Use a standard normal distribution table to find the percentage of area above the standard normal curve that is specified.Above z Equal 1.49What is the percentage off the area above z equal 1.49?(Round to the nearest tenth as needed.) Using the browser's print will lead to an undesirable print-out. Use the Print item from the "Quesprint-outAIAAzA497zAzA2.80z00488224000498563694541.121.682.81014880042122162.25498574551.692.821.13 3710200822649858 2194551.142.931.7037303050122.27592221.153754561.712.84040167772.28602264571.163771.722.85051:3003020132.29612291.173794581.732.86060242.30622321.183814591.742.8749807028632.312361.193834601.752.88498060322.32642391.203852321.76 46149819592.8909Este0362.33.651.213871.77 4622.9049810 04066 2452341.223891.78 463498112910442.35672491.23 3914631.792.9249812048682521.243932.361.804642.9349813052.69 2552.371.253941.81 46549814 056.702581.26 3961.822.38466498152.95060712611.273981.832.39466160642.96 499722641.28 4002.401.84 467.17499068297.732671.294021.85468241IS0712.98499742701.304031.86469242190752.99499.752731.314051.87469200794993.00.762761.32 4071.8847021083773.012794991.33 4081.89 471087.78 2821.344991.903.0241047123091.792851.35 4121.914724993.03095SO2881.364131.92 4733.0449925099$1 2911.374151.934733.0549926103$22941.38 4161.944743.064992710683 2971.394181.954743.0749928 110$41.40 4191.964753.0849930030229 114851.41 4211.97 4763.0949986 3051.42 4221.984763.1049930 11831 122$7 3081.434241.994773.1149932 1263111.444252.00 4773.12 499SS89 31333 1291.45 4273.1349934 13390 316201 4782.022.03 4794783.141.46 4281.47 42949935137913193.1549936 141923211.48 4313.164992.04 4792.05 48037 144933241.494323.1749938 148943261.504332.06 4803.1849939 15295 329207 4813.19 4994015596 3321.51 4351.52 4361.53 4372.08 4813.20 49915997 3342.09 4823.2149916398 3371.54 4382.10 4823.2249916699 3391.55 439211 4833.234991701.003411.56 4412.124833.244991741.01 3441.57 4422.134833.254991771.023462144843.264991.03 3492.154843.275002.16 4851.58 4431.591.60 4451.611.62 4473.285001.04 3511.05 3531.06 35521747 18148 18418850 19251 19552 19953 2024853.29 5002.184853.305001.07 3581.63 4492.194863.315001.08 3601.64 4502.204863.325001.09 3621.65 451221 4873.335003644522.2248754 20555 2091.101.11 3671.661.67 4532.234874143444546184464884884894894894904904904904914914914914924924924924932.444932.454932.464932474932.48 4932.494942.504942.51 4942.524942.534942.54 4952.554952.564952.57 4952.58 4952.594952.604952.61 4962.624962.63 4962.64 4962.65 4962.66 4962.67 4962.684962.69 4962.70 4972.714972.72 4972.73 4972.74 4972.75 4972.76 4972.77 4972.784972.79 4972.43498498498

Answered: Image /qna-images/answer/334516a8-cddd-421e-bd8e-a4980dccd684.jpg

Use a standard normal distribution table to find the percentage of area above the standard normal curve that is specified. Above z Equal 1.49 What is the percentage off the area above z equal 1.49? (Round to the nearest tenth as needed

Use a standard normal distribution table to find the percentage of area above the standard normal curve that is specified. Above z Equal 1.49 What is the percentage off the area above z equal 1.49? (Round to the nearest tenth as needed

College Algebra

1st Edition

ISBN:9781938168383

Author:Jay Abramson

Publisher:Jay Abramson

Chapter9: Sequences, Probability And Counting Theory

Section9.7: Probability

Problem 2SE: What is a sample space?

