For a data set of the pulse rates for a sample of adult females, the lowest pulse rate is 34 beats per minute, the mean of the listed pulse rates is x = 77.0 beats per minute, and theirstandard deviation is s= 13.6 beats per minute,a. What is the difference between the pulse rate of 34 beats per minute and the mean pulse rate of the females?b. How many standard deviations is that [the difference found in part (a)]?c. Convert the pulse rate of 34 beats per minutes to a z score.d. If we consider pulse rates that convert to z scores between - 2 and 2 to be neither significantly low nor significantly high, is the pulse rate of 34 beats per minute significant?b. The difference is standard deviations.(Round to two decimal places as needed.)c. The z score is z=(Round to two decimal places as needed.)d. The lowest pulse rate isHelp me solve thisView an exampleGet more help -Next3:44 AM For a data set of the pulse rates for a sample of adult females, the lowest pulse rate is 34 beats per minute, the mean of the listed pulse rates is xstandard deviation is s= 13.6 beats per minute.77.0 beats per minute, and theira. What is the difference between the pulse rate of 34 beats per minute and the mean pulse rate of the females?b. How many standard deviations is that [the difference found in part (a)]?c. Convert the pulse rate of 34 beats per minutes to a z score.d. If we consider pulse rates that convert to z scores between - 2 and 2 to be neither significantly low nor significantly high, is the pulse rate of 34 beats per minute significant?a. The difference isbeats per minute.(Type an integer or a decimal. Do not round.)b. The difference isstandard deviations.(Round to two decimal places as needed.)c. The z score is z=

HERE GIVEN Mean=X¯=77 Beats S=13.6 Beats

Answered: standard deviation is s = 13.6 beats…

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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A data set lists weights (lb) of plastic discarded by households. The highest weight is 5.74 lb, the mean of all of the weights is x = 1.866 lb, and the standard deviation of the weights is s = 2.472 lb. a. What is the difference between the weight of 5.74 lb and the mean of the weights? b. How many standard deviations is that [the difference found in part (a)]? c. Convert the weight of 5.74 lb to a z score. d. If we consider weights that convert to z scores between -2 and 2 to be neither significantly low nor significantly high, is the weight of 5.74 lb significant?
Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 78.7 Mbps. The complete list of 50 data speeds has a mean of x overbar equals17.78 Mbps and a standard deviation of sequals 37.85 Mbps. a. What is the difference between carrier's highest data speed and the mean of all 50 data speeds? b. How many standard deviations is that [the difference found in part (a)]? c. Convert the carrier's highest data speed to a z score. d. If we consider data speeds that convert to z scores between minus 2 and 2 to be neither significantly low nor significantly high, is the carrier's highest data speed significant?
Q7
The measure of dispersion which shows the variation about the mean is A. Interquartile range B.Standard deviation C.Coefficient of variation D.Range
a quantitative data set has mean 25 and standard deviation 2. approximately what percentage of the observations lie between 21 and 29?
Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 72.5 Mbps. The complete list of 50 data speeds has mean of x = 15.24 Mbps and a standard deviation of s= 34.08 Mbps. a. What is the difference between carrier's highest data speed and the mean of all 50 data speeds? b. How many standard deviations is that [the difference found in part (a)]? c. Convert the carrier's highest data speed to a z score. d. If we consider data speeds that convert to z scores between -2 and 2 to be neither significantly low nor significantly high, is the carrier's highest data speed significant? a. The difference is Mbps (Type an integer or a decimal. Do not round.) b. The difference is standard deviations. (Round to two decimal places as needed.) c. The z score is z = (Round to two decimal places as needed.) d. The carrier's highest data speed is C
c. IQ scores have a mean of 100 and a standard deviation of 16. Albert Einstein reportedly had an Q of 160. Arnav Sharma, a 12-year old boy gained a score of 162. 14 i. Convert Einstein's IQ score and Sharma's score to their corresponding Z-scores. ii. What is the difference between the Z-score of Einstein and Z-score of Sharma? iii. How many standard deviations are Einstein's and Sharma's Z-scores?
Plz explain. Ty
A value of 104.2 would be how many standard deviations above the mean, given a mean of 100, and a standard deviation of 5.6?
For a data set of the pulse rates for a sample of adult females, the lowest pulse rate is 39 beats per minute, the mean of the listed pulse rates is x overbar equals 77.0 beats per minute, and their standard deviation is sequals 21.2 beats per minute. a. What is the difference between the pulse rate of39 beats per minute and the mean pulse rate of the females? b. How many standard deviations is that [the difference found in part (a)]? c. Convert the pulse rate of39 beats per minutes to a z score. d. If we consider data speeds that convert to z scores between minus 2 and 2 to be neither significantly low nor significantly high, is the pulse rate of 39 beats per minute significant?
Use z scores to compare the given values. The tallest living man at one time had a height of 233 cm. The shortest living man at that time had a height of 89.9 cm. Heights of men at that time had a mean of 176.34 cm and a standard deviation of 5.47 cm. Which of these two men had the height that was more extreme? Since the z score for the tallest man is z= height that was more extreme. (Round to two decimal places.) and the z score for the shortest man is z = the man had the
Find the percentage of area under a normal curve between the mean and the given number of standard deviations from the mean. This problem can be solved using technology or the table for the standard normal curve from your text. −0.4 The percentage of area under the normal curve between the mean and −0.4 standard deviations from the mean is

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