P Reference Li... V My Citation...© 5 Emerging...O Final Reflec...1) 7 ways to de...- Using Past L...Course HomeO TakMath 109 - 5th editionMichelle Frank &| 12/14/21.Standard normal distribution table (page 1)Standard Normal (z) Distribution: Cumulative Area from the LEFT.00.01.02.03.0405.0607.08.09-3.50 and.0001lower-3,50 andlower-3.4.0003.0003.0003.0003.0003.0003.0003.0003.0003.0002-3.4-3.3.0005.0005.0005.0004.0004.0004.0004.0004.0004.0003-3.3-3.2.0007.0007.00060006.0006.0006.0006.0005.0005.0005-3.2b. R-3.1.0010.0009.0009.0009.0008.0008.0008.0008.0007.0007-3.1-3.0.0013.0013.0013.00120012.0011.0011.0011c. Q.0010.0010-3.0-2.900190018.0018.0017.0016.0016.0015.0015.0014.0014-2.9-2.80026.0025.0024.0023.0023.0022.0021.0021.0020.0019-2.8-2.7.00350034.0033.0032.0031.0030.0029.0028.0027.0026-2.7-2.600470045.0044.0043.0041.0040.0039.0038.0037.0036-2.6-2.5.00620060.0069.0067.0065.0064.0062.0051.0049.0048-2.5-2.4.00820080.0078.0075.0073.0071.0069.0068.00660064-24-2.30107.0104.0102.0099.0096.0094.0091.0089.0087.0084-2.3-2.201390136.0132.0129.0125.01220119.0116.0113.0110-2.2-2.1.0179.0174.0170.0166.0162.0158.0154.0150.0146.0143-2.1PrintDoneNext The test statistic of z = 0.96 is obtained when testing the claim that p>0.8.a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed.b. Find the P-value.c. Using a significance level of a = 0.10, should we reject Ho or should we fail to reject H,?Click here to view page 1 of the standard normal distribution table.Click here to view page 2 of the standard normal distribution table.a. This is atestb. P-value =(Round to three decimal places as needed.)c. Choose the correct conclusion below.O A. Fail to reject Ho. There is not sufficient evidence to support the claim that p> 0.8.O B. Reject Ho. There is sufficient evidence to support the claim that p > 0.8.O C. Fail to reject Ho. There is sufficient evidence to support the claim that p>0.8.O D. Reject Ho. There is not sufficient evidence to support the claim that p>0.8.NextCHRG BATT PAIR

Answered: Math 109 - 5th edition Michelle Frank…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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please answer the question in the photo, charts also attacthed in second photo
In a survey of women in a certain country (ages 20 - 29), the mean height was 63.6 inches with a standard deviation of 2.73 inches. Answer the following questions about the specified normal distribution (a) What height represents the 95th percentile? (b) What height represents the first quartile? Click to view page 1 of the table. Click to view page 2 of the table. (a) The height that represents the 95th percentile is (Round to two decimal places as needed.) inches. Enter your answer in the answer box and then click Check Answer. Help Me Solve This View an Example Get More Help - Clear All MacBook Air 吕口 F3 esc F7 F1 F2 @ $ %
In a survey, 38% of the respondents stated that they talk to their pets on the telephone. A veterinarian believed this result to be too high, so she randomly selected 170 pet owners and discovered that 61 of them spoke to their pet on the telephone. Does the veterinarian have a right to be skeptical? Use the a = 0.1 level of significance. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). Because npo (1- Po) = 40.1 > 10, the sample size is less than 5% of the population size, and the sample is given to be random, the requirements for testing the hypothesis are satisfied. (Round to one decimal place as needed.) What are the null and alternative hypotheses? Ho: | versus H,: (Type integers or decimals. Do not round.) 1 Standard Normal Distribution Table (page 1) Area Standard Normal Distribution 0.00 0.01 0.02 0.03 0.04 0,05 0.06 0,07 0.08 0,09 -34 -33 -32 -31 -30 0.0003 0.0005 0.0007 0.0010 0.0013 0.0003…
