All of the full questions (including methods to measure central tendency) are in the attached document.© BrainMass Inc. brainmass.com October 25, 2018, 8:58 am ad1c9bdddf
I have solved the problems in a generic manner, because their specific answers depend on your personal and/or professional environment. So, please consider these solutions as guidance for you not as final answers. Read the solutions carefully and translate them as per your environment(s).
a. Median is used where there some extreme value data exist, e.g, in a sport most of the sports persons earn in a well defined range but some of the top players earn extremely large money in comparison to normal players. Hence in these cases, mean will answer a biased value. Therefore, median is a better measure.
Mean is used when data are normal distributed, e.g., in a class room, age of students is normally distributed.
Mode is used when frequency is need to be count, e.g., in a given class room, number of students from the same city.
Mean: Average age of associates of a company.
Median: Median of salary of associates in a company ( because, associates at top post have salaries much larger than normal associates)
Mode: Associates, working in my company belong to same city.
If there are n persons in the group.
First we estimate probability of NO TWO persons have birthday on same day.
Take one by one person. 1st person has birth day on one of the 365 days. Consider that day as a reference day.
Hence, probability of 1st person birthday on the given day = 1 = (365 - (1-1))/365
For next (2nd) person, probability of birthday not on same day = 364/365 = (365 - (2-1))/365
For next (3rd) person, probability of birthday not matching with previous 2 persons = 363/365 = (365 - (3-1))/365
For next (4th) person, probability of birthday not matching with previous 3 persons = 362/365 = (365 - (4-1))/365
For last (nth) person, probability of birthday not matching with previous (n-1) persons = (365 - (n-1))/365
Hence, probability of no two persons having birthday on same day,
q = 365/365 * 364/365 * 363/365 * 362/365 * ... * (365 - (n-1))/365
=> q = (365*364*363*...*(365-n+1))/365^n
=> q = 365!/(365^n * (365-n)!)
=> q = P(365, n)/365^n
Here, P for permutation, and ! means factorial.
Hence, probability of at least 2 persons having birthday on same day,
p = 1 - q = 1 - P(365,n)/365^n
Note: in case n is given to you please ...
A few statistics definitions and a few statistics problems are solved here.
Measures of Central Tendency
This Discussion will give students the opportunity to calculate or identify the three measures of central tendency. Select an appropriate real life situation where one measure would be more appropriate than the other two measures of center.
a.Select a common topic or hypothesis and record the topic in your posting.
"What is the average number of times you order out for dinner on a weekly basis"
b. Sample at least 15 people and record their data in a simple table or chart
9 Males 9 Females, 18 All together
Name of person Number of times take out is ordered weekly
c. You can gather your data at work, on the phone, or via some other method. This is your "Sampling Design." Which of the four sampling techniques best describes your design?
d.Explain in moderate detail the method you used to gather your data. In statistics this venture is called the "Methodology."
d. Make sure you break your sample into classes or groups, such as: males/females or ages or time of day, etc
Males and Females, age does not matter
e.Calculate the mean, median, and mode for your data as a whole.
f. Now calculate the mean, median, and mode of each of your classes or groups.
g.Indicate which measure of central tendency BEST describes your data and WHY.View Full Posting Details