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# Choosing a Method to Measure Central Tendency

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All of the full questions (including methods to measure central tendency) are in the attached document.

https://brainmass.com/statistics/descriptive-statistics/choosing-method-measure-central-tendency-559585

#### Solution Preview

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).

1.
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.

b.
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.

2.a.
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 ...

#### Solution Summary

A few statistics definitions and a few statistics problems are solved here.

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