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# 1. determine the area under the standard normal curve that

1. determine the area under the standard normal curve that lies to the right of (a)Z=0.04,(b)Z=-0.39,(c)Z=1.76 and(d)Z=-1.39
Hi there,

A) 0.484
B) 0.652
C) 0.039
D) 0.918

Thanks!
Martin, Physicist
Category: Math Homework
Satisfied Customers: 781
Experience: 20+ years of research, engineering and teaching
Customer: replied 4 years ago.

"for marty" find the indicated probability of the standard normal random variable Z.

P(-2.07 < or equal to Z< -0.81) =

Customer: replied 4 years ago.

P(-2.07 < or equal to Z< -0.81) = 0.1897

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Customer: replied 4 years ago.

6. steel rods are manufactured with a mean length of 28 cm. because of variability in the manufacturing process, the lengths of the rods are approximatley normally distributed with a standard deviation of 0.06 cm

(a) what proportion of rods has a length less than 27.9 cm?

(round to four decimal places as needed)

(b) any rods that are shorter than 27.86 cm or lonnger than 28.48 m are discarded. what proportion of rods will be discarded?

(c)using the results of part (b) if 500 rods are manufactured in a day how many should the plant manager expet to discard?

(d) if an order comes in for 10,000 steel rods how many rods should the plant manager expect to manufacture if the order states that all rods must e between 27.9 cm and 28.1 cm

7. suppose the length of the pregnancies of animals are approxiately normal distributed with the mean = 211 days and standrad deviation= 11 days.

(a) what is the probability of randomly selected pregnancy lasts less than 207 days?

(b) what is the porobability that a random sample of 19 pregnancies has a man gestation period of 207 days or less?

(c) what is the porbabilty that a random sample of 49 pregnancies has a mean gestation period of 207 days or less?

9. find the Z score such that the area under the standard normal curve to the left is 0.91

6a.) 0.0478
6b.) 0.0098
6c.) approx 5 rods
6d.) approx 11,057 rods

7a.) 0.3581
7b.) 0.0565
7c.) 0.0055

Please Leave a POSITIVE FEEDBACK to Credit me. BONUS is welcome.Thanks.

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Customer: replied 4 years ago.

9. find the Z score such that the area under the standard normal curve to the left is 0.91

8. The shape of the didstribution of the time reguired to get an oil change at a 15-minute oil change facicility is unknown. however, records indicate that the mean time is 16.9 minutes, and thee standard deviation is 4.9 minnutes. (a) to compare probabilities regarding thr sample mean using the nnoral model, what ssize sample w9ould be required? (b) what is the probability that a random sample of n=40 oil changes results in a sample mean time less than 15 minutes?
(a) choose the requred sample size: a. any sample size could be used. b.the sample size needs to be less than 30. c. the normal model cannot be used if the shape of the distribution is unknown. d. the sample size needs to be greaterr than 30.
the probability is approximatley?

9.) z = 1.3408 or approx 1.34

8a.) option D (sample size should be greater than 30)
8b.) 0.0071

How many more questions do you have?

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Customer: replied 4 years ago.

15

ok please post all 15 questions at once
Customer: replied 4 years ago.

10. the mean incubation time for a type of fertilized egg kept at 100.3 degreesF IS 21 DAYs. suppose that the incubaton times are approximatley normally distributed with a standard deviation of 1 day.
a.what is the probability that a randomly selcted fertilized egg hatches in less than 20 days?
b.what is the probability that a randomly selcted fertilized egg takes over 22 days to hatch?
c.what is the probability that a randomly selcted fertilized egg hatches between 19 and 21 days?
d. would it be unusualfor ann egg to hatch in less than 18 days why?

