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2(4 - x)> 8 and 5-5x>4(3-x)
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SNHU Median Housing Price Prediction Model Project
Overview
Recall that samples are used to generate a statistic, which businesses use to estimate the population parameter. ...
SNHU Median Housing Price Prediction Model Project
Overview
Recall that samples are used to generate a statistic, which businesses use to estimate the population parameter. You have learned how to take samples from populations and use them to produce statistics. For two quantitative variables, businesses can use scatterplots and the correlation coefficient to explore a potential linear relationship. Furthermore, they can quantify the relationship in a regression equation.
Prompt
This assignment picks up where the Module Two assignment left off and will use components of that assignment as a foundation.
You have submitted your initial analysis to the sales team at D.M. Pan Real Estate Company. You will continue your analysis of the provided Real Estate Data spreadsheet using your selected region to complete your analysis. You may refer back to the initial report you developed in the Module Two Assignment Template to continue the work. This document and the National Statistics and Graphs spreadsheet will support your work on the assignment.
Note: In the report you prepare for the sales team, the dependent, or response, variable (y) should be the listing price and the independent, or predictor, variable (x) should be the square feet.
Using the Module Three Assignment Template, specifically address the following:
Regression Equation: Provide the regression equation for the line of best fit using the scatterplot from the Module Two assignment.
Determine r: Determine r and what it means. (What is the relationship between the variables?)
Determine the strength of the correlation (weak, moderate, or strong).
Discuss how you determine the direction of the association between the two variables.
Is there a positive or negative association?
What do you see as the direction of the correlation?
Examine the Slope and Intercepts: Examine the slopeb1b1 and intercept b0b0
Draw conclusions from the slope and intercept in the context of this problem.
Does the intercept make sense based on your observation of the line of best fit?
Determine the value of the land only.
Note: You can assume, when the square footage of the house is zero, that the price is the value of just the land. This happens when x=0, which is the y-intercept. Does this value make sense in context?
Determine the R-squared Coefficient: Determine the R-squared value.
Discuss what R-squared means in the context of this analysis.
Conclusions: Reflect on the Relationship: Reflect on the relationship between square feet and sales price by answering the following questions:
Is the square footage for homes in your selected region different than for homes overall in the United States?
For every 100 square feet, how much does the price go up (i.e., can you use slope to help identify price changes)?
What square footage range would the graph be best used for?
Probability and Statistics
1. Read Chapter 4, and then describe, in your own words, another method of finding the mean by using the sample space (lis ...
Probability and Statistics
1. Read Chapter 4, and then describe, in your own words, another method of finding the mean by using the sample space (list of possible values) and probabilities (the technique is in the book). 2. Create a list of seven, 2-digit numbers (with no duplicates) and another set of seven probabilities (with no duplicates). The probabilities must add to 1. Open R, and manually enter those numbers and their corresponding probabilities to calculate the mean using only addition and multiplication (in other words, enter only the numbers, the plus sign, and the * for multiplication, like on the bottom of Yakir, 2011, p. 57). Paste all the R output into your journal.3. Describe in your own words what your calculation is doing and what the answer means.
draw the conclusion for upcoming presidential election
Suppose that there are two (2) candidates (i.e., Jones and Johns) in the upcoming presidential election. Sara notes that ...
draw the conclusion for upcoming presidential election
Suppose that there are two (2) candidates (i.e., Jones and Johns) in the upcoming presidential election. Sara notes that she has discussed the presidential election candidates with 15 friends, and 10 said that they are voting for candidate Jones. Sara is therefore convinced that candidate Jones will win the election because Jones gets more than 50% of votes.Answer the following questions in the space provided below:Based on what you now know about statistical inference, is Sara’s conclusion a logical conclusion? Why or why not?How many friend samples Sara should have in order to draw the conclusion with 95% confidence interval? Why?How would you explain your conclusion to Sara without using any statistical jargon? Why?
Create visual displays of data
You will submit one Word document for this assignment. You will create this Word document by cutting and pasting SPSS outp ...
