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Chapter 5: Q49E (page 237)

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

Suppose the amount of liquid dispensed by a certain machine is uniformly distributed with lower limit \({\rm{A = 8oz}}\) and upper limit\({\rm{B = 10oz}}\). Describe how you would carry out simulation experiments to compare the sampling distribution of the (sample) fourth spread for sample sizes\({\rm{n = 5,10,20}}\), and\({\rm{30}}\).

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a. What is the approximate distribution of \({\rm{\bar X ? of \bar Y?}}\)

b. What is the approximate distribution of \({\rm{\bar X - \bar Y}}\)? Justify your answer.

c. Calculate (approximately) \(P( - 1拢\bar X - \bar Y拢1)\)

d. Calculate. If you actually observed , would you doubt that \({{\rm{\mu }}_{\rm{1}}}{\rm{ - }}{{\rm{\mu }}_{\rm{2}}}{\rm{ = 5?}}\)

Return to the situation described in Exercise \({\rm{3}}\).

a. Determine the marginal pmf of \({{\rm{X}}_{\rm{1}}}\), and then calculate the expected number of customers in line at the express checkout.

b. Determine the marginal pmf of \({{\rm{X}}_{\rm{2}}}\).

c. By inspection of the probabilities \({\rm{P(}}{{\rm{X}}_{\rm{1}}}{\rm{ = 4),P(}}{{\rm{X}}_{\rm{2}}}{\rm{ = 0),}}\) and \({\rm{P(}}{{\rm{X}}_{\rm{1}}}{\rm{ = 4,}}{{\rm{X}}_{\rm{2}}}{\rm{ = 0),}}\) are \({{\rm{X}}_{\rm{1}}}\) and \({{\rm{X}}_{\rm{2}}}\) independent random variables? Explain

It is known that\({\rm{80\% }}\)of all brand A extremal hard drives work in a satisfactory manner throughout the warranty period (are "successes"). Suppose that\({\rm{n = 15}}\)drives are randomly selected. Let\({\rm{X = }}\)the number of successes in the sample. The statistic\({\rm{X/n}}\)is the sample proportion (fraction) of successes. Obtain the sampling distribution of this statistic. (Hint: One possible value of\({\rm{X/n is }}{\rm{.2}}\), corresponding to\({\rm{X = 3}}\). What is the probability of this value (what kind of\({\rm{rv}}\)is\({\rm{X}}\))?)

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a. If the distribution is normal, what is the probability that the sample mean hardness for a random sample of \({\rm{9}}\) pins is at least \({\rm{51}}\)?

b. Without assuming population normality, what is the (approximate) probability that the sample mean hardness for a random sample of \({\rm{40 }}\) pins is at least \({\rm{51}}\)?

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