Question #1 A statics professor plans class so carefully that the lengths of her classes are uniformly distributed between 47.0 and 57.0 minutes. Find the probability that a given class period runs between 51.25 and 51.5 minutes.
Find the probability of selecting a class that runs between 51.25 and 51.5 minutes. Round to three decimals places. I don’t need to see the work only the answer.
Question #2 Find the area of shaded region. The graph depicts the standard distribution of bone density scores with a mean 0 and standard deviation 1. Curve shown has Z=-1.17 with shading to the right. The question is the area of the shaded region is _______ (Round to four decimal places).
Question #3 – Assume that a randomly selected subject is given a bone density test. Bone density test scores are normally distributed with a mean of 0 and a standard deviation of 1. Draw a graph and find P11, the 11th percentile. This is the bone density score separating the bottom11% form the top 89%. The question is The density score corresponding to P11 is _____ they show four graphs, A is shaded from left to right to p11(0 to 3/4 of area, or B from left to P11 shaded from left to (10%) shaded. C Shaded from 1/4 to the entire right, and D at 3/4 to remainder of curve to the right.
Question #4 – Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. The curve shown is shade at middle with 104 with shade going to right to 130, curve continues unshaded. The question is the Area of the shaded region is _____ Round to four places.
Question#5 – Assume tha tadults have IQ scores that are normally disctrbuted wiht a meand of U = 105 and a standard deviation of O = 20. Find the probability that a randomly selected adult has an IQ of less than 133. The question is The probability that a randomly adult has IQ less than 133 is _____ Round to four.
Question #7 – Assume that women’s heights are normally disctributed with a meand given U = 62.6 in, and standard deviation O = 1.9 in.A – If 1 women is randomly selected, find the probability tha her height is less than 63 in.
B – If 47 women are randomly selected, find the probability that they have a mean height less than 63 in.Question – A = The probability is approximately _____ Round to four.
Question #10 – use the sample data and confidence level given below to complete parts A through D.
A research institute poll asked respondents if they acted to annoy a bad driver. In the poll, n=2442 and x= 955 who said that they honked, Use a 95% confience level.A – Find the best point estimate of the population proportion p = ______ Round to three places.
B – Identify the value of the margin of error E = ____ round to four places.
C- Construct the Confidence interval _____ < p < _____ round to three places.
D- Write a statement that correctly interprets the confidence interval. Choose one. I’ll type the selection in a second.
For D –
1 – One has 95% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.
2. – 95% of sample proportions will fail between the lower bound and the upper bound.
3. One has 95% confidence that the sample proportion is equal to the population proportion.
4. There is a 95% chance that the turve valvue of the population will fail between the lower bound and the upper bound.
Question #11 – During a period of 11 years 784 of the people selected for grand jury were sampled, and 68% of them were immigrants. Use the sample dated to construct a 99% confidence interval estimate of the proportion of grand jury members who were immigrants. Given the amount the people eligible for jury duty, 69.5% of them were immigrants, does it appear that the jury selection process was somehow biased against immigrants?
____ < p < _____ Round to three.Does it appear that the jury selection process was somehow biased against immigrants?
1. No, the confidence interval includes the true percentage of immigrants.
2. Yes, the confidence interval doesn not include the true percentage of immigrants.
Question #12 – Many states are carefully considering steps that would help them collect sales taxes on items purchased through the internet. How many randomly selected sales transactions must be surveyed to determine the percentage that transpired over the internet? Assume that we want to be 90% confident that the sample percentage is within two percentage points of the true population percentage for all sales transactions. Question is n= _____ Round up to the nearest integer.
Question #14 – To construct a confidence interval using the given confidence level, do whichever of the following is appropriate. (a) Find the Critical value Za/2. (b) find the ciritcal value ta/2, or (c) state that neither the normal nor the t distribution applies.90%; n=100, o= 17.0; population appears to be skewed. Answers are A to E
A – ta/2 = 1.984
B- za/2 = 1.96
C – Za/2 = 1.645
D- tx/2 = 1.660
E. Neither the normal nor the t distribution applies.
Question #15- Twelve different video games showing substance use were observed and the duration times of game play (in seconds) are listed below. The design of the study justifies the assumption, that the sample can be treated as a simple random sample. Use the data to construct a 99% confidence interval estimate of u. the mean duration of game play.4051, 3883, 3853,4031, 4316, 4816, 4650, 4042, 4995, 4824, 4328, 4324.Question is: What is the confidence interval estimate of th population mean u?____ < u < ____ Round to one place as needed.