Create visual displays of data
You will submit one Word document for this assignment. You will create this Word document by cutting and pasting SPSS output into Word and naming the file johnsoncPSY7107-5.doc Begin by downloading the following SPSS data sets:Downloadfestival.savChickflick.savHiccups.savTextmessages.savExamAnxiety.savPart A. Creating Visual Displays of Data. For this assignment, you will copy and paste output you created while working in Chapter 4 into a Word document. Please read the instructions below to ensure you are pasting the correct material into your assignment (This chapter has you create many charts and not all are required for this assignment).Using the data set: DownloadFestival.sav, calculate the mean, standard deviation, and minimum and maximum values for the variable Day1. You should notice that there is an outlier for this variable and you should change the value of the outlier you have identified to 2.02. Next, you will create a boxplot for males and females for the Day1 variable. Be sure to save the data set with a new name, indicating it is the corrected data set (outlier identified and corrected). Save this boxplot, with an appropriate title in your assignment document. Using the data set: Chick-Flick.sav, create a simple bar chart for independent means. The variables you will use are: Arousal, Film, and Gender (grouping variable). Be sure to display error bars and save your chart with an appropriate title in your assignment document.Using the data set: Hiccups.sav, create a bar chart for related means. The variables you will use are: Baseline, Tongue Pulling, Carotid Artery Massage, Digital Rectal Massage. Be sure to display error bars, include labels for the X- and Y-axis, and save your chart with an appropriate title in your assignment document.Using the data set: Text Messages.sav (note: you may see an additional data set with the same name: TextMessages.sav – either will create the correct output), create a clustered bar chart for mixed designs. The variables you will use are: Time1, Time2, and Group. Be sure to display error bars, include labels for the X- and Y-axis, and save your chart with an appropriate title in your assignment document.Using the data set: Exam Anxiety.sav, create a scatterplot that includes a regression line. The variables you will use are: Exam Performance and Exam Anxiety. Be sure to include the regression line and save your chart with an appropriate title in your assignment document.Part B. Why Exploratory Data Analysis? Write a short paragraph that highlights your understanding of why exploratory data analysis is a critical part of any analytical strategy (300-500 words). You should support your answer to this question by citing and referencing the course materials or other scholarly resources. This answer is worth 5 points (half the assigned points for this assignment). To receive full credit you must show a high level of understanding the importance of exploring data visually. You should now have the following file:lastnamefirstinitialPSY7107-5.doc
21 pages
20171212003730sym 506 Bus Info 1
Refer to the Buena School District bus data. Select the variable referring c. Draw a cumulative frequency distribution. Fo ...
20171212003730sym 506 Bus Info 1
Refer to the Buena School District bus data. Select the variable referring c. Draw a cumulative frequency distribution. Forty percent of the buses
statistics problems
Statistics 10 questions Girls x P(x) Open in StatCrunch + ...
statistics problems
Statistics 10 questions Girls x P(x) Open in StatCrunch + Copy to Clipboard + Open in Excel + 0 0.0040.004 1 0.0210.021 2 0.1060.106 3 0.2040.204 4 0.3300.330 5 0.2040.204 6 0.1060.106 7 0.0210.021 8 0.0040.004 The accompanying table describes results from groups of 8 births from 8 different sets of parents. The random variable x represents the number of girls among 8 children. Complete parts (a) through (d) below.LOADING...Click the icon to view the table.a. Find the probability of getting exactly 1 girl in 8 births.nothing (Type an integer or a decimal. Do not round.)b. Find the probability of getting 1 or fewer girls in 8 births.nothing (Type an integer or a decimal. Do not round.)c. Which probability is relevant for determining whether 1 is a significantly low number of girls in 8 births: the result from part (a) or part (b)?A.Since the probability of getting 0 girls is less likely than getting 1 girl, the result from part (a) is the relevant probability.B.Since the probability of getting more than 1 girl is the complement of the result from part (b), this is the relevant probability.C.Since the probability of getting 1 girl is the result from part (a), this is the relevant probability.D.Since getting 0 girls is an even lower number of girls than getting 1 girl, the result from part (b) is the relevant probability.d. Is 1 a significantly low number of girls in 8 births? Why or why not? Use 0.05 as the threshold for a significant event.A.No, since the appropriate probability is greater than 0.05, it is not a significantly low number.B.No, since the appropriate probability is less than 0.05, it is not a significantly low number.C.Yes, since the appropriate probability is greater than 0.05, it is a significantly low number.D.Yes, since the appropriate