Question

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

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

Step-by-step solution

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

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 \(1.28\).

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: \(x = 70 + 1.28(12)\)

Calculate the test completion time for 90% of the students

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

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.

Key concepts

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.