MATH302 WEEK4 HOMEWORK PRACTICE

docx

School

American Military University *

*We aren’t endorsed by this school

Course

302

Subject

Statistics

Date

Jan 9, 2024

Type

docx

Pages

Uploaded by carelessvicki

Question 1 1 / 1 point Find P(1.31 < Z < 2.15). Round answer to 4 decimal places. Answer: ___.0793___ Hide question 1 feedback In Excel, =NORM.S.DIST(2.15,TRUE)-NORM.S.DIST(1.31,TRUE) Question 2 1 / 1 point Find the area under the standard normal distribution to the left of z = -1.05. Round answer to 4 decimal places. Answer: ___.1469___ View question 2 feedback Question 3 1 / 1 point Find P(-1.96 ≤ Z ≤ 1.96). Round answer to 2 decimal places. Answer: ___.95___ Hide question 3 feedback In Excel, =NORM.S.DIST(1.96,TRUE)-NORM.S.DIST(-1.96,TRUE) Question 4 0 / 1 point Which type of distribution does the graph illustrate? Disribution graph Incorrect Response Poisson Distribution

Normal Distribution Correct Answer Uniform Distribution Right skewed Distribution Question 5 1 / 1 point A population has the distribution given by the following histogram with a mean of 0.6 and standard deviation of 0.2. population chart starts at 0.0 goes to 1.0 Select the expected distribution of mean for samples of size 45 selected from the above population. sample mean distribution graph Sample mean distribution graph Question 6 1 / 1 point Suppose that the longevity of a light bulb is exponential with a mean lifetime of eight years. Find the probability that a light bulb lasts less than one year.

0.1175 0.1859 9.6318 0.3034 0.3682 Hide question 6 feedback P(x < 1) In Excel, =EXPON.DIST(1,1/8,TRUE) Question 7 1 / 1 point The average lifetime of a certain new cell phone is three years. The manufacturer will replace any cell phone failing within two years of the date of purchase. The lifetime of these cell phones is known to follow an exponential distribution. The decay rate is:

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

0.1666 0.6666 0.3333 0.5000 Hide question 7 feedback 1/3 = .3333 Question 8 1 / 1 point The average lifetime of a set of tires is three years. The manufacturer will replace any set of tires failing within two years of the date of purchase. The lifetime of these tires is known to follow an exponential distribution. What is the probability that the tires will fail within two years of the date of purchase? 0.8647 0.2212 0.4866

0.9997 Hide question 8 feedback P(x < 2) In Excel, =EXPON.DIST(2,1/3,TRUE) Question 9 1 / 1 point The average lifetime of a certain new cell phone is three years. The manufacturer will replace any cell phone failing within two years of the date of purchase. The lifetime of these cell phones is known to follow an exponential distribution. What is the median lifetime of these phones (in years)? 5.5452 1.3863 0.1941 2.0794

Hide question 9 feedback Median Lifetime is the 50th percentile. Use .50 in the equation and the rate of decay is 1/3 Question 10 1 / 1 point Miles per gallon of a vehicle is a random variable with a uniform distribution from 25 to 35. The probability that a random vehicle gets between 28 and 33 miles per gallon is: Answer: (Round to two decimal place) ___.50___ Hide question 10 feedback Interval goes from 25 < x < 35 P(28 < x < 33) = Question 11 1 / 1 point A local grocery delivery time has a uniform distribution over 0 to 60 minutes. What is the probability that the grocery delivery time is more than 15 minutes on a given day? Answer: (Round to 2 decimal places.) ___.75___ Hide question 11 feedback Interval goes from 0 < x < 60 P(x > 15) = Question 12 1 / 1 point

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

The mail arrival time to a department has a uniform distribution over 0 to 60 minutes. What is the probability that the mail arrival time is more than 20 minutes on a given day? Answer: (Round to 2 decimal places.) ___.67___ Hide question 12 feedback Interval goes from 0 < x < 60 P( 20 > x) = Question 13 1 / 1 point The waiting time for a bus has a uniform distribution between 0 and 8 minutes. What is the 90th percentile of this distribution? (Recall: The 90th percentile divides the distribution into 2 parts so that 90% of area is to the left of 90th percentile) _______ minutes Answer: (Round answer to one decimal place.) ___7.2___ Hide question 13 feedback P(X < x) = .90 .90 = .90*8 = x 7.2 = x Question 14 1 / 1 point The waiting time for a train has a uniform distribution between 0 and 10 minutes. What is the probability that the waiting time for this train is more than 4 minutes on a given day? Answer: (Round to two decimal place.) ___.60___