probability is less than 0.05, it is a significantly low number.2. Assume that random guesses are made for 66 multiple-choice questions on a test with 55 choices for each question, so that there are n equals=66 trials, each with probability of success (correct) given by p equals=0.200.20. Find the probability of no correct answers.LOADING...The probability of no correct answers is nothing . (Round to three decimal places as needed.)3. A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 4747 tablets, then accept the whole batch if there is only one or none that doesn't meet the required specifications. If one shipment of 30003000 aspirin tablets actually has a 33% rate of defects, what is the probability that this whole shipment will be accepted? Will almost all such shipments be accepted, or will many be rejected? The probability that this whole shipment will be accepted is nothing .(Round to four decimal places as needed.), The company will acceptsnothing % of the shipments and will reject nothing % of the shipments, so (Round to two decimal places as needed.)4. For bone density scores that are normally distributed with a mean of 0 and a standard deviation of 1, find the percentage of scores that area. significantly high (or at least 2 standard deviations above the mean).b. significantly low (or at least 2 standard deviations below the mean).c. not significant (or less than 2 standard deviations away from the mean).a. The percentage of bone density scores that are significantly high is nothing %.(Round to two decimal places as needed.)b. The percentage of bone density scores that are significantly low is nothing %.(Round to two decimal places as needed.)c. The percentage of bone density scores that are not significant is nothing %.(Round to two decimal places as needed.)5. Assume that an adult female is randomly selected. Suppose females have pulse rates that are normally distributed with a mean of 74.074.0 beats per minute and a standard deviation of 12.512.5 beats per minute. Find the probability of a pulse rate between 6060 beats per minute and 7070 beats per minute. (Hint: Draw a graph.)The probability is nothing .(Round to four decimal places as needed.)6. An airliner carries 150150 passengers and has doors with a height of 7575 in. Heights of men are normally distributed with a mean of 69.069.0 in and a standard deviation of 2.82.8 in. Complete parts (a) through (d).a. If a male passenger is randomly selected, find the probability that he can fit through the doorway without bending.The probability is (Round to four decimal places as needed.) b. If half of the 150150 passengers are men, find the probability that the mean height of the 7575 men is less than 7575 in.The probability is (Round to four decimal places as needed.) c. When considering the comfort and safety of passengers, which result is more relevant: the probability from part (a) or the probability from part (b)? Why?A.The probability from part (b) is more relevant because it shows the proportion of male passengers that will not need to bend.B.The probability from part (a) is more relevant because it shows the proportion of flights where the mean height of the male passengers will be less than the door height.C.The probability from part (b) is more relevant because it shows the proportion of flights where the mean height of the male passengers will be less than the door height.D.The probability from part (a) is more relevant because it shows the proportion of male passengers that will not need to bend.d. When considering the comfort and safety of passengers, why are women ignored in this case?A.There is no adequate reason to ignore women. A separate statistical analysis should be carried out for the case of women.B.Since men are generally taller than women, a design that accommodates a suitable proportion of men will necessarily accommodate a greater proportion of women.C.Since men are generally taller than women, it is more difficult for them to bend when entering the aircraft. Therefore, it is more important that men not have to bend than it is important that women not have to bend.7.Aclinical trial tests a method designed to increase the probability of conceiving a girl. In the study 664664 babies were born, and 332332 of them were girls. Use the sample data to construct a 9999% confidence interval estimate of the percentage of girls born. Based on the result, does the method appear to be effective?nothing less than< p less than<nothing (Round to three decimal places as needed.)Does the method appear to be effective?No,the proportion of girls is notis not significantly different from 0.5.Yes, the proportion of girls is significantly different from 0.5.8. In the week before and the week after a holiday, there were 10 comma 00010,000 total deaths, and 49404940 of them occurred in the week before the holiday.a. Construct a 9090% confidence interval estimate of the proportion of deaths in the week before the holiday to the total deaths in the week before and the week after the holiday. b. Based on the result, does there appear to be any indication that people can temporarily postpone their death to survive the holiday?a. nothing less than<pless than<nothing (Round to three decimal places as needed.)b. Based on the result, does there appear