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Chapter 6: Problem 14

There are 40 students in an elementary statistics class. On the basis of years of experience, the instructor knows that the time needed to grade a randomly chosen first examination paper is a random variable with an expected value of \(6 \mathrm{~min}\) and a standard deviation of \(6 \mathrm{~min}\). a. If grading times are independent and the instructor begins grading at \(6: 50\) p.m. and grades continuously, what is the (approximate) probability that he is through grading before the 11:00 p.m. TV news begins? b. If the sports report begins at \(11: 10\), what is the probability that he misses part of the report if he waits until grading is done before turning on the TV?

Short Answer

Expert verified
a. 9.34% probability of finishing before news. b. 85.31% probability of missing sports report.

Step by step solution

01

Identify the total grading time needed

Since there are 40 students and the expected grading time per student is 6 minutes, calculate the total expected grading time. This is given by \( E(T) = 40 \times 6 = 240 \) minutes.
02

Calculate the standard deviation of the total grading time

The standard deviation of the grading time for one paper is 6 minutes. Since grading times are independent, the standard deviation of the total grading time for 40 papers is \( \sigma_T = \sqrt{40} \times 6 = 6 \sqrt{40} \approx 37.95 \) minutes.
03

Calculate the available grading time until 11:00 p.m.

The instructor starts grading at 6:50 p.m. and has until 11:00 p.m. to finish. This gives him \(4 \times 60 + 10 = 190\) minutes to finish grading.
04

Determine when grading is completed within available time

Convert the problem to a standard normal distribution problem; we seek \(P(T < 190)\). Convert the time to a standard normal variable using the formula: \( Z = \frac{X - E(T)}{\sigma_T} = \frac{190 - 240}{37.95} \approx -1.32 \).
05

Find probability for completion before 11:00 p.m

Use the standard normal distribution table to find \(P(Z < -1.32)\). The value is approximately 0.0934, meaning there is a 9.34% chance to finish before 11:00 p.m.
06

Determine the grading time available until 11:10 p.m.

Now calculate the available time for grading until 11:10 p.m. The instructor has \(200\) minutes from 6:50 p.m. to 11:10 p.m.
07

Calculate Probability for missing sports report

Identify \(P(T >= 200)\) as \(1 - P(T < 200)\). Convert 200 minutes to the standard normal variable: \( Z = \frac{200 - 240}{37.95} \approx -1.05 \). So, \(P(Z < -1.05) \approx 0.1469\). Thus, \(P(T \geq 200) \approx 1 - 0.1469 = 0.8531\).
08

Conclusion

The probability that the instructor misses part of the sports report if he waits until grading is done before turning on the TV is approximately 85.31%.

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Key Concepts

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

Expected Value
When we talk about expected value, it's all about understanding the average result you can anticipate from a random event. Think of it as the long-term average or mean. In the context of our exercise, the expected value tells us the average grading time per paper, which is given as 6 minutes. This is important because it helps us predict how long all grading will take on average for 40 students.
  • The expected value is calculated as the sum of all possible values, each multiplied by its probability.
  • In this problem, we multiply the average time per paper by the number of papers: \(E(T) = 40 \times 6\).
  • This gives us a total expected grading time of 240 minutes.
Understanding expected value reduces uncertainty by providing a numerical basis for predicting outcomes.
Standard Deviation
Standard deviation measures how spread out numbers are in a data set. In other words, it tells us how much the grading time for each paper might deviate from the average grading time. For our scenario, the standard deviation for grading one paper is 6 minutes.
  • The larger the standard deviation, the more spread out or variable the data points are.
  • Since each grading time is independent, finding the standard deviation for the total grading time involves the formula: \( \sigma_T = \sqrt{40} \times 6\).
  • This results in a standard deviation of approximately 37.95 minutes for grading all papers.
Standard deviation is crucial in this scenario because it helps us understand the range of time that grading may take overall, a key ingredient in predicting probabilities related to the completion of grading.
Normal Distribution
The normal distribution, sometimes called a bell curve, is a probability distribution that is symmetric about the mean. Its shape is determined by the mean and standard deviation. In this exercise, the total grading time is assumed to fit a normal distribution because we're dealing with a large number of independent random events.
  • The normal distribution is crucial because it allows us to use Z-scores to calculate probabilities for the given grading times.
  • In our exercise, this helps determine the likelihood of completing grading by certain times, such as before 11:00 p.m. or 11:10 p.m.
  • Conversion to standard normal involves calculating Z-scores: \( Z = \frac{X - E(T)}{\sigma_T} \).
This reveals how the instructor can expect deviations in total grading time and compute probabilities for completion thanks to the properties of the normal distribution.
Independent Random Variables
Independent random variables do not affect each other. In our example, the grading time for one student's paper is independent of another's. This independence means we can confidently calculate the collective effect, such as total expected time and standard deviation, by simple addition and multiplication.
  • Independence ensures that the occurrence of one event (grading one paper) doesn't impact another.
  • For our calculation, independence allows us to compute the total expected time and cumulative standard deviation across all papers effectively.
  • This property simplifies the use of formulas for expected value and standard deviation, making statistical predictions more manageable.
Understanding independence is essential as it underpins our ability to apply statistical methods like those used for this exercise's predictive analysis accurately.

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Most popular questions from this chapter

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