MATLAB: An Introduction with Applications
6th Edition
ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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- Page 2 of 5 B. Organizing and Working with Data. 1. Find the z-score for the weight of a 155-pound woman. The mean weight of woman age 20 or older is 170.6 pounds with a standard deviation of 5.4 pounds. Round your answer to two decimal places.arrow_forwardYou want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately = 60.5. You would like to be 95% confident that your estimate is within 5 of the true population mean. How large of a sample size is required? Round up to nearest whole number. n= Question Help: Video Written Example Message instructor Submit Question Jump to Answer 80 F1 F2 F3 Q A N @ 2 W S #3 X E D $ F4 R or dº % 5 F5 T MacBook Air F6 6 Y & 7 A4 F7 U * 00 8 J DII F8 | ( 9 K F9 0 ) оarrow_forward❎❎❎❎✅✅✅✅✅✅✅arrow_forward
- Cereals sodium values have a mean of 167 and a standard deviation of 77.3. Find the z-score for the cereal that has a sodium value of 0. How would you interpret this z-score?arrow_forwardThe frequency distribution was obtained using a class width of 0.5 for data on cigarette tax rates. Use the frequency distribution to approximate the population mean and population standard deviation. Compare these results to the actual mean μ = $1.638 and standard deviation o = $1.052. Click the icon to view the frequency distribution for the tax rates. The population mean is $ (Round to three decimal places as needed.) book Get more help. C *** Frequency distribution of cigarette tax rates G Search or type URL Tax Rate 0.00-0.49 0.50-0.99 1.00-1.49 1.50-1.99 2.00-2.49 2.50-2.99 3.00-3.49 3.50-3.99 4.00-4.49 Print Frequency O 7 12 6 6 6 6 3 3 1 Done - X Clear all Check answerarrow_forwardPlease answer d and e.arrow_forward
- Most people think that the "normal" adult body temperature is 98.6degrees°F. In a more recent study, researchers reported that a more accurate figure may be 98.198.1degrees°F. Furthermore, the standard deviation appeared to be around 0.60.6degrees°F. Assume that a Normal model is appropriate. Complete parts a through c below. Question content area bottom Part 1 a) In what interval would you expect most people's body temperatures to be? Explain. Select the correct choice below and fill in the answer box(es) to complete your choice. A. Using the 68-95-99.7 Rule, about 95% of the body temperatures are expected to be less than enter your response heredegrees°F. (Round to one decimal place as needed.) B. Using the 68-95-99.7 Rule, about 95% of the body temperatures are expected to be between enter your response heredegrees°F and enter your response heredegrees°F. (Use ascending order. Round to one decimal place as needed.) C. Using the…arrow_forwardPlease help me solve thisarrow_forwardA normal population has mean u = 50 and standard deviation o = 5. What is the 80th percentile of the population? Use the TI-84 Plus calculator. Round the answer to at least one decimal place.arrow_forward
- You are asked to find some descriptive statistics for the heartrate data. Use the methods in the practice activity as a guide to help you find the mean, median, and standard deviation for the heartrate data. Mean: 74.1 Median: 74 Standard Deviation: 6.5 Which measure of central tendency is the most appropriate for the heartrate data?arrow_forwardAugusta National Golf Course. Earlier in this section, we found that the population standard deviation of the lengths of the holes at the Augusta National Golf Club is 132.0 yd. In this context, is the number 132.0 a parameter or a statistic? Explain your answer.arrow_forward
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