to be any indication that people can temporarily postpone their death to survive the holiday?YesYes, because the proportion could notcould not easily equal 0.5. The interval isis substantially less than 0.5 the week before the holiday.NoNo, because the proportion couldcould easily equal 0.5. The interval is notis not less than 0.5 the week before the holiday.9. A data set includes 103103 body temperatures of healthy adult humans having a mean of 98.398.3degrees°F and a standard deviation of 0.730.73degrees°F. Construct a 9999% confidence interval estimate of the mean body temperature of all healthy humans. What does the sample suggest about the use of 98.6degrees°F as the mean body temperature?Click here to view a t distribution table. LOADING...Click here to view page 1 of the standard normal distribution table. LOADING...Click here to view page 2 of the standard normal distribution table. LOADING...What is the confidence interval estimate of the population mean muμ?nothing degrees°Fless than<muμless than<nothing degrees°F(Round to three decimal places as needed.)What does this suggest about the use of 98.6degrees°F as the mean body temperature?A.This suggests that the mean body temperature could be lower thanbe lower than 98.6degrees°F.B.This suggests that the mean body temperature could very possibly bevery possibly be 98.6degrees°F.C.This suggests that the mean body temperature could be higher thanbe higher than 98.6degrees°F.10. accompanying data set and construct a 9090% confidence interval estimate of the mean pulse rate of adult females; then do the same for adult males. Compare the results.LOADING...Click the icon to view the pulse rates for adult females and adult males.Construct a 9090% confidence interval of the mean pulse rate for adult females.nothing bpmless than<muμless than<nothing bpm(Round to one decimal place as needed.)Construct a 9090% confidence interval of the mean pulse rate for adult males.nothing bpmless than<muμless than<nothing bpm(Round to one decimal place as needed.)Compare the results.A.The confidence intervals do not overlap, so it appears that there is no significant difference in mean pulse rates between adult females and adult males.B.The confidence intervals overlap, so it appears that there is no significant difference in mean pulse rates between adult females and adult males.C.The confidence intervals do not overlap, so it appears that adult females have a significantly higher mean pulse rate than adult males.D.The confidence intervals overlap, so it appears that adult males have a significantly higher mean pulse rate than adult females. Please wait...
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Most Popular Content
SNHU Median Housing Price Prediction Model Project
Overview
Recall that samples are used to generate a statistic, which businesses use to estimate the population parameter. ...
SNHU Median Housing Price Prediction Model Project
Overview
Recall that samples are used to generate a statistic, which businesses use to estimate the population parameter. You have learned how to take samples from populations and use them to produce statistics. For two quantitative variables, businesses can use scatterplots and the correlation coefficient to explore a potential linear relationship. Furthermore, they can quantify the relationship in a regression equation.
Prompt
This assignment picks up where the Module Two assignment left off and will use components of that assignment as a foundation.
You have submitted your initial analysis to the sales team at D.M. Pan Real Estate Company. You will continue your analysis of the provided Real Estate Data spreadsheet using your selected region to complete your analysis. You may refer back to the initial report you developed in the Module Two Assignment Template to continue the work. This document and the National Statistics and Graphs spreadsheet will support your work on the assignment.
Note: In the report you prepare for the sales team, the dependent, or response, variable (y) should be the listing price and the independent, or predictor, variable (x) should be the square feet.
Using the Module Three Assignment Template, specifically address the following:
Regression Equation: Provide the regression equation for the line of best fit using the scatterplot from the Module Two assignment.
Determine r: Determine r and what it means. (What is the relationship between the variables?)
Determine the strength of the correlation (weak, moderate, or strong).
Discuss how you determine the direction of the association between the two variables.
Is there a positive or negative association?
What do you see as the direction of the correlation?
Examine the Slope and Intercepts: Examine the slopeb1b1 and intercept b0b0
Draw conclusions from the slope and intercept in the context of this problem.
Does the intercept make sense based on your observation of the line of best fit?
Determine the value of the land only.
Note: You can assume, when the square footage of the house is zero, that the price is the value of just the land. This happens when x=0, which is the y-intercept. Does this value make sense in context?
Determine the R-squared Coefficient: Determine the R-squared value.
Discuss what R-squared means in the context of this analysis.
