The following is a set of data for a population with N = 10. 6 7 14 2 15 11 12 10 6 10 O a. Compute the population mean. b. Compute the population standard deviation. a. The population mean is µ =D. (Type an integer or a decimal. Do not round.) b. The population standard deviation is a =| (Round to two decimal places as needed.)

Find the mean of the data summarized in the given frequency distribution. Compare the computed mean to the actual mean of 58.1 degrees. Which of the following best describes the relationship between the computed mean and the actual​ mean?   Low Temperature ​(◦​F) 40−44 45−49 50−54 55−59 60−64   Frequency 2 6 11 6 3

one z-score unit is equivalent to ______ of the distribution.  a. the variance b. the range c. the standard deviation d. the mean e. the correlation coefficient

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.)

A. Population ? B. Population mean ? C. Median ? F. IQR ? G. Population standard deviation ?H. population variance ?

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

Below is a sample of 10 quiz grades for a student in a Quantitative Reasoning class. Calculate the Standard Deviation by finding the deviations for each quiz grade as well as the listed statistics. Round to 2 decimal places as needed. Quiz Grade 70 90 70 50 Deviation 80 Listen 50 70 90 70 90 Squared Deviation Mean Quiz Grade Min Quiz Grade Max Quiz Grade Range of Quiz Grades The sum of DEVIATIONS is always equal to Sum of Squared Deviations Quiz Count Degrees of Freedom Variance Standard Deviation which is why the squared deviation is needed.

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.