The method of tree ring dating gave the following years A.D. for an archaeological excavation site. Assume that the population of x values has anapproximately normal distribution.1306 1264 1313 1187 1268 1316 1275 1317 1275(a) Use a calculator with mean and standard deviation keys to find the sample mean year x and sample standard deviation s. (Round youranswers to the nearest whole number.)A.D.yrX=S=(b) Find a 90% confidence interval for the mean of all tree ring dates from this archaeological site. (Round your answers to the nearest wholenumber.)lower limitupper limitA.D.A.D.

Answered: The method of tree ring dating gave the…

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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Suppose an interval estimate for the population mean was 62.84 to 69.46. The population standard deviation was assumed to be 6.50, and a sample of 100 observations was used. The mean of the sample was:
Show your handwritten work on the paper. The following is a frequency distribution of ages of residents of a senior citizens' home in New York. Fill in the midpoints, then calculate the mean and the standard deviation. Round to two decimal positions. You may assume the data represents a population. Ages of Residents Midpoints Frequency 50-62 63 75 76--88 a. Mean (u)3D b. Standard deviation (o) = Round the standard deviation up to 2 decimal places if necessary. Hint: Use the midpoint for each class (ages of residents). 2.
What is the population standard deviation using the following formula
Find mean & standard deviation using table: X Frequency 2 20% 5 10% 8 15% 10 30% 12 25% For the mean, i get 7.4 & 20%. Std dev i get 4.459362. Just wanted to check my numbers.
Use the following data to answer the following questions. You can use excel or any app. as calculator.46 93 5630 39 4524 48 4560 55 4547 62 7063 72 6145 72 6574 48 6373 55 4872 55 44 1. The mean value is __________. a. 55.8833 b. 55.5500 c. 45 d. 55.8333 2. The median value is __________. a. 55.8833 b. 55.5 c. 55 d. 45 3. The mode is __________. a. 55.8833 b. 45 c. 55 d. 55.8333 4. The range is __________. a. 67 b. 68 c. 69 d. 70 5. The first quartile is __________. a. 43 b. 44 c. 40 d. 45
Please answer the question in the picture below to answer the question in the other picture
for a population with a standard deviation of 4.2 and mean of 32.5, draw a bell shaped curve, marking location of the mean and the numbers that correspond with all the standard deviations Type your answer here:
Suppose baby kittens' weights are normally distributed with a mean of 11.4 and a standard deviation of 2.1. The Z-score tells you how many units above the average (if Z-score is positive) or below the average (if Z- Score is negative) any particular baby kitten's weight is. Find the baby kitten weight that corresponds to the following Z-scores. Use the formula Z where u is the mean, o is the standard deviation, and X is the baby kitten weight. a. Z = = 0.07, X = b. Z = 1.63, X =
Use the following set of values for #10 - 13. Round answers to nearest hundredth, if necessary. .3, .6, 9, 1.3, .4, .6, 1.2, 1.4, 1.1, .2, .2 10. Find the mean, median, and mode(s) 11. Find the range 12. Find the interquartile range 13. Find the standard deviation
Find the mean of the data summarized in the given frequency distribution. Compare the computed mean to the actual mean of 52.5 degrees. Low Temperature (o F) 40 - 44 45 - 49 50 - 54 55 - 59 60 – 64 D Frequency 2 9. ... The mean of the frequency distribution is degrees. (Round to the nearest tenth as needed.) Which of the following best describes the relationship between the computed mean and the actual mean? O A. The computed mean is close to the actual mean because the difference between the means is more than 5% of the actual mean. B. The computed mean is not close to the actual mean because the difference between the means is less than 5% of the actual mean. C. The computed mean is not close to the actual mean because the difference between the means is more than 5% of the actual mean. D. The computed mean is close to the actual mean because the difference between the means is less than 5% of the actual mean.
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