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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 students to complete the exam?

Short Answer

Expert verified
a. Approximately 15.87% of students would be unable to finish the exam in the allotted time of 50 minutes. b. The exam time should be about 51.4 minutes for 90% of the students to be able to finish. c. The fastest 25% of the students complete the exam within approximately 42.63 minutes.

Step by step solution

01

(Step 1: Find the z-score for 50 minutes)

Since the distribution of the exam completion time is normal, we should standardize the value of 50 minutes to its corresponding z-score. The z-score is calculated by the formula: \(z = \frac{x - \mu}{\sigma}\), where x is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. In this case, \(\mu = 45\) minutes and \(\sigma = 5\) minutes. So for 50 minutes, we have: \(z = \frac{50 - 45}{5} = 1\).
02

(Step 2: Find the proportion of students finishing after 50 minutes)

We will now use the standard normal distribution table (z-table) or a calculator to find the proportion of students finishing after 50 minutes. We need to find the area to the right of z=1. Check the z-table or use a calculator to find the area to the left of z=1, which is 0.8413. Since the total area under the normal curve is 1, the area to the right of z=1 is: \(1 - 0.8413 = 0.1587\). Therefore, about 15.87% of students would be unable to finish the exam in 50 minutes.
03

(Step 3: Find the z-score for the 90th percentile)

To find the time allowed so that 90% of the students can finish the exam, we first need to find the z-score that corresponds to the 90th percentile. Check the z-table or use a calculator to find the z-score that gives an area to the left of 0.9. The z-score that corresponds to the 90th percentile is approximately 1.28.
04

(Step 4: Find the exam time for the 90th percentile)

Using the z-score found in the previous step, we can now calculate the time allowed for the exam so that 90% of the students can finish. Recall the z-score formula: \(z = \frac{x - \mu}{\sigma}\). We know that z=1.28, \(\mu = 45\), and \(\sigma = 5\). Rearrange the formula to solve for x: \(x = \mu + z\sigma = 45 + 1.28(5) = 51.4\). Therefore, the exam time should be about 51.4 minutes for 90% of the students to be able to finish.
05

(Step 5: Find the z-score for the 25th percentile)

To find the time required for the fastest 25% of the students to finish the exam, we first need to find the z-score that corresponds to the 25th percentile. Check the z-table or use a calculator to find the z-score that gives an area to the left of 0.25. The z-score corresponds to about -0.674.
06

(Step 6: Find the exam time for the 25th percentile)

Using the z-score found in the previous step, we can now calculate the time required for the fastest 25% of students to finish the exam. Use the z-score formula: \(z = \frac{x - \mu}{\sigma}\). We know that z=-0.674, \(\mu = 45\), and \(\sigma = 5\). Rearrange the formula to solve for x: \(x = \mu + z\sigma = 45 - 0.674(5) = 42.63\). Therefore, the fastest 25% of the students complete the exam within approximately 42.63 minutes.

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Key Concepts

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

Z-score
Understanding the Z-score is crucial when dealing with normal distributions. The Z-score tells us how many standard deviations away a particular data point is from the mean of the distribution. In simpler terms, it helps us determine where a specific point lies in relation to the average.

To calculate the Z-score, use the formula:
  • d \( z = \frac{x - \mu}{\sigma} \)where:
    • \(x\) is the data point you're examining.
    • \(\mu\) (mu) is the mean of the distribution.
    • \(\sigma\) (sigma) is the standard deviation.
For example, if a student's time to complete an exam is 50 minutes, the Z-score would tell us how this time compares to an average time of 45 minutes with a standard deviation of 5 minutes. Using the formula, we get a Z-score of 1, meaning the time is 1 standard deviation above the mean.

This standardized approach allows us to use a common scale for understanding probabilities and percentiles, regardless of the original units of measure.
Percentile
Percentiles are a way of expressing how a particular value compares to other values in the dataset. A percentile indicates the relative standing of a value within a distribution. Essentially, if a score is in the 90th percentile, it means that the score is higher than 90% of the other scores.

To find the percentile corresponding to a particular Z-score, you use a Z-table or statistical software to find the cumulative probability. For instance, a Z-score of 1 corresponds to the 84th percentile, which means that the value is greater than 84% of the distribution. In our exercise, to ensure 90% of students finish an exam within the allowed time, we find the Z-score for the 90th percentile, which is approximately 1.28.

Understanding percentiles is key to decision-making processes, like determining how much exam time to allocate if a certain percentage of students should finish on time.
Standard Deviation
Standard Deviation (SD) is a measure of the amount of variation or dispersion in a set of values. In the context of a normal distribution, it tells us about the spread of the data points around the mean. A small SD means data points are generally closer to the mean, whereas a larger SD indicates they are more spread out.

If an exam completion time has a mean of 45 minutes with an SD of 5 minutes, most students complete the exam within this range:
  • 68% of students will complete it within one standard deviation (40 to 50 minutes).
  • 95% will do so within two standard deviations (35 to 55 minutes).
  • 99.7% will be within three standard deviations (30 to 60 minutes).
This concept of standard deviation helps us understand potential variability in data and plays a vital role in calculating Z-scores and percentiles. It's a fundamental piece of how we interpret and manage statistical data.

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Most popular questions from this chapter

Suppose that your statistics professor tells you that the scores on a midterm exam were approximately normally distributed with a mean of 78 and a standard deviation of 7 . The top \(15 \%\) of all scores have been designated A's. Your score is \(89 .\) Did you earn an \(\mathrm{A}\) ? Explain.

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