/*! 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} Q. 6.2 XA glass act In a process for ma... [FREE SOLUTION] | 魅影直播

魅影直播

XA glass act In a process for manufacturing glassware, glass stems are sealed by heating them in a flame. Let X be the temperature (in degrees Celsius) for a randomly chosen glass. The mean and standard deviation of X are X=550Cand X=5.7C.

a. Is tempe谐ature a discrete 芯谐 continuous 谐andom variable? Explain 褍芯u谐 answer.

b. The target temperature is 550C. What are the mean and standard deviation of the number of degrees off target, D=X-550?

c. A manager asks for results in degrees Fahrenheit. The conversion of X into degrees Fahrenheit is given by Y=95X+32Y=95X+32. What are the mean Yand the standard deviation Yof the temperature of the flame in the Fahrenheit scale?

Short Answer

Expert verified

(a)The temperature is a continuous random variable because it takes on decimal values i.e. 30.7C

(b)Mean of D,

D=X-550=X-550=550-550=0C

Standard deviation of D ,

D=X-550=X=5.7C

(c)The standard deviation is multiplied by 95.

Y=95X=95(5.7)=10.26F

Step by step solution

01

Part (a) Step 1: Given Information

X : temperature (in degrees Celsius) for a randomly chosen glass

For X :

Mean,

X=550C

Standard deviation,

X=5.7C

02

Part (a) Step 2: Simplification

Discrete data are restricted to define separate values.

For example,

Integers or counts.

Whereas,

Continuous data are not restricted to define separate values

For example,

Decimal, rational or real numbers.

In this case,

The temperature is a continuous random variable because it takes on decimal values i.e. 30.7C.

03

Part (b) Step 1: Given Information

X : temperature (in degrees Celsius) for a randomly chosen glass

For X:

Mean,

X=550C

Standard deviation,

X=5.7C

Number of degrees off target,

D=X-550

04

Part (b) Step 2: Simplification

Property mean:

aX+b=aX+b

Property standard deviation:

aX+b=|a|X

Now,

We have

D=X-550

Thus,

Mean of D,

D=X-550=X-550=550-550=0C

Standard deviation of D,

D=X-550=X=5.7C

05

Part (c) Step 1: Given Information

X : temperature (in degrees Celsius) for a randomly chosen glass

For X:

Mean,

X=550C

Standard deviation,

X=5.7C

Conversion of X nto degrees Fahrenheit:

Y=95X+32

Where,

Y: temperature in degrees Fahrenheit

06

Part (c) Step 2: Simplification

For temperature conversion (degree Celsius to degree Fahrenheit):

Y=95X+32

Then

Every data value in the distribution of Y is multiplied by the same constant 95and increased by the same constant 32

If every data value is added by the same constant, the center of the distribution is also increased by the same constant.

Also,

If every data value is multiplied by the same constant, the center of the distribution is also multiplied by the same constant.

We know that

The mean is the measure of the center.

Thus,

The mean is multiplied by 95and increased by 32 .

Y=95X+32=95(550)+32=990+32=1022F

If every data value is added by the same constant, the spread of the distribution is unaffected.

Also,

If every data value is multiplied by the same constant, the spread of the distribution is also multiplied by the same constant.

We know that

The standard deviation is the measure of the spread.

Thus,

The standard deviation is multiplied by 95.
Y=95X=95(5.7)=10.26F

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 魅影直播!

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

Working outExercise 10 described a large sample survey that asked a sample of people aged 19 to 25 years, 鈥淚n the past seven days, how many times did you go to an exercise or fitness center or work out?鈥 The response Y for a randomly selected survey respondent has the probability distribution shown here. From Exercise 10, E(Y)=1.03. Find the standard deviation of Y. Interpret this value.

Spoofing (4.2) To collect information such as passwords, online criminals use "spoofing" to direct Internet users to fraudulent websites. In one study of Internet fraud, students were warned about spoofing and then asked to log into their university account starting from the university's home page. In some cases, the log-in link led to the genuine dialog box. In others, the box looked genuine but, in fact, was linked to a different site that recorded the ID and password the student entered. The box that appeared for each student was determined at random. An alert student could detect the fraud by looking at the true Internet address displayed in the browser status bar, but most just entered their ID and password.

a. Is this an observational study or an experiment? Justify your answer.

b. What are the explanatory and response variables? Identify each variable as categorical or quantitative.

Cranky mower To start her old lawn mower, Rita has to pull a cord and hope for some luck. On any particular pull, the mower has a 20%chance of starting.

a. Find the probability that it takes her exactly 3 pulls to start the mower.

b. Find the probability that it takes her more than 6 pulls to start the mower.

Taking the train According to New Jersey Transit, the 8:00A.M. weekday train from Princeton to New York City has a 90%chance of arriving on time on a randomly selected day. Suppose this claim is true. Choose 6 days at random. Let localid="1654594369074" Y=the number of days on which the train arrives on time.

Each entry in a table of random digits like Table Dhas probability $$ of being a 0 , and the digits are independent of one another. Each line of Table D contains 40 random

digits. The mean and standard deviation of the number of 0 s in a randomly selected line will be approximately

a. mean =0.1, standard deviation =0.05.

b. mean =0.1, standard deviation =0.1.

c. mean =4, standard deviation =0.05.

d. mean =4, standard deviation =1.90.

e. mean =4, standard deviation =3.60.

See all solutions

Recommended explanations on Math Textbooks

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.