Hide question 14 feedback Interval goes from 0 < x < 10 P(x > 4) = Question 15 1 / 1 point Miles per gallon of a vehicle is a random variable with a uniform distribution from 25 to 35. The probability that a random vehicle gets between 25 and 34 miles per gallon is: Answer: (Round to one decimal place) ___.9___ Hide question 15 feedback Interval goes from 25 < x < 35 P(25 < x < 34) = Question 16 1 / 1 point The average amount of water in randomly selected 16-ounce bottles of water is 16.1 ounces with a standard deviation of 0.5 ounces. If a random sample of thirty-six 16-ounce bottles of water are selected, what is the probability that the mean of this sample is less than 15.9 ounces of water? Answer: (round to 4 decimal places) ___.0082___ Hide question 16 feedback New SD = .5/SQRT(36) P(x < 15.9), in Excel =NORM.DIST(15.9,16.1,0.08333,TRUE)

Question 17 1 / 1 point The time a student sleeps per night has a distribution with mean 6.1 hours and a standard deviation of 0.6 hours. Find the probability that average sleeping time for a randomly selected sample of 36 students is more than 6 hours per night. Answer: (round to 4 decimal places) ___.8413___ Hide question 17 feedback New SD = .6/SQRT(36) P(x > 6), in Excel =1-NORM.DIST(6,6.1,0.1,TRUE) Question 18 0 / 1 point The final exam grade of a mathematics class has a skewed distribution with mean of 78 and standard deviation of 7.2. If a random sample of 38 students selected from this class, then what is the probability that the average final exam grade of this sample is between 75 and 80? Answer: (round to 4 decimal places) ___.9514___Incorrect Response(0.9515, .9515) Hide question 18 feedback New SD = 7.2/SQRT(38) P(75 < x < 80), in Excel =NORM.DIST(80,78,1.1679943,TRUE)-NORM.DIST(75,78,1.1679943,TRUE) Question 19 1 / 1 point The time a student sleeps per night has a distribution with mean 6.3 hours and a standard deviation of 0.6 hours. Find the probability that average sleeping time for a randomly selected sample of 42 students is more than 6.5 hours per night. Answer: (round to 4 decimal places) ___.0154___ Hide question 19 feedback

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

New SD = .6/SQRT(42) P(x > 6.5), in Excel =1-NORM.DIST(6.5,6.3,0.092582,TRUE) Question 20 1 / 1 point The average amount of a beverage in randomly selected 16-ounce beverage can is 15.8 ounces with a standard deviation of 0.5 ounces. If a random sample of thirty-six 16-ounce beverage cans are selected, what is the probability that the mean of this sample is less than 15.7 ounces of beverage? Answer: (round to 4 decimal places) ___.1151___ Hide question 20 feedback New SD =.5/SQRT(36) P(x <15.7), in Excel =NORM.DIST(15.7,15.8,0.08333,TRUE) 1 / 1 point Find P(1.31 < Z < 2.15). Round answer to 4 decimal places. Answer: ___.0793 ___ k RUE)-NORM.S.DIST(1.31,TRUE) 1 / 1 point Find the area under the standard normal distribution to the left of z = -1.05. Round answer to 4 decimal places. Answer: ___.1469 ___

k 1 / 1 point Find P(-1.96 ≤ Z ≤ 1.96). Round answer to 2 decimal places. Answer: ___.95 ___ k RUE)-NORM.S.DIST(-1.96,TRUE) 0 / 1 point Which type of distribution does the graph illustrate? Poisson Distribution Normal Distribution Uniform Distribution Right skewed Distribution Question 5 1 / 1 point A population has the distribution given by the following histogram with a mean of 0.6 and standard deviation of 0.2.

Select the expected distribution of mean for samples of size 45 selected from the above population.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Question 6 1 / 1 point Suppose that the longevity of a light bulb is exponential with a mean lifetime of eight years. Find the probability that a light bulb lasts less than one year. 0.1175 0.1859 9.6318 0.3034 0.3682 Hide question 6 feedback P(x < 1) In Excel, =EXPON.DIST(1,1/8,TRUE ) 1 / 1 point The average lifetime of a certain new cell phone is three years. The manufacturer will replace any cell phone failing within two years of the date of purchase. The lifetime of these cell phones is known to follow an exponential distribution. The decay rate is:

