There are 40 students in an elementary statistics class. On...

There are 40 students in an elementary statistics class. On the basis of years of experience, the instructor knows that the needed to grade a randomly chosen paper from the first exam is a random variable with an expected value of 6 $\min$ and a standard deviation of 6 $\mathrm{min.}$ (a) If grading times are independent and the instructor begins grading at $6 : 50 \mathrm{p.m.}$ 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?

There are 40 students in an elementary statistics class. On the basis of years of experience, the
instructor knows that the needed to grade a randomly chosen paper from the first exam is a random variable with an expected value of 6 $\min$ and a standard deviation of 6 $\mathrm{min.}$
(a) If grading times are independent and the instructor begins grading at $6 : 50 \mathrm{p.m.}$ 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?

Key Concepts

Central Limit Theorem

The Central Limit Theorem states that the sum (or average) of a large number of independent, identically distributed random variables tends toward a normal distribution, regardless of the shape of the original distribution. This concept is fundamental when approximating the distribution of the sum of individual grading times, allowing us to use normal probability methods even if the underlying distribution is not normal.

Sum of Independent Random Variables

When dealing with independent random variables, the mean of their sum is the sum of their individual means, and the variance is the sum of their individual variances. This principle is crucial when calculating the total expected time and overall variability for grading multiple papers, as it permits the aggregation of individual average times and deviations into a collective distribution.

Standardization

Standardization involves converting a random variable into a z-score by subtracting the mean and dividing by the standard deviation. This process transforms the variable into the standard normal distribution, enabling easy computation of probabilities using standard normal tables or software, which is essential for determining the likelihood of grading being completed before a specified time.

Normal Distribution Probability Calculation

Once a problem is reduced to a normal distribution through the Central Limit Theorem and standardization, probability calculations can be carried out using the cumulative distribution function (CDF) of the standard normal distribution. This method allows for the computation of the likelihood that a sum of random variables falls below or above a certain threshold, as required in time-based scenarios.

Transcript

00:01 So we're given that this teacher has 40 students.

00:04 The expected value that it takes to grade one paper is six minutes with the standard deviation of also six minutes.

00:11 And what we're trying to find is the probability that he's able to grade all these papers before the news starts.

00:18 So it says that he'll start at 6 .50 p .m.

00:23 Grading and go till 11 p .m.

00:27 So if we convert this to minutes, well, this is four hours in ten minutes, so this is four times 60 plus 10 minutes, which is equal to four times 60 is 240 plus 10 is 250.

00:48 So the expected time that it's supposed to take is 250 minutes.

00:54 Sorry, not the expected time.

00:56 This is the amount of time he has to finish the grading in order to be done before the news starts.

01:04 So now if we take our expected value, which is six, and if we multiply this by the number of students, which is 40, we'll get 240.

01:17 And this is the expected value, or the expected amount of time, it's supposed to take him to grade all 40 students.

01:24 So this will be helpful later when we calculate z scores.

01:30 And the last thing i'm going to calculate before we go to the z score is the standard deviation over all 40 students.

01:39 And we can find that by taking our standard deviation of 60 and multiplying by the square root of our end value or the 40 students in this case.

01:50 So this is equal to 6 times square root of 40, which is equal to 6 times, well, you can take a 2 out of there.

02:02 And 2 squared is 4.

02:03 So if we do abide by 4, we have the square root of 10.

02:06 So we have 6 times 2 square roots of 10, which is equal to 12 square roots of 10.

02:14 Now what we're going to do is we want to find this probability of x being like.

02:20 Than are equal to 250 minutes and where x is the amount of time it takes to grade all 40 students.

02:28 So we're going to use a z score to then be able to use the standard normal distribution table.

02:36 So we're going to have z is equal to our x minus our expected value over all 40 students, which i'll just say is u divided by our standard deviation overall 40 students.

02:50 So this is equal to, in this case, if we're going to do it for 250, it would be equal to 250 minus 240, since that's the expected amount of time it's supposed to take to grade all 40 students, divided by 12 squareds of 10.

03:09 So if we simplify this, 250 minus 240 is 10, and we're dividing by 12 squareths of 10...

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