11. a researcher wishes to estimate the average blood alchol concentration (BAC) for drivers involved in fatal accidents who are found to have positive BAC values. he randomly selects reecordsfrom 1100 suchh drivers and dtermind the smple mean BAC to be 0.17g/dl
(a) assuming that the populati0ion standard deviation for the BAC is 0.07 g/dl, construct an interperta 90% confidence interval for the mean BAC.
A

interpret a 90% confident inervalfor the mean bac. choose the correct answer. a) the researcher is 90% confident that the population mean bac is in the confidence interval for drivers involved in fatal accidents who have a positive bac value.
b) the researcher is10% confident that the population mean bac is in the confidence interval involved in fatal accidents who have a positive bac value. c) the researcher is 90% confident that the population mean bac is not in the confidence interval for drivers involved in fatal accidents ho have a positive bac value..
(B) all states of a countryh use a bac of 0.10%g/dl as the legal intoxication level . is it possible that the mean bac of all drivers involved in fatal accidents who are found to have a positive bac values is less than the legal intoxication level? yes or no
12) a poll of 547adults fro a certain countryh employed full or part time asked "how uch total time in minutes do you spend commuting to and from work in a typical day?" survey results indicate that x=45.7 minutes and s=30.9minutes.complete a and b. a0construct and interpret a 90% confidence interval for the mean commute time of adult from this country emloyed full or part time.the lower bound is --- minutes. the upper bound is ----minutes.
interpret a 90% confidence interval for the mean commte time of adults employed full or part time. choose correct answer..a) there is a 90% confidence that ythe mean commute time of adults from this country, employed full or part time lies betweeen the lower and upper bounds of the interval.b)there is a 90% confidence that the mean commute time iis greater than the upper bound of the interval. c) there is 90% confidence that the mean commute time is less than the lower bound of the interval.
(b) is it possible that the mmean coute time is less than 40 min? yes or no
is it likely? yes or no
13) people were polled on how many books they read the previous year. ho many subjects are needed to estimate the number of books read the previous year within one book with a 90% confidence? initial survey results indicate that standARD DIVIATION EQUALS 11.5 BOOKS. 905 CONFIDENCE LEVEL REQUIRES ? SUBJECTS.
14) A SIMPLE RANDOM SAMPLE OF SINE N=40 is drawn from a population. the sample mean is found to be x=120.5 and the sample standard deviation is found to be s=12.7 construct a 99% confidence interval for thepopulation mean. the lower bound is?
the upper bound is ?

15) a random variable x is norally distributed with mean =17 and standard diviation=5

A) compute Z1=14-17/5
b)Z2=20-17/5
C) THE AREA UNDER THE NORMAL CURVE BETWEEN X1=14 and x2=20 is 0.451. what is the area between z1 andz2?
(a) z1=
b)z2=
c)the area is?

i will send the rest in one messsage! thank you

1.)
A) 0.484
B) 0.652
C) 0.039
D) 0.918
2.) Answer: P(-2.07 < or equal to Z< -0.81) = 0.1897
6a.) 0.0478
6b.) 0.0098
6c.) approx 5 rods
6d.) approx 11,057 rods
7a.) 0.3581
7b.) 0.0565
7c.) 0.0055
9.) z = 1.3408 or approx 1.34
8a.) option D (sample size should be greater than 30)
8b.) 0.0071
10a.) 0.1587
10b.) 0.1587
10c.) 0.4772
10d.) Yes since the probability of less than 18 days is 0.0013
11a.) 90% CI for population mean = (0.1665, 0.1735) Option A
11b.) No since the probability that the mean of 1100 such drivers is almost zero
12a.) 90% CI for population mean = (43.53, 47.87)
12b.) Is it possible that the mean is less than 40 No since P(z no since the probability is so small and almost zero
13.) 358
14.) 99% CI for population mean = (115.33, 125.67) lower bound is 115.33 and upper bound is 125.67
15a.) z1 = -0.6
15b.) z2 = 0.6 15c.) area = 0.4515
Bonus will be highly appreciated :)
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BONUS is welcome.Thanks. If you need my help asap, just write in the first part of your question: "For Mr Glenn G. only!" Thanks again
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Customer: replied 4 years ago.