Conclusions: Reflect on the Relationship: Reflect on the relationship between square feet and sales price by answering the following questions:
Is the square footage for homes in your selected region different than for homes overall in the United States?
For every 100 square feet, how much does the price go up (i.e., can you use slope to help identify price changes)?
What square footage range would the graph be best used for?
Probability and Statistics
1. Read Chapter 4, and then describe, in your own words, another method of finding the mean by using the sample space (lis ...
Probability and Statistics
1. Read Chapter 4, and then describe, in your own words, another method of finding the mean by using the sample space (list of possible values) and probabilities (the technique is in the book). 2. Create a list of seven, 2-digit numbers (with no duplicates) and another set of seven probabilities (with no duplicates). The probabilities must add to 1. Open R, and manually enter those numbers and their corresponding probabilities to calculate the mean using only addition and multiplication (in other words, enter only the numbers, the plus sign, and the * for multiplication, like on the bottom of Yakir, 2011, p. 57). Paste all the R output into your journal.3. Describe in your own words what your calculation is doing and what the answer means.
draw the conclusion for upcoming presidential election
Suppose that there are two (2) candidates (i.e., Jones and Johns) in the upcoming presidential election. Sara notes that ...
draw the conclusion for upcoming presidential election
Suppose that there are two (2) candidates (i.e., Jones and Johns) in the upcoming presidential election. Sara notes that she has discussed the presidential election candidates with 15 friends, and 10 said that they are voting for candidate Jones. Sara is therefore convinced that candidate Jones will win the election because Jones gets more than 50% of votes.Answer the following questions in the space provided below:Based on what you now know about statistical inference, is Sara’s conclusion a logical conclusion? Why or why not?How many friend samples Sara should have in order to draw the conclusion with 95% confidence interval? Why?How would you explain your conclusion to Sara without using any statistical jargon? Why?
Create visual displays of data
You will submit one Word document for this assignment. You will create this Word document by cutting and pasting SPSS outp ...
Create visual displays of data
You will submit one Word document for this assignment. You will create this Word document by cutting and pasting SPSS output into Word and naming the file johnsoncPSY7107-5.doc Begin by downloading the following SPSS data sets:Downloadfestival.savChickflick.savHiccups.savTextmessages.savExamAnxiety.savPart A. Creating Visual Displays of Data. For this assignment, you will copy and paste output you created while working in Chapter 4 into a Word document. Please read the instructions below to ensure you are pasting the correct material into your assignment (This chapter has you create many charts and not all are required for this assignment).Using the data set: DownloadFestival.sav, calculate the mean, standard deviation, and minimum and maximum values for the variable Day1. You should notice that there is an outlier for this variable and you should change the value of the outlier you have identified to 2.02. Next, you will create a boxplot for males and females for the Day1 variable. Be sure to save the data set with a new name, indicating it is the corrected data set (outlier identified and corrected). Save this boxplot, with an appropriate title in your assignment document. Using the data set: Chick-Flick.sav, create a simple bar chart for independent means. The variables you will use are: Arousal, Film, and Gender (grouping variable). Be sure to display error bars and save your chart with an appropriate title in your assignment document.Using the data set: Hiccups.sav, create a bar chart for related means. The variables you will use are: Baseline, Tongue Pulling, Carotid Artery Massage, Digital Rectal Massage. Be sure to display error bars, include labels for the X- and Y-axis, and save your chart with an appropriate title in your assignment document.Using the data set: Text Messages.sav (note: you may see an additional data set with the same name: TextMessages.sav – either will create the correct output), create a clustered bar chart for mixed designs. The variables you will use are: Time1, Time2, and Group. Be sure to display error bars, include labels for the X- and Y-axis, and save your chart with an appropriate title in your assignment document.Using the data set: Exam Anxiety.sav, create a scatterplot that includes a regression line. The variables you will use are: Exam Performance and Exam Anxiety. Be sure to include the regression line and save your chart with an appropriate title in your assignment document.Part B. Why Exploratory Data Analysis? Write a short paragraph that highlights your understanding of why exploratory data analysis is a critical part of any analytical strategy (300-500 words). You should support your answer to this question by citing and referencing the course materials or other scholarly resources. This answer is worth 5 points (half the assigned points for this assignment). To receive full credit you must show a high level of understanding the importance of exploring data visually. You should now have the following file:lastnamefirstinitialPSY7107-5.doc
21 pages
20171212003730sym 506 Bus Info 1
Refer to the Buena School District bus data. Select the variable referring c. Draw a cumulative frequency distribution. Fo ...
