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There are \({\rm{40}}\) 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 \({\rm{6}}\)min and a standard deviation of \({\rm{6}}\)min.

a. If grading times are independent and the instructor begins grading at \({\rm{6:50}}\) p.m. and grades continuously, what is the (approximate) probability that he is through grading before the \({\rm{11:00}}\) p.m. TV news begins?

b. If the sports report begins at \({\rm{11:10,}}\) 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) \(P\left( {{T_0}拢250} \right) = 0.6026\)

b) \({\rm{P}}\left( {{{\rm{T}}_{\rm{0}}}{\rm{ > 260}}} \right){\rm{ = 0}}{\rm{.2981}}\)

Step by step solution

01

Definition of standard deviation

The square root of the variance is the standard deviation of a random variable, sample, statistical population, data collection, or probability distribution. It is less resilient in practice than the average absolute deviation, but it is algebraically easier.

02

Determining the (approximate) probability that he is through grading before the \({\rm{11:00}}\) p.m. TV news begins

There is a total of 250 minutes between 6:50 PM and \({\rm{11:00PM}}\). The sum of the 40 random variables provided in the exercise can be used to indicate total grading time (mean 6 minutes, standard deviation 6 minutes). Define it as follows:

\({{\rm{T}}_{\rm{0}}}{\rm{ = }}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{2}}}{\rm{ + \ldots + }}{{\rm{X}}_{{\rm{40}}}}{\rm{.}}\)

This random variable's mean value is

\({{\rm{\mu }}_{{{\rm{T}}_{\rm{0}}}}}{\rm{ = n \times \mu = 40 \times 6 = 240,}}\)

as well as the standard deviation

\({{\rm{\sigma }}_{{{\rm{T}}_{\rm{0}}}}}{\rm{ = }}\sqrt {\rm{n}} {\rm{ \times \sigma = }}\sqrt {{\rm{40}}} {\rm{ \times 6 = 37}}{\rm{.95}}{\rm{.}}\)

It is simple to determine the requested probability by determining the mean and standard deviation of the random variable \({{\rm{T}}_{\rm{0}}}\) as follows:

\(\begin{array}{*{20}{c}}{{\rm{P}}\left( {{{\rm{T}}_{\rm{0}}}{\rm{拢 250}}} \right)}&{{\rm{ = P}}\left( {\frac{{{{\rm{T}}_{\rm{0}}}{\rm{ - }}{{\rm{\mu }}_{{{\rm{T}}_{\rm{0}}}}}}}{{{{\rm{\sigma }}_{{{\rm{T}}_{\rm{0}}}}}}}{\rm{拢 }}\frac{{{\rm{250 - 240}}}}{{{\rm{37}}{\rm{.95}}}}} \right){\rm{ = P(Z拢 0}}{\rm{.26)}}}\\{}&{\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{0}}{\rm{.6026}}}\end{array}\)

(\({\rm{1}}\)): from the appendix's normal probability table. Software can also be used to calculate the likelihood.

03

Determining the probability that he misses part of the report if he waits until grading is done before turning on the TV

There are \({\rm{260}}\) minutes left until the sports report starts. The following statement is correct:

\(\begin{array}{*{20}{c}}{{\rm{P}}\left( {{{\rm{T}}_{\rm{0}}}{\rm{ > 260}}} \right){\rm{ = P}}\left( {\frac{{{{\rm{T}}_{\rm{0}}}{\rm{ - }}{{\rm{\mu }}_{{{\rm{T}}_{\rm{0}}}}}}}{{{{\rm{\sigma }}_{{{\rm{T}}_{\rm{0}}}}}}}{\rm{ > }}\frac{{{\rm{260 - 240}}}}{{{\rm{37}}{\rm{.95}}}}} \right){\rm{ = P(Z > 0}}{\rm{.53)}}}\\{\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{0}}{\rm{.2981}}{\rm{.}}}\end{array}\)

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Most popular questions from this chapter

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