1. the mean incubation time for a type of fertilized egg kept at 100.5 degreesF IS 20 DAYs. suppose that the incubaton times are approximatley normally distributed with a standard deviation of 1 day.
a.what is the probability that a randomly selcted fertilized egg hatches in less than 19 days?
b.what is the probability that a randomly selcted fertilized egg takes over 21 days to hatch?
c.what is the probability that a randomly selcted fertilized egg hatches between 18 and 20 days?
d. would it be unusualfor ann egg to hatch in less than 17.5 days why?

2.
find the indicated probability of the standard normal random variable Z.

P(-2.01 < or equal to Z< -1.09) =

3. find the Z score such that the area under the standard normal curve to the left is 0.95

4. A) compute Z1=19-20/3
b)Z2=21-20/3
C) THE AREA UNDER THE NORMAL CURVE BETWEEN X1=14 and x2=20 is 0.451. what is the area between z1 andz2?
(a) z1=
b)z2=
c)the area is?

6. people were polled on how many books they read the previous year. ho many subjects are needed to estimate the number of books read the previous year within one book with a 90% confidence? initial survey results indicate that standARD DIVIATION EQUALS 14.3 BOOKS. 90% CONFIDENCE LEVEL REQUIRES how many SUBJECTs

7. a researcher wishes to estimate the average blood alchol concentration (BAC) for drivers involved in fatal accidents who are found to have positive BAC values. he randomly selects reecordsfrom 1400 suchh drivers and dtermind the smple mean BAC to be 0.16g/dl
(a) assuming that the populati0ion standard deviation for the BAC is 0.06 g/dl, construct an interperta 90% confidence interval for the mean BAC.
A

interpret a 90% confident inervalfor the mean bac. choose the correct answer. b. the researcher is 90% confident that the population mean bac is in the confidence interval for drivers involved in fatal accidents who have a positive bac value.
a. the researcher is 10% confident that the population mean bac is in the confidence interval involved in fatal accidents who have a positive bac value. c) the researcher is 90% confident that the population mean bac is not in the confidence interval for drivers involved in fatal accidents ho have a positive bac value..
(B) all states of a countryh use a bac of 0.10%g/dl as the legal intoxication level . is it possible that the mean bac of all drivers involved in fatal accidents who are found to have a positive bac values is less than the legal intoxication level? yes or no

8. steel rods are manufactured with a mean length of 29 cm. because of variability in the manufacturing process, the lengths of the rods are approximatley normally distributed with a standard deviation of 0.08 cm

(a) what proportion of rods has a length less than 28.9 cm?

(round to four decimal places as needed)

(b) any rods that are shorter than 28.83 cm or lonnger than 29.17 m are discarded. what proportion of rods will be discarded?

(c)using the results of part (b) if 5000 rods are manufactured in a day how many should the plant manager expet to discard?

(d) if an order comes in for 10,000 steel rods how many rods should the plant manager expect to manufacture if the order states that all rods must e between 28.9 cm and 29.1 cm

10. The shape of the didstribution of the time reguired to get an oil change at a 15-minute oil change facicility is unknown. however, records indicate that the mean time is 16.7 minutes, and thee standard deviation is 4.6 minnutes. (a) to compare probabilities regarding thr sample mean using the nnoral model, what ssize sample would be required? (b) what is the probability that a random sample of n=35
oil changes results in a sample mean time less than 15 minutes?
(a) choose the requred sample size: a. any sample size could be used. b.the sample size needs to be less than 30. c. the normal model cannot be used if the shape of the distribution is unknown. d. the sample size needs to be greaterr than 30.
the probability is approximatley?

10. suppose the length of the pregnancies of animals are approxiately normal distributed with the mean = 298 days and standrad deviation= 28 days.

(a) what is the probability of randomly selected pregnancy lasts less than 287 days?