20171212003730sym 506 Bus Info 1
Refer to the Buena School District bus data. Select the variable referring c. Draw a cumulative frequency distribution. Forty percent of the buses
statistics problems
Statistics 10 questions Girls x P(x) Open in StatCrunch + ...
statistics problems
Statistics 10 questions Girls x P(x) Open in StatCrunch + Copy to Clipboard + Open in Excel + 0 0.0040.004 1 0.0210.021 2 0.1060.106 3 0.2040.204 4 0.3300.330 5 0.2040.204 6 0.1060.106 7 0.0210.021 8 0.0040.004 The accompanying table describes results from groups of 8 births from 8 different sets of parents. The random variable x represents the number of girls among 8 children. Complete parts (a) through (d) below.LOADING...Click the icon to view the table.a. Find the probability of getting exactly 1 girl in 8 births.nothing (Type an integer or a decimal. Do not round.)b. Find the probability of getting 1 or fewer girls in 8 births.nothing (Type an integer or a decimal. Do not round.)c. Which probability is relevant for determining whether 1 is a significantly low number of girls in 8 births: the result from part (a) or part (b)?A.Since the probability of getting 0 girls is less likely than getting 1 girl, the result from part (a) is the relevant probability.B.Since the probability of getting more than 1 girl is the complement of the result from part (b), this is the relevant probability.C.Since the probability of getting 1 girl is the result from part (a), this is the relevant probability.D.Since getting 0 girls is an even lower number of girls than getting 1 girl, the result from part (b) is the relevant probability.d. Is 1 a significantly low number of girls in 8 births? Why or why not? Use 0.05 as the threshold for a significant event.A.No, since the appropriate probability is greater than 0.05, it is not a significantly low number.B.No, since the appropriate probability is less than 0.05, it is not a significantly low number.C.Yes, since the appropriate probability is greater than 0.05, it is a significantly low number.D.Yes, since the appropriate probability is less than 0.05, it is a significantly low number.2. Assume that random guesses are made for 66 multiple-choice questions on a test with 55 choices for each question, so that there are n equals=66 trials, each with probability of success (correct) given by p equals=0.200.20. Find the probability of no correct answers.LOADING...The probability of no correct answers is nothing . (Round to three decimal places as needed.)3. A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 4747 tablets, then accept the whole batch if there is only one or none that doesn't meet the required specifications. If one shipment of 30003000 aspirin tablets actually has a 33% rate of defects, what is the probability that this whole shipment will be accepted? Will almost all such shipments be accepted, or will many be rejected? The probability that this whole shipment will be accepted is nothing .(Round to four decimal places as needed.), The company will acceptsnothing % of the shipments and will reject nothing % of the shipments, so (Round to two decimal places as needed.)4. For bone density scores that are normally distributed with a mean of 0 and a standard deviation of 1, find the percentage of scores that area. significantly high (or at least 2 standard deviations above the mean).b. significantly low (or at least 2 standard deviations below the mean).c. not significant (or less than 2 standard deviations away from the mean).a. The percentage of bone density scores that are significantly high is nothing %.(Round to two decimal places as needed.)b. The percentage of bone density scores that are significantly low is nothing %.(Round to two decimal places as needed.)c. The percentage of bone density scores that are not significant is nothing %.(Round to two decimal places as needed.)5. Assume that an adult female is randomly selected. Suppose females have pulse rates that are normally distributed with a mean of 74.074.0 beats per minute and a standard deviation of 12.512.5 beats per minute. Find the probability of a pulse rate between 6060 beats per minute and 7070 beats per minute. (Hint: Draw a graph.)The probability is nothing .(Round to four decimal places as needed.)6. An airliner carries 150150 passengers and has doors with a height of 7575 in. Heights of men are normally distributed with a mean of 69.069.0 in and a standard deviation of 2.82.8 in. Complete parts (a) through (d).a. If a male passenger is randomly selected, find the probability that he can fit through the doorway without bending.The probability is (Round to four decimal places as needed.) b. If half of the 150150 passengers are men, find the probability that the mean height of the 7575 men is less than 7575 in.The probability is (Round to four decimal places as needed.) c. When considering the comfort and safety of passengers, which result is more relevant: the probability from part (a) or the probability from part (b)? Why?A.The probability