Question

There are 46 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 5 min and a standard deviation of 4 min. (Round your answers to four decimal places.)
(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?
1

(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?
2

Answer 1

Concepts and reason

The Z-score is measure how much standard deviation below or above the population mean.

The mean of the sampling distribution of the sample mean is the mean of the population from which the scores were sampled.

The standard error of the mean is the standard deviation of the sampling distribution of the sample mean.

Fundamentals

The formula for z-score is,

$z = \frac{{x - \mu }}{\sigma }$

Here, $\mu$ be the population mean and $\sigma$ be the population standard deviation.

The formula for sampling distribution of the sample mean is ${\mu _{\bar x}} = \mu$

The formula for standard deviation of the sample mean is,

${\sigma _{\bar x}} = \frac{\sigma }{{\sqrt n }}\,$

Here, $\sigma$ is the population standard deviation and $n$ is the sample size.

The formula for z-score is,

$z = \frac{{\bar x - {\mu _{\bar x}}}}{{\frac{\sigma }{{\sqrt n }}}}$

The formula for mean is,

$\mu = E\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)$

The formula for standard deviation is,

$\sigma = \sqrt {V\left( {\sum\limits_{i = 1}^n {{X_i}} } \right)}$

(a)

It is given that there are 46 students in an elementary class.

That is, $n = 46$

The instructor knows that the time needed to grade a randomly chosen first examination paper is a random variable with the expected value of 5 min and a standard deviation of 4 min.

That is, the time need to grade a paper follows normal distribution with mean 5 and standard deviation 4.

The mean time need to grade 46 students is,

$\begin{array}{c}\\{\mu _T} = E\left( {\sum\limits_{i = 1}^{46} {{X_i}} } \right)\\\\ = \sum\limits_{i = 1}^{46} {E\left( {{X_i}} \right)} \\\\ = 46\left( 5 \right)\\\\ = 230\\\end{array}$

The standard deviation of time needed to grade 46 students is,

$\begin{array}{c}\\{\sigma _T} = SD\left( {\sum\limits_{i = 1}^{46} {{X_i}} } \right)\\\\ = \sqrt {V\left( {\sum\limits_{i = 1}^{46} {{X_i}} } \right)} \\\\ = \sqrt {\sum\limits_{i = 1}^{46} {V\left( {{X_i}} \right)} } \\\end{array}$

$\begin{array}{l}\\ = \sqrt {46\left( {16} \right)} \\\\ = 27.129\\\end{array}$

If grading times are independent and the instructor begins grading at 6:50 P.M. and grades continuously then the probability that he is through grading before the 11:00 P.M. TV news begins.

Here, it can be observed that from 6:50 P.M to 11:00 PM we have 250 min.

He will completely grade all 46 papers in $10 + 4\left( {60} \right) = 250\min$

The required probability is,

$\begin{array}{c}\\P\left( {T < 250} \right) = P\left( {\frac{{T - {\mu _T}}}{{{\sigma _T}}} < \frac{{250 - 230}}{{27.129}}} \right)\\\\ = P\left( {Z < 0.7372} \right)\\\\ = 0.7695\\\end{array}$

(b)

If the sports report begins at 11:10 then the probability that he misses part of the report if he waits until grading is done before turning on the TV.

Here, it can be observed that from 6:50 P.M to 11:10 PM we have 260 min.

He will completely grade all 46 papers in $10 + 4\left( {60} \right) + 10 = 260\min$

The required probability is,

$\begin{array}{c}\\P\left( {T > 260} \right) = 1 - P\left( {T \le 260} \right)\\\\ = 1 - P\left( {\frac{{T - {\mu _T}}}{{{\sigma _T}}} < \frac{{260 - 230}}{{27.129}}} \right)\\\\ = 1 - P\left( {Z < 1.105828} \right)\\\end{array}$

$\begin{array}{c}\\ = 1 - 0.865599\\\\ = 0.134401\\\end{array}$

Ans: Part a

The probability that he through grading before the 11:00 P.M. TV news begins is 0.7695.

Part b

The probability that he waits until grading is done before turning on the TV is 0.1344.