The following is a sample of sanitation scores for 20 cruise ships.91 91 81 87 93 92 89 92 86 98 87 85a. Find the mean and standard deviation of the sanitation scores.x = 89.15 (Round to two decimal places as needed.)s = 4.34(Round to two decimal places as needed.)b. Calculate the intervals x ±s, x ±2s, x ± 3s.X+S=(Round to two decimal places as needed.)92 91 90819283 89 93

We have given that, We have from the above, x̄ = 89.15 and s= 4.34

Answered: The following is a sample of sanitation…

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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The following data are the distances from the workplace (in miles) for the 5 employees of a small business. 14, 3, 8, 13, 2 Assuming that these distances constitute an entire population, find the standard deviation of the population. Round your answer to two decimal places.
Find the mean of the data summarized in the given frequency distribution. Compare the computed mean to the actual mean of 57.4 degrees. The mean of the frequency distribution is (Round to the nearest tenth as needed.)
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 =
For the frequency data: x f 15 31 25 111 30 82 35 25 40 14 Calculate the following statistics (round your answers to the nearest hundredth): n = mean = Q = s = median = mode = range = (enter a single value - not an interval) If n > 30, then SD = ; if n < 30, then SD = s. Based the the data above the Standard Deviation is
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.
2. The table below shows the recorded speeds of several cars on a busy street: Recorded Speeds (mph) 35 38 41 35 36 55 a) Calculate the mean absolute deviation both with and without the data value of 55. Round to the nearest hundredth if necessary.

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