(b) what is the porobability that a random sample of 18 pregnancies has a man gestation period of 287 days or less?

(c) what is the porbabilty that a random sample of 37 pregnancies has a mean gestation period of 287 days or less?

11. determine the area under the standard normal curve that lies to the right of (a)Z=0.83,(b)Z=-0.08,(c)Z=-1.88 and(d)Z=0.03

12. A SIMPLE RANDOM SAMPLE OF SIzE N=40 is drawn from a population. the sample mean is found to be x=120.5 and the sample standard deviation is found to be s=13.4 construct a 99% confidence interval for thepopulation mean. the lower bound is?
the upper bound is ?

14. a poll of 547adults fro a certain countryh employed full or part time asked "how uch total time in minutes do you spend commuting to and from work in a typical day?" survey results indicate that x=45.5 minutes and s=30.9minutes.complete a and b. a0construct and interpret a 90% confidence interval for the mean commute time of adult from this country emloyed full or part time.the lower bound is --- minutes. the upper bound is ----minutes.
interpret a 90% confidence interval for the mean commte time of adults employed full or part time. choose correct answer..a) there is a 90% confidence that ythe mean commute time of adults from this country, employed full or part time lies betweeen the lower and upper bounds of the interval.b)there is a 90% confidence that the mean commute time iis greater than the upper bound of the interval. c) there is 90% confidence that the mean commute time is less than the lower bound of the interval.
(b) is it possible that the mmean coute time is less than 40 min? yes or no
is it likely? yes or no

-sorry to send same problems with different nymbers but i had a time limit on my homework and missed it by 45 seconds. so it sht off, thank you. i also have one more set of 15 questions to send you but i cant until this is done.

please accept now answers since you have already posted so many questions and yet has not accept any answers, Note these are new set of questions since the given numbers were already changed and different from the previous questions

1a.) 0.1587
1b.) 0.1587
1c.) 0.4772
1d.) Yes since the probability of less than 17.5 days is 0.0062

2.
find the indicated probability of the standard normal random variable Z.

P(-2.01 < or equal to Z< -1.09) = 0.1156

3.) z = 1.6449

4a.) z1 = -0.3333
4b.) z2 = 0.3333
4c.) area = 0.2611

6.) 554

remaining answers for a few more mins

Customer: replied 4 years ago.
ok no problem i will do all the rates at the end :]
please accept now answers since you have already posted so many questions and yet has not accept any answers, Note these are new set of questions since the given numbers were already changed and different from the previous questions

1a.) 0.15871b.) 0.15871c.) 0.47721d.) Yes since the probability of less than 17.5 days is 0.0062

2.find the indicated probability of the standard normal random variable Z.
P(-2.01 < or equal to Z< -1.09) = 0.1156

3.) z = 1.6449

4a.) z1 = -0.33334b.) z2 = 0.33334c.) area = 0.2611
6.) 554

7a.) 90% confidence interval for population mean = (0.1574, 0.1626)7b.) interpretation:
The researcher is 90% confident that the population mean bac is in the confidence interval for drivers involved in fatal accidents who have a positive bac value.
8a.) 0.10568b.) 0.03368c.) 1688d.) 12679

10a.) Option D (greater than 30)10b.) 0.0144
Note there are two question #10
10a.) 0.347210b.) 0.047810c.) 0.0084

11a.) 0.203311b.) 0.531911c.) 0.969911d.) 0.4880

12a.)lower bound = 115.04upper bound = 125.96

14a.)90% CI for population mean = (43.33, 47.67)lower bound = 43.33upper bound = 47.6714b.) Yes but the probability is almost zero, it is almost or nearly impossible
Is it likely --> No

Please Leave a POSITIVE FEEDBACK to Credit me. BONUS is welcome.Thanks.
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Trigeoscal, Masters Degree
Category: Math Homework
Satisfied Customers: 530
Experience: Mathematics (Pre Calculus and Calculus, Statistics, etc.)
Customer: replied 4 years ago.
im very confused why dont i see anything
Customer: replied 4 years ago.

