/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q51E The time taken by a randomly sel... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Q51E (page 237)

The time taken by a randomly selected applicant for a mortgage to fill out a certain form has a normal distribution with mean value \({\rm{10}}\) min and standard deviation \({\rm{2}}\) min. If five individuals fill out a form on one day and six on another, what is the probability that the sample average amount of time taken on each day is at most \({\rm{11}}\) min?

Short Answer

Expert verified

\({\rm{P(\bar X拢 11) = 0}}{\rm{.772}}\)

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 probability that the sample average amount of time taken on each day is at most \({\rm{11}}\) min

The time taken has a normal distribution with a standard deviation of \({\rm{\sigma = 2}}\)minutes and a \({\rm{\mu = 10}}\)minute mean. \({{\rm{\mu }}_{{\rm{\bar X}}}}{\rm{ = 10}}\)is the expected value of random variable \({\rm{X}}\) (sample average), and the standard deviation is

\(\begin{array}{*{20}{c}}{{{\rm{\sigma }}_{{\rm{\bar X}}}}{\rm{ = }}\frac{{\rm{1}}}{{\sqrt {\rm{n}} }}{\rm{ \times \sigma = }}\frac{{\rm{1}}}{{\sqrt {\rm{5}} }}{\rm{ \times 2 = 0}}{\rm{.894,}}\;\;\;{\rm{\;when\;n = 5}}}\\{{{\rm{\sigma }}_{{\rm{\bar X}}}}{\rm{ = }}\frac{{\rm{1}}}{{\sqrt {\rm{n}} }}{\rm{ \times \sigma = }}\frac{{\rm{1}}}{{\sqrt {\rm{6}} }}{\rm{ \times 2 = 0}}{\rm{.816,}}\;\;\;{\rm{\;when\;n = 6}}}\end{array}\)

For the first day (\({\rm{n = 5}}\)individuals), the requested probability is

\(\begin{aligned}{{{\rm{P}}_{\rm{1}}}{\rm{(\bar X拢 11) &= P\left( {\frac{{{\rm{\bar X - }}{{\rm{\mu }}_{{\rm{\bar X}}}}}}{{{{\rm{\sigma }}_{{\rm{\bar X}}}}}}{\rm{拢 }}\frac{{{\rm{11 - 10}}}}{{{\rm{0}}{\rm{.894}}}}} \right){\rm{ = P(Z拢 1}}{\rm{.12)}}}\\&= {\rm{0}}{\rm{.8686,}}}\end{aligned}\)

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

For the second day (\({\rm{n = 6}}\)individuals), the requested probability is

\(\begin{aligned}{{{\rm{P}}_{\rm{2}}}{\rm{(\bar X拢 11) &= P\left( {\frac{{{\rm{\bar X - }}{{\rm{\mu }}_{{\rm{\bar X}}}}}}{{{{\rm{\sigma }}_{{\rm{\bar X}}}}}}{\rm{拢 }}\frac{{{\rm{11 - 10}}}}{{{\rm{0}}{\rm{.816}}}}} \right){\rm{ = P(Z拢 1}}{\rm{.22)}}}\\ &= {\rm{0}}{\rm{.8888,}}}\end{aligned}\)

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

\(\begin{aligned} P(\bar X拢 11) &= {{\rm{P}}_{\rm{1}}}{\rm{(\bar X拢 11) \times }}{{\rm{P}}_{\rm{2}}}{\rm{(\bar X拢 11)}}}\\ &= 0 {\rm{.8686 \times 0}}{\rm{.8888}}}\\&= 0 {\rm{.772}}}\end{aligned}\)

Because the two days' outcomes are unrelated, the request probability (for both days) is the product of the computed probabilities.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A shipping company handles containers in three different sizes:\(\left( 1 \right)\;27f{t^3}\;\left( {3 \times 3 \times 3} \right)\)\(\left( 2 \right) 125 f{t^3}, and \left( 3 \right)\;512 f{t^3}\). Let \({X_i}\left( {i = \;1, 2, 3} \right)\)denote the number of type i containers shipped during a given week. With \({\mu _i} = E\left( {{X_i}} \right)\)and\(\sigma _i^2 = V\left( {{X_i}} \right)\), suppose that the mean values and standard deviations are as follows:

\(\begin{array}{l}{\mu _1} = 200 {\mu _2} = 250 {\mu _3} = 100 \\{\sigma _1} = 10 {\sigma _2} = \,12 {\sigma _3} = 8\end{array}\)

a. Assuming that \({X_1}, {X_2}, {X_3}\)are independent, calculate the expected value and variance of the total volume shipped. (Hint:\(Volume = 27{X_1} + 125{X_2} + 512{X_3}\).)

b. Would your calculations necessarily be correct if \({X_i} 's\)were not independent?Explain.