Question # 16 – In order to estimate the mean amount of time computer users spend on the internet each month, how many computer, users, must be surveyed in order to be 95% confident that your sample mean is within 10 minutes of the population mean? Assume that the standard deviation of the population of month time spent on the internet is 214 minutes. What is a major obstacle to getting a good estimate of the population mean? Use technology to find the estimate minimum requires sample size.The minimum sample size required ____ computer uses. Round up to the nearest whole number.What is a major obstacle to getting a good estimate of the population mean?
1. The data does not provide information what the computer users did while on the internet.
2. It is difficult to precisely measure the amount of time spent on the internet, invalidate some data values.
3. There may not be 1,760 computer users to survey.
4. There are no obstacles to getting a good estimate of the population mean.
Question 17 – Randomly selected students participated in an experiment to test their ability to determine when one minute (or sixty seconds) has passed. Forty students yielded a sample mean of 60.3 seconds, Assuming the o=11.6 seconds, construct a interpret a 99% confidence interval estimate of the population mean of all students. Question What is the 99% confidence interval for the population mean u?____ < u < ____ Round to one place as needed. Based on the results, is it likely that the student’s estimates have a mean that is reasonable to sixty seconds? I’ll type the sections is an sec. Question #18 For the given claim, complete parts (a) and (b).Claim: At least 26% of internet users pay bills online. A recent survey of 385 users indicated that 22% pay their bills online.A = u, o, or p block =, does not =, >or=, or < or =, block 26, 0.22, 22, 0.26B = Identify the null and the alternative hypotheses. Ho = u, o, p block, =, > , does not equal, or < block, 26, 0.22, 0.26 block H1 = u, o, p block, =, >, <, does not equal block, 0.26, 22, 26, 0.22 block
Question #19 – The claim is that the proportion of adults who smoked a cigarette in the past week is less than 0.35, and the sample statistics include n=1391 subjects with 501 saying that they smoked a cigarette in the past week. Find the value of the test statistic. The test statistic is _____ Round to two places.
Question 20- assume that the significance level is a=0.1. Use the given information to find the P-value and the critical value(s). The test statistic of z= -1.77 is obtained when testing the claim that p < 0.3Question is: What is the P-valve?
P-Value = _____ Found to four places as needed. The critical value(s) is / are: ____ Round to two places ans needed, use a coma to separate answers as needed.
Question #22 –
In a Harris poll, adults were asked if they are in favor of abolishing the penny, Amoung the responces, 1281 answered NO, 487 answered Yes, and 392 had no opinion. What is the sample proportion of yes responses, and what is used to represent it?
Choose the correct answer:
A = p with arrow on top = 0.275. the symple p with arrow on top is used to represent a sample proportion.
B = p = 0.275. the symbol p is used to represent a sample proportion.
C = p = 0.225 The symbole p is used to represent a sample proportion.
D = p with arrow on top = 0.225. The symbole p with arrow on top is used to represent a sample proportion.
Question 23 –
A certain drug is used to treat asthma. In a clinical trial of the drug, 25 of 284 treated subjects experienced headaches (based on data from the manufacturer). The accompany calculator display shows results from a test on the claim that less than 8% of treated subjects experienced headaches. Use the normal distribution, as an approximation of the binomial distribution and assuming a 0.01 significance level to complete parts A to E.Window of data shows:
p with arrow on top = 0.0880281690
n= 284Question A – Is the test two-tailed, left tailed, or right tailed?
_ Two TailedB. What is the test statistics?
z = _____ Round to two places.C. What is the P-value?
P-value = ______ Round to four places.I’ll type D and E in a sec.
d. What is the null hypothesis and what do you conclude about it?
1. Ho: p< 0.08 2. Ho: p>0.08
3. Ho: p does not equal 0.08
4. Ho: p=0.08Decide weather to reject the null hypothesis. Choose
1. Fail to reject the null hypothesis becuase the P-valve is less than or equal to the significance level, a.
2. Reject the null hypothesis because the P-value is greater than the significance level, a.
3. Reject the null hypothesis because the P-valve is less than or equal ot the significance leve, a.
4. Fail to reject the null hypothesis because the P-value is greater than the significace level, a.E. What is the final conclusion? I’ll type in a sec.
4. Fail to reject the null hypothesis because the P-value is greater than the significance level, a.
E. What is the final conclusion?
A – There is not sufficient evidence to warrant rejection of the claim that less than 8% of treated subjects experienced headaches.
B – There is sufficient evidence to warrant rejection of the claim that less than 8% of treated subjects experienced headaches.
C. There is not sufficient evidence to support the claim that less than 8% of treated subjects experienced headaches.
D. There is sufficient evidenct support the claim that less than 8% of teated subjects experienced headaches.