0.1666 0.6666 0.3333 0.5000 Hide question 7 feedback 1/3 = .3333 1 / 1 point The average lifetime of a set of tires is three years. The manufacturer will replace any set of tires failing within two years of the date of purchase. The lifetime of these tires is known to follow an exponential distribution. What is the probability that the tires will fail within two years of the date of purchase? 0.8647 0.2212 0.4866 0.9997 Hide question 8 feedback P(x < 2) In Excel, =EXPON.DIST(2,1/3,TRUE ) 1 / 1 point The average lifetime of a certain new cell phone is three years. The manufacturer will replace any cell phone failing within two years of the date of purchase. The lifetime of these

cell phones is known to follow an exponential distribution. What is the median lifetime of these phones (in years)? 5.5452 1.3863 0.1941 2.0794 Hide question 9 feedback Median Lifetime is the 50th percentile. Use .50 in the equation and the rate of decay is 1/3 − �� (1−.5)13 1 / 1 point Miles per gallon of a vehicle is a random variable with a uniform distribution from 25 to 35. The probability that a random vehicle gets between 28 and 33 miles per gallon is: Answer: (Round to two decimal place) ___.50 ___ ck x < 35 ) ∗ 1(35−25) 1 / 1 point A local grocery delivery time has a uniform distribution over 0 to 60 minutes. What is the probability that the grocery delivery time is more than 15 minutes on a given day? Answer: (Round to 2 decimal places.) ___.75 ___

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

ck < 60 0−0) 1 / 1 point The mail arrival time to a department has a uniform distribution over 0 to 60 minutes. What is the probability that the mail arrival time is more than 20 minutes on a given day? Answer: (Round to 2 decimal places.) ___.67 ___ ck < 60 60−0) 1 / 1 point The waiting time for a bus has a uniform distribution between 0 and 8 minutes. What is the 90th percentile of this distribution? (Recall: The 90th percentile divides the distribution into 2 parts so that 90% of area is to the left of 90th percentile) _______ minutes Answer: (Round answer to one decimal place.) ___7.2 ___ ck

1 / 1 point The waiting time for a train has a uniform distribution between 0 and 10 minutes. What is the probability that the waiting time for this train is more than 4 minutes on a given day? Answer: (Round to two decimal place.) ___.60 ___ ck < 10 0) 1 / 1 point Miles per gallon of a vehicle is a random variable with a uniform distribution from 25 to 35. The probability that a random vehicle gets between 25 and 34 miles per gallon is: Answer: (Round to one decimal place) ___.9 ___ ck x < 35 ) ∗ 1(35−25) 1 / 1 point The average amount of water in randomly selected 16-ounce bottles of water is 16.1 ounces with a standard deviation of 0.5 ounces. If a random sample of thirty-six 16-ounce bottles of water are selected, what is the probability that the mean of

this sample is less than 15.9 ounces of water? Answer: (round to 4 decimal places) ___.0082 ___ ck 1,0.08333,TRUE) 1 / 1 point The time a student sleeps per night has a distribution with mean 6.1 hours and a standard deviation of 0.6 hours. Find the probability that average sleeping time for a randomly selected sample of 36 students is more than 6 hours per night. Answer: (round to 4 decimal places) ___.8413 ___ ck 0.1,TRUE) 0 / 1 point The final exam grade of a mathematics class has a skewed distribution with mean of 78 and standard deviation of 7.2. If a random sample of 38 students selected from this class, then what is the probability that the average final exam grade of this sample is between 75 and 80? Answer: (round to 4 decimal places) ___.9514 ___ (0.9515, .9515)

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

ck ) 1679943,TRUE)-NORM.DIST(75,78,1.1679943,TRUE) 1 / 1 point The time a student sleeps per night has a distribution with mean 6.3 hours and a standard deviation of 0.6 hours. Find the probability that average sleeping time for a randomly selected sample of 42 students is more than 6.5 hours per night. Answer: (round to 4 decimal places) ___.0154 ___ ck 3,0.092582,TRUE) 1 / 1 point The average amount of a beverage in randomly selected 16- ounce beverage can is 15.8 ounces with a standard deviation of 0.5 ounces. If a random sample of thirty-six 16-ounce beverage cans are selected, what is the probability that the mean of this sample is less than 15.7 ounces of beverage? Answer: (round to 4 decimal places) ___.1151 ___ Hide question 20 feedback New SD =.5/SQRT(36) P(x <15.7), in Excel =NORM.DIST(15.7,15.8,0.08333,TRUE)

MATH302 WEEK4 HOMEWORK PRACTICE

Related Documents