Test the hypothesis using the P-value approach. Be sure to verify the requirements of the test Ho: p=0.5 versus H₁: p > 0.5 n = 250; x = 145; a = 0.1 Click here to view page 1 of the table Click here to view page 2 of the table Calculate the test statistic, Zo Zo (Round to two decimal places as needed.) = DE COOK
Assume the readings on thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. Find the probability that a randomly selected thermometer reads between -1.71 and -0.31 and draw a sketch of the region. Click to view page 1 of the table. Click to view page 2 of the table. Standard Normal Table (Page 1) Sketch the region. Choose the correct graph below. OA. -1.71 -0.31 The probability is (Round to four decimal places as needed.) OB. NEGATIVE Z Scores Standard Normal (z) Distribution: Cumulative Area from the LEFT .00 .01 02 .03 .04 .05 .06 .07 .08 .09 -1.71 -0.31 -3.50 and lower .0001 -3.4 0003 0003 -3.3 .0005 .0005 .0003 .0005 .0003 .0004 .0003 0004 .0003 .0003 .0003 .0003 .0002 .0004 .0004 .0004 .0004 .0003 -3.2 .0007 .0007 .0006 .0006 .0006 .0006 .0006 .0005 .0005 .0005 -3.1 .0010 .0009 .0009 .0009 .0008 .0008 .0008 .0008 .0007 .0007 -3.0 .0013 0013 .0013 0012 0012 .0011 .com .0011 0010 0010 -2.9 .0019 .0018 0018 .0017 0016 .0016 0015 .0015 0014…
Standard Normal Distribution Table (page 1) Standard Normal Distribution Table (page 2) Area Area Standard Normal Distribution Standard Normal Distribution 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0003 0.0005 0.0007 0.0003 0.0005 0.0007 0.0003 0.0005 0.0006 0.0003 0.0004 0.0006 0.0003 0.0004 0.0006 0.0003 0.0004 0.0006 0.0003 0.0004 0.0006 0.0003 0.0004 0.0005 0.0003 0.0004 0.0005 0.0007 0.0010 0.0002 0.0003 0.0005 0.0007 0.0010 0.5040 0.5438 0.5832 0.6217 0.6591 0.5080 0.5478 0.5871 0.6255 0.6628 0.5120 0.5517 0.5910 0.6293 0.6664 0.5199 0.5596 0.5987 0.6368 0.6736 0.5279 0.5675 0.6064 0.6443 0.6808 0.5319 0.5714 0.6103 0.5160 0.5557 0.5239 0.5636 0.6026 0.5359 0.5753 0.6141 -3.4 0.5000 0.5398 0.5793 0.0 -3.3 -3.2 0.1 0,2 0.5948 -3.1 -3.0 0.0010 0.0013 0.0009 0.0013 0.0009 0.0013 0.0009 0.0012 0.0008 0.0012 0.0008 0.0011 0.3 0.4 0.6179 0.6554 0.6480 0.6844 0.6517 0.6879 0.0008 0.0008 0.0011 0.6331 0.6406 0.6772 0.0011 0.6700…
A sports reporter conducts a study of the fans' overall satisfaction with the sporting event after the event is completed. He surveys 165 randomly selected fans on their way out of the event and asks them to rate their satisfaction with the event on a scale from 0 to 10. The average satisfaction rating is 6.9. It is known from previous studies of this type that the standard deviation in satisfaction level is 2.5. Calculate the margin of error and construct the 95% confidence interval for the true mean satisfaction level for the sporting event. Standard Normal Distribution Table ΕΞ Dound in th SAMSUNG SUBMIT QUESTION THES SAVE PROGRESS SUBMIT
Please help give me correct answer
About _________% of the area under the curve of the standard normal distribution is between z=−0.928z=-0.928 and z=0.928z=0.928 (or within 0.928 standard deviations of the mean).
Over the years, the mean customer satisfaction rating at a local restaurant has been 85. The restaurant was recently remodeled, and now the management claims the mean customer rating, u, is not equal to 85. In a sample of 42 customers chosen at random, the mean customer rating is 83.1. Assume that the population standard deviation of customer ratings is 5.3. Is there enough evidence to support the claim that the mean customer rating is different from 85? Perform a hypothesis test, using the 0.05 level of significance. (a) State the null hypothesis H, and the alternative hypothesis H,. H,: I OO H: 0 D=D0 (b) Perform a Z-test and find the p-value. Here is somne information to help you with your Z-test. • The value of the test statistic is given by

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