Let X1, X2, and X3 represent the times necessary to perform three successive repair tasks at a certain service facility. Suppose they are independent, normal rv鈥檚 with expected values \({\mu _1}, {\mu _2}, and {\mu _3}\)and variances \(\sigma _1^2 , \sigma _2^2, and \sigma _3^2 \), respectively. a. If \(\mu = {\mu _2} = {\mu _3} = 60\)and\(\sigma _1^2 = \sigma _2^2 = \sigma _3^2 = 15\), calculate \(P\left( {{T_0} \le 200} \right)\)and\(P\left( {150 \le {T_0} \le 200} \right)\)? b. Using the \(\mu 's and \sigma 's\)given in part (a), calculate both \(P\left( {55 \le X} \right)\)and \(P\left( {58 \le X \le 62} \right)\).c. Using the \(\mu 's and \sigma 's\)given in part (a), calculate and interpret\(P\left( { - 10 \le {X_1} - .5{X_2} - .5{X_3} \le 5} \right)\). d. If\({\mu _1} = 40, {\mu _1} = 50, {\mu _1} = 60,\),\( \sigma _1^2 = 10, \sigma _2^2 = 12, and \sigma _3^2 = 14\) calculate \(P\left( {{X_1} + {X_2} + {X_3} \le 160} \right)\)and also \(P\left( {{X_1} + {X_2} \ge 2{X_3}} \right).\)

a. Compute the covariance for \({\rm{X}}\) and \({\rm{Y}}\).

b. Compute \({\rm{\rho }}\) for \({\rm{X}}\) and \({\rm{Y}}\).

A stockroom currently has \({\rm{30}}\) components of a certain type, of which \({\rm{8}}\) were provided by supplier \({\rm{1,10}}\)by supplier \({\rm{2}}\) , and \({\rm{12}}\) by supplier \({\rm{3}}\). Six of these are to be randomly selected for a particular assembly. Let \({\rm{X = }}\) the number of supplier l's components selected, \({\rm{Y = }}\) the number of supplier \({\rm{2}}\) 's components selected, and \({\rm{p(x,y)}}\) denote the joint pmf of \({\rm{X}}\) and\({\rm{Y}}\).

a. What is \({\rm{p(3,2)}}\) ? (Hint: Each sample of size \({\rm{6}}\) is equally likely to be selected. Therefore, \({\rm{p(3,2) = }}\) (number of outcomes with \({\rm{X = 3}}\) and \({\rm{Y = 2)/(}}\) total number of outcomes). Now use the product rule for counting to obtain the numerator and denominator.)

b. Using the logic of part (a), obtain \({\rm{p(x,y}}\) ). (This can be thought of as a multivariate hypergeometric distribution-sampling without replacement from a finite population consisting of more than two categories.)

Answer

We have seen that if\(E\left( {{X_1}} \right) = E\left( {{X_2}} \right) = . . . = E\left( {{X_n}} \right) = \mu \), then\(E\left( {{X_1} + . . . + {X_n}} \right) = n\mu \). In some applications, the number of\({X_i} 's\)under consideration is not a fixed number n but instead is a rv N. For example, let N\(5\)be the number of components that are brought into a repair shop on a particular day, and let Xi denote the repair shop time for the\({i^{th}}\)component. Then the total repair time is\({X_1} + {X_2} + . . . + {X_n}\), the sum of a random number of random variables. When N is independent of the\({X_i} 's\), it can be shown that

\(E\left( {{X_1} + . . . + {X_N}} \right) = E\left( N \right) \times \mu \)

a. If the expected number of components brought in on a particular day is\({\bf{10}}\)and the expected repair time for a randomly submitted component is\({\bf{40}}\)min, what is the expected total repair time for components submitted on any particular day?

b. Suppose components of a certain type come in for repair according to a Poisson process with a rate of\(5\)per hour. The expected number of defects per component is\(3.5\). What is the expected value of the total number of defects on components submitted for repair during a four-hour period? Be sure to indicate how your answer follows from the general result just given.
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.