• im very confused why dont i see anything

You still don't see anything? but in my laptop, I see everything I post :((
Please Leave a POSITIVE FEEDBACK to Credit me. BONUS is welcome.Thanks.

If you need my help asap, just write in the first part of your question: "For Mr Glenn G. only!" Thanks again

Thanks. Don't forget to accept my answers if you are satisfied only, so that I will be compensated in helping you, thanks again!
Can I ask for some bonus since I have already actually answered two sets of 15 questions each and it was so tiring

Did you see now the answers? just click the link in the last answer post (before this)

Customer: replied 4 years ago.
how do i give the bonus before im done? will i loose the website
Customer: replied 4 years ago.

1test the hypothesis using the classical approach and thee P-VALUE APPROACH
Ho:p=0.55 versus h1:p<0.55
n=150, x=78 a=0.10
(a) perform the test using the classical approach choose the correct answer
there is not enough info to test
do not reject the null hypothesis
reect the null hypothesis

3(b) PERFORM THE TEST USING THE P-VALUE APPROCH
P-VALUE=?
NOT ENOUGH TO TEST HYPOTHESE NOT REEECT NULL
REJECT NULL

TO TEEST Ho U=107 VERSUS h1 U DOES NOT EQUAL 107 A SIMPLE RANDOM SAMPLE OF SSSIZE N=35 IS OBTAINED.
A. DOES THE OPOULATION HAVE TO BE NORAY DISTRIBUTED TO TEST THIS HYPOTHESIS? WHY?
A. NO0 BECAUS THE TEST IS TWO TAILED.
BYES BECAUSE THE SAMPLE IS SRANDOM
C.YES BECASE N IS GRETHER THAN OR EQAL TO 30
D. NO BECAYSE N IS GREAATHER THAN OR EQUA TO 30

(B) F X=103.8 AND S=5.9 COMPUTE THE TEST STATOSIC. THE TEST STATSITIC IS To = round to neaessrt thoushanth.

4.to test ho; U=50 versus H1, u is less than 50 a random sample size n=22 is obtaoed from a population that is knownb tyo be norallyy distributed ith stand deviation=11
(A)if the sample mean s determond to be x =46.9 cokpute the test statistc round to two decmial places.
(b) if the researcher ecides to test ths hypothiese at the a=0.04 leve of signficane determinne the criticla value. round to 2 decial places use a comma to sperte nswers as needed.

5.the mean consumotion of fruit 3 yrs ago waas 98.8 pounds a dietion believes tha the fruit consumption has rsen ssince then.
\(a) determine the null and alternntive hypotheisis. which of the follwoing is correct
a. ho u=98.8 h1, u<98.8
b.ho u=98.8 h1, u=>98.8
c. ho u=98.8 h1= not equal to 98.8

(b)suppose saple data indicates thge the null hypoothesis should be rejected. state the conclusion of the rsearcher. which of the following is the conclusion than can be reached
A. there is sufficent evidence that the mean consmpution of fruit stayed the sae
b.there is not sufficent evidence that consumotion of fruit has rissen
c there is sufficent evidence that consumpton of fruit risen
d. there is nont suffiecent eveidence that c9onsmpution of fruit stayed the same.

(C) SUPPOSE IN FACT THE MEA CONSUMOTON IS 98.58 POUNDS DID THE RESEARCHER COMITT A TYPOE ONE OR TWO ERROR

8. IF THE CONSEQUENCE OF AKING A TYPE 1 ERROR IS SEVVERE WOULD YOU CHOSE THE LEVELL OF SIGNIFICANCE a, to equal 0.01, 0.05, or 0.10

please accept first the answer or the moderator might block you for not accepting despite sending so many questions

also after accepting the answer, give a high rating, and if you still have new set of questions, please post it in a new thread other than here

