/*! 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} Q53E Rockwell hardness of pins of a c... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Q53E (page 237)

Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of \({\rm{1}}{\rm{.2}}{\rm{.}}\)

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 \({\rm{51}}\)?

Short Answer

Expert verified

a.${\text{P(\bar X51) = 0}}{\text{.0062}}$

b.${\text{P(\bar X51) = 0}}$

Step by step solution

01

Definition of probability

the proportion of the total number of conceivable outcomes to the number of options in an exhaustive collection of equally likely outcomes that cause a given occurrence.

02

Determining the probability that the sample mean hardness for a random sample of \({\rm{9}}\) pins is at least \({\rm{51}}\)

Let \({\rm{X}}\)be a random variable with a \({\rm{\mu = 50}}\)mean and a \({\rm{\sigma = 1}}{\rm{.2}}\) standard deviation. The sample average's (mean) \({\rm{\bar X}}\)is \({{\rm{\mu }}_{{\rm{\bar X}}}}{\rm{ = \mu = 50}}\), while the sample average's (mean) standard deviation is

\({{\rm{\sigma }}_{{\rm{\bar X}}}}{\rm{ = }}\frac{{\rm{1}}}{{\sqrt {\rm{n}} }}{\rm{ \times \sigma = }}\frac{{\rm{1}}}{{\sqrt {\rm{9}} }}{\rm{ \times 1}}{\rm{.2 = 0}}{\rm{.4,}}\;\;\;{\rm{n = 9\;pins}}{\rm{.\;}}\)

The following statement is correct:

\[{{\text{P(\bar X51) = P}}\left( {\frac{{{\text{\bar X - }}{{\text{\mu }}_{{\text{\bar X}}}}}}{{{{\text{\sigma }}_{{\text{\bar X}}}}}}{\text{}}\frac{{{\text{51 - 50}}}}{{{\text{0}}{\text{.4}}}}} \right){\text{ = P(Z2}}{\text{.5)}}}\]

\[{{\text{ = 1 - P(Z < 2}}{\text{.5)}}\mathop {\text{ = }}\limits^{{\text{(1)}}} {\text{1 - 0}}{\text{.9938}}}\]

\[{{\text{ = 0}}{\text{.0062,}}}\]

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

03

 Determining the probability that the sample mean hardness for a random sample of \({\rm{40}}\) pins is at least \({\rm{51}}\)

Let \({\rm{X}}\)be a random variable with a \({\rm{\mu = 50}}\)mean and a \({\rm{\sigma = 1}}{\rm{.2}}\) standard deviation. The sample average's (mean) \({\rm{\bar X}}\)is \({{\rm{\mu }}_{{\rm{\bar X}}}}{\rm{ = \mu = 50}}\), while the sample average's (mean) standard deviation is

\({{\rm{\sigma }}_{{\rm{\bar X}}}}{\rm{ = }}\frac{{\rm{1}}}{{\sqrt {\rm{n}} }}{\rm{ \times \sigma = }}\frac{{\rm{1}}}{{\sqrt {{\rm{40}}} }}{\rm{ \times 1}}{\rm{.2 = 0}}{\rm{.19,}}\;\;\;{\rm{n = 40\;pins}}{\rm{.\;}}\)

The following statement is correct:

\[{{\text{P(\bar X51) = P}}\left( {\frac{{{\text{\bar X - }}{{\text{\mu }}_{{\text{\bar X}}}}}}{{{{\text{\sigma }}_{{\text{\bar X}}}}}}{\text{}}\frac{{{\text{51 - 50}}}}{{{\text{0}}{\text{.19}}}}} \right){\text{ = P(Z5}}{\text{.27)}}}\]

\[{{\text{ = 1 - P(Z < 5}}{\text{.27)}}\mathop {\text{ = }}\limits^{{\text{(1)}}} {\text{1 - 1}}}\]

\[{{\text{ = 0}}}\]

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

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

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

a. Recalling the definition of \({{\rm{\sigma }}^{\rm{2}}}\) for a single \({\rm{rv X}}\), write a formula that would be appropriate for computing the variance of a function \({\rm{h(X,Y)}}\) of two random variables. (Hint: Remember that variance is just a special expected value.)

b. Use this formula to compute the variance of the recorded score\({\rm{h(X,Y)( = max(X,Y))}}\).

Carry out a simulation experiment using a statistical computer package or other software to study the sampling distribution of \({\rm{\bar X}}\) when the population distribution is Weibull with \({\rm{\alpha = 2}}\) and\({\rm{\beta = 5}}\), as in Example\({\rm{5}}{\rm{.20}}\).[A1] Consider the four sample sizes, and\({\rm{30}}\), and in each case use \({\rm{1000}}\) replications. For which of these sample sizes does the \({\rm{\bar X}}\) sampling distribution appear to be approximately normal?

A company maintains three offices in a certain region, each staffed by two employees. Information concerning yearly salaries (\({\rm{1000}}\)s of dollars) is as follows:

\(\begin{array}{*{20}{c}}{{\rm{ Office }}}&{\rm{1}}&{\rm{1}}&{\rm{2}}&{\rm{2}}&{\rm{3}}&{\rm{3}}\\{{\rm{ Employee }}}&{\rm{1}}&{\rm{2}}&{\rm{3}}&{\rm{4}}&{\rm{5}}&{\rm{6}}\\{{\rm{ Salary }}}&{{\rm{29}}{\rm{.7}}}&{{\rm{33}}{\rm{.6}}}&{{\rm{30}}{\rm{.2}}}&{{\rm{33}}{\rm{.6}}}&{{\rm{25}}{\rm{.8}}}&{{\rm{29}}{\rm{.7}}}\end{array}\)

a. Suppose two of these employees are randomly selected from among the six (without replacement). Determine the sampling distribution of the sample mean salary\({\rm{\bar X}}\).

b. Suppose one of the three offices is randomly selected. Let\({{\rm{X}}_{\rm{1}}}\)and\({{\rm{X}}_{\rm{2}}}\)denote the salaries of the two employees. Determine the sampling distribution of\({\rm{\bar X}}\).

c. How does \({\rm{E(\bar X)}}\) from parts (a) and (b) compare to the population mean salary\({\rm{\mu }}\)?

A health-food store stocks two different brands of a certain type of grain. Let \(X = \)the amount (lb) of brand A on hand and \(Y = \)the amount of brand B on hand. Suppose the joint pdf of X and Y is

\(f(x,y) = \left\{ {\begin{array}{*{20}{c}}{kxy\;\;\;\;\;\;\;x \ge 0,\;y \ge 0,\;20 \le x + y \le 30}\\{0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;otherwise}\end{array}} \right\}\)

a. Draw the region of positive density and determine the value of k.

b. Are X and Y independent? Answer by first deriving the marginal pdf of each variable.

c. Compute \(P\left( {X + Y \le 25} \right)\).

d. What is the expected total amount of this grain on hand?

e. Compute \(Cov\left( {X, Y} \right)\)and\(Corr\left( {X, Y} \right)\).

f. What is the variance of the total amount of grain on hand?
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation

Decision Maths

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.