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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

Consider the following 10 observations on the lifetime (in hours) for a certain type of power supply: $$ \begin{array}{lllll} 152.7 & 172.0 & 172.5 & 173.3 & 193.0 \\ 204.7 & 216.5 & 234.9 & 262.6 & 422.6 \end{array} $$ Construct a normal probability plot, and comment on whether it is reasonable to think that the distribution of power supply lifetime is approximately normal. (The normal scores for a sample of size 10 are -1.539,-1.001,-0.656 , \(-0.376,-0.123,0.123,0.376,0.656,1.001,\) and \(1.539 .)\)

Industrial quality control programs often include inspection of incoming materials from suppliers. If parts are purchased in large lots, a typical plan might be to select 20 parts at random from a lot and inspect them. Suppose that a lot is judged acceptable if one or fewer of these 20 parts are defective. If more than one part is defective, the lot is rejected and returned to the supplier. Find the probability of accepting lots that have each of the following (Hint: Identify success with a defective part): a. \(5 \%\) defective parts b. \(10 \%\) defective parts c. \(20 \%\) defective parts

An appliance dealer sells three different models of freezers having \(13.5,15.9,\) and 19.1 cubic feet of storage space. Let \(x=\) the amount of storage space purchased by the next customer to buy a freezer. Suppose that \(x\) has the following probability distribution: $$ \begin{array}{lrrr} x & 13.5 & 15.9 & 19.1 \\ p(x) & 0.2 & 0.5 & 0.3 \end{array} $$ a. Calculate the mean and standard deviation of \(x\). (Hint: See Example 6.15\()\) b. Give an interpretation of the mean and standard deviation of \(x\) in the context of observing the outcomes of many purchases.

A box contains four slips of paper marked \(1,2,3,\) and 4. Two slips are selected without replacement. List the possible values for each of the following random variables: a. \(x=\) sum of the two numbers b. \(y=\) difference between the first and second numbers (first - second) c. \(z=\) number of slips selected that show an even number d. \(w=\) number of slips selected that show a 4

Suppose \(x=\) the number of courses a randomly selected student at a certain university is taking. The probability distribution of \(x\) appears in the following table: $$ \begin{array}{lccccccc} x & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ p(x) & 0.02 & 0.03 & 0.09 & 0.25 & 0.40 & 0.16 & 0.05 \end{array} $$ a. What is \(P(x=4)\) ? b. What is \(P(x \leq 4)\) ? c. What is the probability that the selected student is taking at most five courses? d. What is the probability that the selected student is taking at least five courses? More than five courses? e. Calculate \(P(3 \leq x \leq 6)\) and \(P(3 < x < 6)\). Explain in words why these two probabilities are different.
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