Customer: replied 4 years ago.
i dont understand

Just rate first the answers I posted (note give a high rating so that I will be compensated in helping you)

Then if you still have new set of questions, post them in a another post/thread just like what you did in posting the very first question here

Customer: replied 4 years ago.
i dont know what youo mean by accepting the answers.

then after rating my answer, please, if you still have new set of questions, post them in a separate thread, not in here since they will require another set of payments

I am now waiting for you to rate my answers (please give a high rating only)

thanks
Thank you for giving my answer a high rating

If you still have some new set of questions please post them in a separate or new post/thread and at the start of your question, please write for "Mr. Glenn G. only" so that I can help you the fastest way

thanks again :)

Please Leave a POSITIVE FEEDBACK to Credit me. BONUS is welcome.Thanks.

If you need my help asap, just write in the first part of your question: "For Mr Glenn G. only!" Thanks again

Thanks. Don't forget to accept my answers if you are satisfied only, so that I will be compensated in helping you, thanks again!
Do you still need help in answering these new set of questions of yours?

1test the hypothesis using the classical approach and thee P-VALUE APPROACH
Ho:p=0.55 versus h1:p<0.55
n=150, x=78 a=0.10
(a) perform the test using the classical approach choose the correct answer
there is not enough info to test
do not reject the null hypothesis
reect the null hypothesis
3(b) PERFORM THE TEST USING THE P-VALUE APPROCH
P-VALUE=?
NOT ENOUGH TO TEST HYPOTHESE NOT REEECT NULL
REJECT NULL
TO TEEST Ho U=107 VERSUS h1 U DOES NOT EQUAL 107 A SIMPLE RANDOM SAMPLE OF SSSIZE N=35 IS OBTAINED.
A. DOES THE OPOULATION HAVE TO BE NORAY DISTRIBUTED TO TEST THIS HYPOTHESIS? WHY?
A. NO0 BECAUS THE TEST IS TWO TAILED.
BYES BECAUSE THE SAMPLE IS SRANDOM
C.YES BECASE N IS GRETHER THAN OR EQAL TO 30
D. NO BECAYSE N IS GREAATHER THAN OR EQUA TO 30
(B) F X=103.8 AND S=5.9 COMPUTE THE TEST STATOSIC. THE TEST STATSITIC IS To = round to neaessrt thoushanth.
4.to test ho; U=50 versus H1, u is less than 50 a random sample size n=22 is obtaoed from a population that is knownb tyo be norallyy distributed ith stand deviation=11
(A)if the sample mean s determond to be x =46.9 cokpute the test statistc round to two decmial places.
(b) if the researcher ecides to test ths hypothiese at the a=0.04 leve of signficane determinne the criticla value. round to 2 decial places use a comma to sperte nswers as needed.
5.the mean consumotion of fruit 3 yrs ago waas 98.8 pounds a dietion believes tha the fruit consumption has rsen ssince then.
\(a) determine the null and alternntive hypotheisis. which of the follwoing is correct
a. ho u=98.8 h1, u98.8
c. ho u=98.8 h1= not equal to 98.8
(b)suppose saple data indicates thge the null hypoothesis should be rejected. state the conclusion of the rsearcher. which of the following is the conclusion than can be reached
A. there is sufficent evidence that the mean consmpution of fruit stayed the sae
b.there is not sufficent evidence that consumotion of fruit has rissen
c there is sufficent evidence that consumpton of fruit risen
d. there is nont suffiecent eveidence that c9onsmpution of fruit stayed the same.
(C) SUPPOSE IN FACT THE MEA CONSUMOTON IS 98.58 POUNDS DID THE RESEARCHER COMITT A TYPOE ONE OR TWO ERROR

8. IF THE CONSEQUENCE OF AKING A TYPE 1 ERROR IS SEVVERE WOULD YOU CHOSE THE LEVELL OF SIGNIFICANCE a, to equal 0.01, 0.05, or 0.10