66.5 160 67.5 110 68.5 50 69.5 20 70.5 5 71.5-72.5 1,000 Test the normality of the distribution.
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- Carly was collecting data for a science experiment. She collected the data below. She was wondering if the data has a normal distribution. Interpret whether the data has a normal distribution or not. Attach a copy of your histogram with your answer. 12151818131715 13141511181110 18231416111517 131818156 1115 Use the paperclip button below to attach files. * Student can enter max 3500 characters В I U Ω COST +1.71% arch F12 Prt Sc Inse F10 F11 F9Not everyone pays the same price for the same model of a car. The figure - 99.7%- - 95%- +68% illustrates a normal distribution for the prices paid for a particular model of a new car. The mean is $22,000 and the standard deviation is $2000. Use the 68-95-99.7 Rule to find what percentage of buyers paid between $18,000 and $22,000. 22 18 20 Price of a Model of a New Car (Thousands) 24 26 28 %. The percentage of buyers who paid between $18,000 and $22,000 is (Type an exact answer.) Number of Car BuyersThe average retirement age in America is 64 years old. Brianna thinks that yoga teachers owners retire at age that is younger than average. We want to test Brianna's claim at the 0.09 level of significance. The data below shows the results of a survey of yoga teachers about the age where they retired. Assume that the distribution of the population is normal. 53, 61, 57, 66, 64, 68, 60, 68, 52, 65, 49 Download CSV or cut and paste into Desmos . ) For this study, which type of test should we use? Select an answer i) The null and alternative hypotheses would be: Ho: ? Select an answer H: ?v Select an answerv ii) The test statistic is (Please round your answer to 2 decimal places.) ) The p-value - (Please round your answer to 3 decimal places.) iv) What is our conclusion regarding the null hypothesis? Explain. v) What is our final conclusion about Brianna's claim?
- Not everyone pays the same price for the same model of a car. The figure illustrates a normal distribution for the prices paid for a particular model of a new car. The mean is $ 18 comma 000$18,000 and the standard deviation is $ 2000 Use the 68 - 95-99.7 68-95-99.7 Rule to find the percentage of buyers who paid more than $ 20 comma 000$20,000.Need help pleaseUse a standard normal distribution table to find the percent of the total area under the standard normal curve between the following z-scores. z= - 1.3 and z = - 0.75 Click the icon to view the standard normal distribution table. The percent of the total area between z= - 1.3 and z = - 0.75 is %. (Round to the nearest integer.)
- Do students perform the same when they take an exam alone as when they take an exam in a classroom setting? Eight students were given two tests of equal difficulty. They took one test in a solitary room and they took the other in a room filled with other students. The results are shown below. Exam Scores Alone 75 81 78 80 || 74 || 84 84 89 Classroom 82 | 83 || 84 7784 91 93 99 Assume a Normal distribution. What can be concluded at the the a 0.10 level of significance level of significance? For this study, we should use Select an answer a. The null and alternative hypotheses would be: Ho: HA Select an answer Select an answer vSelect an answer V(please enter a decimal) H Select an answer Select an answer Y Select an answer V (Please enter a decimal) b. The test statistic 7v (please show your answer to 3 decimal places.) c. The p-value = (Please show your answer to 4 decimal places.) d. The p-value is 2 v a e. Based on this, we should Select an answer v the null hypothesis. f. Thus, the…Solve this question please and explain your answersA researcher wishes to test the effects of excerise on the ability to complete a basic skills test. He designs a pre-test and post-test to give to each participant. You believe that there was a difference in the scores. You believe the population of the differences is normally distributed, but you do not know the standard deviation. When calculating difference use Post-test minus Pre-test. pre-test post-test 63 71 97 94 66 66 41 32 96 97 41 45 79 81 98 106 76 70 63 70 42 47 86 94 Which of the following are the correct hypotheses? O Ho: Hd 2 0 HA: Ha 0(claim) O Ho: Hd = 0 HA: Hd + 0(claim) Given that a is 0.05 the critical value is 2.201 and -2.201 The test statistic is: (round to 3 places) The p-value is: (round to 3 places) The decision can be made to: O reject Ho O do not reject Ho The final conclusion is that: O There is enough evidence to reject the claim that there was a difference in the scores. O There is not enough evidence to reject the claim that there was a difference in…
- Find the area of the shaded region. The graph depicts the standard normal distribution with mean 0 and standard deviation 1. Click to view page 1 of the table. Click to view page 2 of the table. Standard normal distribution table (page 1) 7=065 .09 .08 .07 .06 .05 .04 .03 .02 .01 .00 The area of the shaded region is. Standard normal distribution table (page 2) -3.4 .0002 .0003 .0003 .0003 .0003 .0003 .0003 .0003 0003 .0003 3.4 - 3.3 - 3.2 - 3.1 - 3.0 - 2.9 .0003 .0004 .0004 .0004 .0004 .0004 .0004 .0005 .0005 .0005 -3.3 (Round to four decimal places as needed.) .0007 .0010 - 3.2 - 3.1 .0005 .0005 .0005 .0006 .0006 .0006 .0006 .0006 .0007 .0007 .0007 .0008 .0008 .0008 .0008 .0009 .0009 .0009 .0010 .0010 .0011 .0011 .0011 .0012 .0012 .0013 .0013 .0013 -3.0 .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 - 2.9 - 2.8 - 2.7 .0014 .0014 .0015 .0015 .0016 .0016 .0017 .0018 .0018 .0019 0.0 5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 5319 5359 0.0 .0026 .0035 -2.8 .0019 .0020 .0021 .0021 .0022 .0023…30 min in the university student's daily time devoted to sports in Turkey. average and 3 min. It is known to show normal distribution with standard deviation. How many of 8 million university students spend 1 hour a day for sports?z Scores LeBron James, one of the most successful basketball players of all time, has a height of 6 feet 8 inches, or 203 cm. Based on statistics from Data Set 1 “Body Data” in Appendix B, his height converts to the z score of 4.07. How many standard deviations is his height above the mean?