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Chapter 5: Problem 4

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. Probability he finishes grading by 11:00 PM is 60.4%. b. Probability he misses part of the sports report is 30.1%.

Step by step solution

01

Determine Total Grading Time Expected Value

First, calculate the total expected time to grade all 40 exams. Since the expected time to grade one exam is 6 minutes, for 40 exams it is: \[ E(T) = 40 \times 6 = 240 \text{ minutes} \]
02

Calculate the Total Time Standard Deviation

Since grading times are independent, the variance of the total grading time is the sum of the variances of individual grading times. The standard deviation of one paper is given as 6 minutes, so the variance for one paper is \( 6^2 = 36 \). For 40 papers, \[ \text{Total Variance} = 40 \times 36 = 1440 \] and thus, \[ \text{Total Standard Deviation} = \sqrt{1440} = 37.95 \text{ minutes} \]
03

Calculate Total Time Available Before 11:00 PM

Grade start time is 6:50 PM, and he needs to finish by 11:00 PM, so the available time is \[ 11:00 \text{ PM} - 6:50 \text{ PM} = 250 \text{ minutes} \]
04

Find the Z-score for the Probability

The Z-score is calculated to determine how likely it is for grading to be completed in 250 minutes. Using the formula, \[ Z = \frac{X - E(T)}{\sigma(T)} = \frac{250 - 240}{37.95} \approx 0.2638 \]
05

Determine Probability from Z-score

Using the standard normal distribution table, find the probability corresponding to \( Z = 0.2638 \). This Z-score indicates the probability of finishing within 250 minutes is approximately 0.604.Thus, the probability of finishing before 11:00 PM is about 0.604 or 60.4%.
06

Calculate Time Available Before 11:10 PM

There are 20 extra minutes available before the sports report which starts at 11:10 PM (additional 10 minutes from 11:00 PM calculated previously). Total available time is thus \[ 250 + 10 = 260 \text{ minutes} \]
07

Find New Z-score for Sports Report Probability

The new Z-score, given 260 available minutes, is: \[ Z = \frac{260 - 240}{37.95} \approx 0.527 \]
08

Determine Sports Report Probability

From the Z-score table, the probability corresponding to \( Z = 0.527 \) is around 0.699. This means the probability of not finishing in 260 minutes is \( 1 - 0.699 = 0.301 \). Thus, the probability that he misses part of the sports report is approximately 30.1%.

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Key Concepts

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

Expected Value
The expected value, often denoted as \( E(X) \), is a fundamental concept in probability theory. It represents the average or mean value of a random variable over a large number of trials. Think of it as the long-term average outcome you would expect if you repeatedly performed the same experiment.
  • In our problem, the expected value for grading a single paper is given as 6 minutes.
  • This means on average, each paper will take 6 minutes to grade.
  • For 40 papers, the total expected grading time is \( 40 \times 6 = 240 \) minutes.
Understanding expected value can help in planning time management and resource allocation, as it provides a baseline expectation of performance.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean, whereas a high standard deviation means the values are spread out over a wider range. For grading time:
  • Each exam has a standard deviation of 6 minutes.
  • This tells us that the time taken to grade a single paper might deviate from the mean by about 6 minutes in either direction.
  • For all 40 papers, we calculate the total standard deviation using the formula: \( \sqrt{40 \times 36} = 37.95 \) minutes.
Knowing the standard deviation helps in understanding the variability and rigidity of the process. It indicates how much the total grading time can vary from the expected time.
Normal Distribution
In probability theory, the normal distribution is a continuous probability distribution that is symmetric around the mean. It's often depicted as a bell-shaped curve. Many real-world phenomena approximate a normal distribution. Thanks to its properties:
  • We can make predictions about the probability of certain outcomes.
  • For example, the Z-score helps us determine the probability that grading will be completed before a certain time based on our calculated expected time and standard deviation.
  • Using the normal distribution, we find that the probability of finishing grading before 11:00 PM is approximately 60.4%.
This makes normal distribution a powerful tool for decision making in situations involving uncertainty.
Random Variables
Random variables are fundamental to probability theory. They are variables whose possible values are numerical outcomes of a random phenomenon. In our context:
  • The time taken to grade a paper is a random variable because it can vary each time a paper is graded.
  • These grading times are independent of each other, meaning the time for one paper doesn't affect another's.
  • Our paper grades are modeled as random variables to incorporate the inherent uncertainty and variability in grading.
By understanding random variables, statisticians can model and analyze scenarios like our grading problem, predicting probabilities for different potential outcomes and better managing processes involving random variability.

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Most popular questions from this chapter

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The number of parking tickets issued in a certain city on any given weekday has a Poisson distribution with parameter \(\mu=50\). a. Calculate the approximate probability that between 35 and 70 tickets are given out on a particular day. b. Calculate the approximate probability that the total number of tickets given out during a 5 -day week is between 225 and 275 . c. Use software to obtain the exact probabilities in (a) and (b) and compare to their approximations.

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