See similar textbooks

Similar questions

What is a sample space?
The median number of part-time employees at fast-food restaurants in a particular city wasknown to be 18 last year. City officials think the use of part-time employees may be increasing.A sample of nine fast-food restaurants showed that seven restaurants were employing morethan 18 part-time employees, one restaurant was employing exactly 18 part-time employees,and one restaurant was employing fewer than 18 part-time employees. Can it be concludedthat the median number of part-time employees has increased? Test using a = .05.
A warehouse manager wants to know if there is an association between the shift worked and being on time for work. To investigate, he selects a random sample of 70 workers and classifies each one according to the shift they worked most recently and whether they were on time for work. He was unable to classify a substantial number of people as being on time, so he classified those individuals as unknown. The data are displayed in the table. The manager would like to know if these data provide convincing evidence of an association between the shift worked and being on time in the large population of all workers at this warehouse. The random and 10% conditions are met. Is the Large Counts condition met? Yes, the smallest expected count is 5, so all expected counts are at least 5. Yes, the smallest expected count is 8.54, so all expected counts are at least 5. No, the smallest expected count is 2.56, so the expected counts are not all at least 5. No, the smallest expected count is…
The median number of part-time employees at fast-food restaurants in a particular city wasknown to be 18 last year. City officials think the use of part-time employees may be increasing. A sample of nine fast-food restaurants showed that seven restaurants were employing more than 18 part-time employees, one restaurant was employing exactly 18 part-time employees, and one restaurant was employing fewer than 18 part-time employees. Can it be concluded that the median number of part-time employees has increased? Test using α = 0.05
A nationwide job recruiting firm wants to compare the annual incomes for childcare workers in Texas and Indiana. Due to recent trends in the childcare industry, the firm suspects that the mean annual income of childcare workers in Texas is less than the mean annual income of childcare workers in Indiana. To see if this is true, the firm selected a random sample of 20 childcare workers from Texas and an independent random sample of 20 childcare workers from Indiana and asked them to report their mean annual income. The data obtained were as follows. Annual income in dollars Texas 37093 , 33960 , 35692 , 29528 , 30978 , 35504 , 26961 , 28380 , 37537 , 25341 , 34837 , 36489 , 33633 , 41229 , 43577 , 27644 , 43364 , 28415 , 28919 , 34096 Indiana 33672 , 46452 , 35900 , 28486 , 46528 , 34252 , 40952 , 42164 , 36298 , 38828 , 32690 , 29157 , 42770 , 35527 , 40804 ,…
A cell phone company knows that the mean length of calls for all of its customers in a certain city is 9.3 minutes. The company is thinking about offering a senior discount to attract new customers, but only wants to do so if the mean length of calls for current customers who are seniors (65 and over) is lower than it is for the general customer pool. The only way to identify seniors is to conduct a survey and ask people whether they are over age 65. Using this method, the company contacts a random sample of 100 seniors and records the length of their last call. The sample mean and standard deviation for the 100 calls are 8.2 minutes and 10 minutes, respectively. (a) Do you think the data collected on the 100 seniors are approximately bell-shaped? Explain. Bell-shaped data would range from about standard deviation(s) below the mean to standard deviation(s) above the mean. With a mean of 8.2 minutes and standard deviation of 10, the values in this range are possible…
Ashik is planning to propose to a girl he has a crush on in a grand way. He has decided to create an animation movie, conveying his love to his crush and he needs to estimate the total production time (labor-hours necessary) to produce the movie. Since he is so rich, he wants to use multiple big screens to project the movie and the mean production time for a random sample of 14 big-screen animated movie is found to be 53,500 labor-hours. Suppose that Ashik has managed to find out that the population standard deviation is 7462 labor-hours and the distribution of the production time is normal. 10 Construct a 98% confidence interval for the mean production time to produce a big-screen animated movie
A psychologist is interested in determining if the proportion of males who are diagnosed with depression during their first year away from home is different from the proportion of females diagnosed with depression their first year. She thinks that females are more emotionally unstable than males and feels like school programs need to target females more effectively. She conducts a study and randomly samples 300 males and 250 females. Of the males, 30 are diagnosed with depression, and of the females, 50 are diagnosed with depression. Is there enough evidence to support the researcher’s hypothesis?
I have the first two parts but can someone help find the claim at 0.01 significance level? Thanks!
As part of their efforts, the executives also surveyed managers across the country to determinestaffing shortages. Managers were encouraged, but not required, to submit the survey. From theresponses they received, 45 out of 52 managers said they were either “severely understaffed” or“understaffed.” Looking at the numbers, one of the executives is outraged. “How can we expect ourstores to be successful if almost 87% are understaffed?” Give a statistical response tothe executive to assure them that the staffing problem is probably not as bad as it seemsin the preliminary numbers.
the HR Office is trying to choose who will be retained among the employees. The portfolio of each was given to the HR Director. For each employee, the HR Director gathers information by trying to verify the years in service, expertise, qualifications, performance evaluation, etc. but some doubt about who among the employees are qualified always remains when a decision to retain or not is to be made. What are the type I and II decision error costs? Which decision error is more likely to be discovered by her boss? How does this affect the HR Director’s decisions on employee retention?
In a recent study; it was found that 62 percent of college students drink heavily and drive. Assuming all conditions are met, a police chief wanting to enact an emergency protocol for DWI/DUI wants to know if the claim is accurate. He took a sample of 70 college students who drank and drove and found that 46 of those students had a high blood alcohol level. Can we claim that more than 62 percent of all college students drink heavily and drive?