/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 67 The average length of time requi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The average length of time required to complete a college achievement test was found to equal 70 minutes with a standard deviation of 12 minutes. When should the test be terminated if you wish to allow sufficient time for \(90 \%\) of the students to complete the test? (Assume that the time required to complete the test is normally distributed.)

Short Answer

Expert verified
Answer: The test should be terminated at approximately 86 minutes to allow 90% of the students to complete it.

Step by step solution

01

Identifying the Relevant Information

From the problem, we are given: - Mean test completion time, \(\mu = 70\) minutes - Standard deviation, \(\sigma = 12\) minutes - The time required to complete the test is normally distributed - We want to find the time for 90% of the students to complete the test
02

Find the z-score corresponding to 90% of the students

Find the z-score that corresponds to 90% of the students completing the test. We can look this up in a standard normal distribution table or use a calculator with normal distribution functions. From a standard normal table or calculator, we find the z-score corresponding to 90% (0.9) is approximately \(1.28\).
03

Use the z-score formula to find the test completion time

We can use the z-score formula, which relates a value from a normal distribution to its corresponding z-score: \(z = \frac{x - \mu}{\sigma}\) We can rearrange the formula to solve for the test completion time, \(x\): \(x = \mu + z\sigma\) Plugging in the values we know: \(x = 70 + 1.28(12)\)
04

Calculate the test completion time for 90% of the students

Now, compute the approximate completion time for 90% of the students: \(x \approx 70 + 1.28(12) \approx 70 + 15.36 = 85.36\)
05

Interpret the results

The test should be terminated at approximately 85.36 minutes to allow 90% of the students to complete the test. It is common to round up this value for practical purposes, so we can say the test should be terminated at 86 minutes.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding Z-Score
When dealing with the normal distribution, a useful measure called the z-score helps determine how far a specific data point is from the mean. Basically, it's a numerical representation that tells you how many standard deviations a particular value is away from the mean of a dataset.
To calculate a z-score, use the formula:
  • \( z = \frac{x - \mu}{\sigma} \)
Where:
  • \( x \) represents the data point you're analyzing.
  • \( \mu \) is the mean of the dataset.
  • \( \sigma \) is the standard deviation of the dataset.
A positive z-score indicates the data point is above the mean, while a negative one signifies it is below the mean. This measure allows us to use standardized tables or calculators to find probabilities and percentages associated with a certain value in a normal distribution.
For example, in our exercise, we calculated a z-score of approximately 1.28, indicating that the required test time is 1.28 standard deviations above the mean, making it possible to predict that 90% of students will complete the test within that timeframe.
Decoding Standard Deviation
Standard deviation is a key concept in statistics, often used to describe the amount of variation or dispersion in a set of values. It effectively measures how spread out the data points are from the mean.
The formula for standard deviation, \( \sigma \), involves a series of steps:
  • Calculating the mean (average) of the dataset.
  • Subtracting the mean from each data point to find the deviations.
  • Squaring each deviation to make all values positive.
  • Finding the average of these squared deviations.
  • Taking the square root of this average to obtain the standard deviation.
In our exercise, the standard deviation was provided as 12 minutes, indicating the typical deviation from the average test completion time of 70 minutes. A smaller standard deviation would suggest that most students finish close to this mean time, whereas a larger one would indicate a broader spread of completion times across students.
Grasping the Mean
The mean, often known as the average, is one of the most commonly used measures in statistics to summarize a set of data. It provides a central value around which the data clusters. To calculate the mean, you sum up all the data points in a set and then divide this total by the number of data points.
  • If you have a set of numbers: \( x_1, x_2, \ldots, x_n \)
  • The mean \( \mu \) can be calculated using:
  • \( \mu = \frac{x_1 + x_2 + \cdots + x_n}{n} \)
In the context of our exercise, the mean test completion time was found to be 70 minutes. This figure tells us that on average, most students finish the achievement test within this timeframe. Understanding the mean helps gauge overall performance and set realistic time limits for events such as exams based on average completion rates.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For a car traveling 30 miles per hour (mph), the distance required to brake to a stop is normally distributed with a mean of 50 feet and a standard deviation of 8 feet. Suppose you are traveling \(30 \mathrm{mph}\) in a residential area and a car moves abruptly into your path at a distance of 60 feet. a. If you apply your brakes, what is the probability that you will brake to a stop within 40 feet or less? Within 50 feet or less? b. If the only way to avoid a collision is to brake to a stop, what is the probability that you will avoid the collision?

The typical American family spends lots of time driving to and from various activities, and lots of time in the drive-thru lines at fast-food restaurants. There is a rising amount of evidence suggesting that we are beginning to burn out! In fact, in a study conducted for the Center for a New American Dream, Time magazine reports that \(60 \%\) of Americans felt pressure to work too much, and \(80 \%\) wished for more family time. \({ }^{7}\) Assume that these percentages are correct for all Americans, and that a random sample of 25 Americans is selected. a. Use Table 1 in Appendix I to find the probability that more than 20 felt pressure to work too much. b. Use the normal approximation to the binomial distribution to aproximate the probability in part a. Compare your answer with the exact value from part a. c. Use Table 1 in Appendix I to find the probability that between 15 and 20 (inclusive) wished for more family time. d. Use the normal approximation to the binomial distribution to approximate the probability in part c. Compare your answer with the exact value from part c.

One method of arriving at economic forecasts is to use a consensus approach. A forecast is obtained from each of a large number of analysts, and the average of these individual forecasts is the consensus forecast. Suppose the individual 2008 January prime interest rate forecasts of economic analysts are approximately normally distributed with the mean equal to \(8.5 \%\) and a standard deviation equal to \(0.2 \% .\) If a single analyst is randomly selected from among this group, what is the probability that the analyst's forecast of the prime rate will take on these values? a. Exceed \(8.75 \%\) b. Be less than \(8.375 \%\)

Find a \(z_{0}\) such that \(P\left(-z_{0}

How does the IRS decide on the percentage of income tax returns to audit for each state? Suppose they do it by randomly selecting 50 values from a normal distribution with a mean equal to \(1.55 \%\) and a standard deviation equal to \(.45 \% .\) (Computer programs are available for this type of sampling.) a. What is the probability that a particular state will have more than \(2.5 \%\) of its income tax returns audited? b. What is the probability that a state will have less than \(1 \%\) of its income tax returns audited?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.