/*! 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 80 Consider the variable \(x=\) tim... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 80

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 \(45 \mathrm{~min}\) and standard deviation \(5 \mathrm{~min}\). a. If \(50 \mathrm{~min}\) 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 we 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 50 minutes. b. Around 51.4 minutes should be allowed for the exam if we want 90% of the students to finish in the allotted time. c. The fastest 25% of students would require approximately 41.65 minutes to complete the exam.

Step by step solution

01

- Calculate Proportion of students are unable to finish in 50 minutes

To solve part a. The first task is to normalize the value of 50 minutes by subtracting the mean and dividing by the standard deviation to find its corresponding z-score, \(z_1\). This is done by the following formula : \(z = (X - 渭) / 蟽 \), which gives us: \( z_1 = (50 - 45) / 5 = 1 \). Using a standard normal distribution table or a calculator, one can determine the proportion of students who have a normalized score less than 1, which corresponds to being able to complete the test in 50 minutes. The proportion is 0.8413. To find out the proportion of students who would not finish the test in 50 minutes, subtract this value from 1. This gives us \(1 - 0.8413 = 0.1587\) or approximately 15.87% of students.
02

- Calculate Time required to finish by 90% of students

In part b, one needs to find out the time that would allow 90% of students to complete the exam. This entails finding the z-score corresponding to the 90th percentile from the standard normal distribution, \(z_{90}\). From the standard normal distribution table or using a calculator, one would find that \( z_{90} = 1.28 \). Then, to calculate the time, use this z-score and the original mean (渭) and standard deviation (蟽) in the formula \(X = 渭 + z_90 * 蟽\). This yields: \(X = 45 + 1.28 * 5 = 51.4\) minutes, which is approximately the time required for 90 percent of all students to complete the exam.
03

- Determine Time required by fastest 25% students

The fastest 25% students implies finding the 25th percentile or lower quartile. By referencing the standard normal distribution table or using a calculator, one finds that the z-score corresponding to the 25th percentile, \(z_{25}\), is approximately -0.67. The actual time can then be calculated by using the formula \(X = 渭 + z_{25} * 蟽\), resulting in: \(X = 45 + (-0.67) * 5 = 41.65 \) minutes, which is the approximate time required for the fastest 25% students to complete the exam.

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