O a. 443.66 < µ 5 527.94. Data appears to be normally distributed with only one outlier959080706050200300400500600700800RainfllO b. NONEO. 443.52 < µ < 528.08. Data appears to be normally distributed with few outliers600700500Rainfall200300400800O d. 443.52 < µ 5 528.08. The data appear to be normally distributed9580200300400500600700800Rainfll..* * **** PAGASA provided the mean annual rainfall (in millimeters) in Cagayan Province from years 2000-2015. Rainfall measurements are as follows:499.2, 555.2, 398.8, 391.9, 453.4, 459.8, 483.7, 417.6, 469.2,452.4, 499.3, 340.6, 522.8, 469.9, 527.2, 565.5, 584.1, 727.3,558.6, 338.6Construct a 95% confidence interval for the mean annual rainfall. Does the data appear normallydistributed? Include your solution.

Answered: a. 443.66 <µ S 527.94. Data appears to…

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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Answer A.
An article contained the following observations on degree of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range: 416 421 422 423 426 428 430 434 437 439 446 447 449 453 456 462 464 Comment on any interesting features. (Select all that apply.) ___The data appears to be centered near 438. ___There are no outliers. ___There is little or no skew. ___There is one outlier. ___The data is strongly skewed. ___The data appears to be centered near 428 Calculate a two-sided 95% confidence interval for true average degree of polymerization. (Round your answers to two decimal places.) (____, ____)
The following data represent the concentration of dissolved organic carbon (mg/L) collected from 20 samples of organic soil. Assume that the population is normally distributed. Complete parts (a) through (c) on the right. 16.87 29.80 8.81 27.10 16.51 22,49 11.90 30.91 20.46 14.90 14.86 33.67 19.80 5.20 15.35 14.86 9.72 8.09 7.40 (a) Find the sample mean. 18.30 The sample mean is 17.35. (Round to two decimal places as needed.) (b) Find the sample standard deviation. The sample standard deviation is 8.14. (Round to two decimal places as needed.) (c) Construct a 98% confidence interval for the population mean u. The 98% confidence interval for the population mean u is ( ,). (Round to two decimal places as needed.)
The following data represent the concentration of dissolved organic carbon (mg/L) collected from 20 samples of organic soil. Assume that the population is normally distributed. Complete parts (a) through (c) on the right. 11.40 29.80 27.10 16.51 20.46 8.81 30.91 19.80 20.46 14.86 14.86 20.46 16.87 33.67 9.72 18.30 14.90 15.35 20.46 8.09 (a) Find the sample mean. The sample mean is (Round to two decimal places as needed.)
The following data represent the concentration of dissolved organic carbon (mg/L) collected from 20 samples of organic soil. Assume that the population is normally distributed. Complete parts (a) through (c) on the right. 16.87 29.80 27.10 16.51 11.90 8.81 22.49 20.46 14.90 33.67 30.91 14.86 5.20 15.35 9.72 19.80 14.86 8.09 7.40 18.30 (a) Find the sample mean. The sample mean is 17.35. (Round to two decimal places as needed.) (b) Find the sample standard deviation. The sample standard deviation is (Round to two decimal places as needed.)
The following data represent the concentration of dissolved organic carbon (mg/L) collected from 20 samples of organic soil. Assume that the population is normally distributed. Complete parts (a) through (c) on the right.
The following data represent the concentration of dissolved organic carbon (mg/L) collected from 20 samples of organic soil. Assume that the population is normally distributed. Complete parts (a) through (c) on the right. 29.80 15.42 15.42 27.10 16.51 20.46 8.81 20.46 14.90 33.67 30.91 14.86 5.20 15.35 9.72 19.80 14.86 8.09 16.87 18.30 (a) Find the sample mean. The sample mean is (Round to two decimal places as needed.) (b) Find the sample standard deviation. The sample standard deviation is (Round to two decimal places as needed.) (c) Construct a 90% confidence interval for the population mean u. The 90% confidence interval for the population mean u is ( | ). (Round to two decimal places as needed.)
A process is monitored for defective items by taking a sample of 200 items each day and calculating the proportion that are defective. Let p; be the proportion of defective items in the ith sample. For the last 30 samples, the sum of the proportions is E P; = 1.64. Calculate the center line and the 30 upper and lower control limits for a p chart.
The following data represent the concentration of dissolved organic carbon (mg/L) collected from 20 samples of organic soil. Assume that the population is normally distributed. Complete parts (a) through (c) on the right. 7.40 29.80 27.10 16.51 14.00 8.81 5.20 20.46 14.90 33.67 30.91 14.86 16.87 15.35 9.72 19.80 14.86 8.09 10.30 18.30 (a) Find the sample mean. The sample mean is (Round to two decimal places as needed.) (b) Find the sample standard deviation. The sample standard deviation is (Round to two decimal places as needed.) (c) Construct a 95% confidence interval for the population mean μ. The 95% confidence interval for the population mean μis () () (Round to two decimal places as needed.)
Example57. From the following cumulative distribution of marks obtained by 22 students calculate mode: Marks below 10 20 30 40 50 No. of students 3 8. 17 20 22 Ma No. G
An article contained the following observations on degree of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range: 416 421 422 423 426 428 430 434 437 439 446 447 449 453 456 462 464 Comment on any interesting features. (Select all that apply.) __The data appears to be centered near 438. __There are no outliers. __There is little or no skew. __There is one outlier. __The data is strongly skewed. __The data appears to be centered near 428. (c) Calculate a two-sided 95% confidence interval for true average degree of polymerization. (Round your answers to two decimal places.) (______,_______)

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a. 443.66 <µ S 527.94. Data appears to be normally distributed with only one outlier 95 90 70 60 50 40 30 20 10 200 500 Rainfal 300 400 600 700 800