0.8\) a. Calculate \(p(4)\), and interpret this probability. b. Calculate \(p(6),\) the probability that all six selected passengers prefer to rest or sleep. c. Calculate \(P(x \geq 4)\).

Suppose that fuel efficiency (miles per gallon, mpg) for a particular car model under specified conditions is normally distributed with a mean value of \(30.0 \mathrm{mpg}\) and a standard deviation of \(1.2 \mathrm{mpg}\). a. What is the probability that the fuel efficiency for a randomly selected car of this model is between 29 and \(31 \mathrm{mpg} ?\) b. Would it surprise you to find that the efficiency of a randomly selected car of this model is less than \(25 \mathrm{mpg} ?\) c. If three cars of this model are randomly selected, what is the probability that each of the three have efficiencies exceeding \(32 \mathrm{mpg}\) ? d. Find a number \(x^{*}\) such that \(95 \%\) of all cars of this model have efficiencies exceeding \(x^{*}\left(\right.\) i.e., \(\left.P\left(x>x^{*}\right)=0.95\right)\).
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