from part (b) is more relevant because it shows the proportion of male passengers that will not need to bend.B.The probability from part (a) is more relevant because it shows the proportion of flights where the mean height of the male passengers will be less than the door height.C.The probability from part (b) is more relevant because it shows the proportion of flights where the mean height of the male passengers will be less than the door height.D.The probability from part (a) is more relevant because it shows the proportion of male passengers that will not need to bend.d. When considering the comfort and safety of passengers, why are women ignored in this case?A.There is no adequate reason to ignore women. A separate statistical analysis should be carried out for the case of women.B.Since men are generally taller than women, a design that accommodates a suitable proportion of men will necessarily accommodate a greater proportion of women.C.Since men are generally taller than women, it is more difficult for them to bend when entering the aircraft. Therefore, it is more important that men not have to bend than it is important that women not have to bend.7.Aclinical trial tests a method designed to increase the probability of conceiving a girl. In the study 664664 babies were born, and 332332 of them were girls. Use the sample data to construct a 9999% confidence interval estimate of the percentage of girls born. Based on the result, does the method appear to be effective?nothing less than< p less than<nothing (Round to three decimal places as needed.)Does the method appear to be effective?No,the proportion of girls is notis not significantly different from 0.5.Yes, the proportion of girls is significantly different from 0.5.8. In the week before and the week after a holiday, there were 10 comma 00010,000 total deaths, and 49404940 of them occurred in the week before the holiday.a. Construct a 9090% confidence interval estimate of the proportion of deaths in the week before the holiday to the total deaths in the week before and the week after the holiday. b. Based on the result, does there appear to be any indication that people can temporarily postpone their death to survive the holiday?a. nothing less than<pless than<nothing (Round to three decimal places as needed.)b. Based on the result, does there appear to be any indication that people can temporarily postpone their death to survive the holiday?YesYes, because the proportion could notcould not easily equal 0.5. The interval isis substantially less than 0.5 the week before the holiday.NoNo, because the proportion couldcould easily equal 0.5. The interval is notis not less than 0.5 the week before the holiday.9. A data set includes 103103 body temperatures of healthy adult humans having a mean of 98.398.3degrees°F and a standard deviation of 0.730.73degrees°F. Construct a 9999% confidence interval estimate of the mean body temperature of all healthy humans. What does the sample suggest about the use of 98.6degrees°F as the mean body temperature?Click here to view a t distribution table. LOADING...Click here to view page 1 of the standard normal distribution table. LOADING...Click here to view page 2 of the standard normal distribution table. LOADING...What is the confidence interval estimate of the population mean muμ?nothing degrees°Fless than<muμless than<nothing degrees°F(Round to three decimal places as needed.)What does this suggest about the use of 98.6degrees°F as the mean body temperature?A.This suggests that the mean body temperature could be lower thanbe lower than 98.6degrees°F.B.This suggests that the mean body temperature could very possibly bevery possibly be 98.6degrees°F.C.This suggests that the mean body temperature could be higher thanbe higher than 98.6degrees°F.10. accompanying data set and construct a 9090% confidence interval estimate of the mean pulse rate of adult females; then do the same for adult males. Compare the results.LOADING...Click the icon to view the pulse rates for adult females and adult males.Construct a 9090% confidence interval of the mean pulse rate for adult females.nothing bpmless than<muμless than<nothing bpm(Round to one decimal place as needed.)Construct a 9090% confidence interval of the mean pulse rate for adult males.nothing bpmless than<muμless than<nothing bpm(Round to one decimal place as needed.)Compare the results.A.The confidence intervals do not overlap, so it appears that there is no significant difference in mean pulse rates between adult females and adult males.B.The confidence intervals overlap, so it appears that there is no significant difference in mean pulse rates between adult females and adult males.C.The confidence intervals do not overlap, so it appears that adult females have a significantly higher mean pulse rate than adult males.D.The confidence intervals overlap, so it appears that adult males have a significantly higher mean pulse rate than adult females. 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