Basic Statistics Exam Tutor

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Introduction to basic statistics exam online tutor:

Statistics, on the one hand, means lists of numerical values; for example, the salaries of the monthly employees of a company, or the SAT scores of the incoming students of a university. Statistics as a science, on the other hand, is the branch of mathematics which organizes, analyzes, and interprets such raw data. Statistical methods are applicable to any area of human endeavor where numerical data are collected for some type of decision-making process.

Measures of central tendency: Mean and Median

There are various ways of giving an overview of data. One way is by the graphical descriptions for example bar graphs and circular graphs. But in this article we shall also be learning numerical descriptions of data. Numbers such as the mean and median are given, in some sense, the central or middle values of the data. Other numbers, such as variance and standard deviation, measure the dispersion or spread of the data about the mean.

The data we discuss will come either from a random sample of a larger population or from the larger population itself. ...
... We distinguish these two cases using different notations as follows:

n = number of items in the sample,

N = number of elements in the population

`barx` = sample mean

$\mu$ = Population mean

S2 = Sample variance

$\sigma ^{2}$ =population variance

basic statistics exam online tutor-Mean, Median

Mean:

Suppose a sample consists of the eight numbers:

7, 11, 11, 8, 12, 7, 6, 6

The sample mean `barx` is defined to be the sum of the values divided by the number of values; that is

`barx` =(7+11+11+8+12+7+6+6)/8= 68/8=8.5

Generally speaking, suppose x1, x2....xn are n numerical values of some sample. Then

Sample mean: `barx` =$\mathbf{\frac{x_{1}+x_{2}+....................x_{n}}{n}=\frac{\sum x_{i}}{n}}$

Now suppose that the data are organized into a frequency table; let there be k distinct numerical values x1, x2, ......xk, occurring with respective frequencies f1, f2,,.....fk. Then the product f1x1 gives the sum of the x1's f2x2 gives the sum of the x2's and so on. Also,

f1+f2+...............+fk=n

The total number of data items. Hence formula can be rewritten as

Sample mean: `barx`= $\mathbf{\frac{f_{1}x_{1}+f_{2}x_{2}+............+f_{k}x_{k}}{f_{1}+f_{2}............+f_{k}}=\frac{\sum f_{i}x_{i}}{\sum f_{i}}}$

Median:

Consider a list x1,x2.........xn of n data values which are sorted in increasing order. The median of the data, denoted by

$\tilde{x}$ is defined to be the " middle value". That is,

Median: $\tilde{x}$={[(n+1)/2]th term when n is odd,

{[(n/2)th term +[(n/2)+1]th term ]/2 when n is even.

Note that $\tilde{x}$ is the average of the (n/2)th and [(n/2)+1]th terms when n is even.

Suppose, for example, the following two lists of sorted numbers are given:

List A: 11, 11, 16, 17, 25

List B: 1, 4, 8, 8, 10, 16, 16, 19

List A has five terms; its median $\tilde{x}$=16, the middle or third term. List B has 8 terms; its median $\tilde{x}$=9, the average of the fourth term(8) and the fifth term(10).

One property of median $\tilde{x}$ is that there are just as many numbers less than $\tilde{x}$ as there are greater than $\tilde{x}$.

The cumulative frequency distribution can be used to find the median of an arbitrary set of data.

Solved problem on basic statistics exam online tutor

1. The owner of a small company has 15 employees. Five employees earn Rs. 25,000 per year, seven earn $30,000, three earn Rs 40,000, and the owner's annual salary is Rs 153,000.

(a) Find the mean and median salaries of all 16 persons in the company.

(b) Find the mean and median salaries if the owner's salary is increased by Rs 80,000.

Solution: (a) The mean salary is

$\bar{x}$= $\frac{5*25,000+7*30,000+3*40,000+153,000}{16}$= 608,000/16= Rs 38,000

Since there are 16 persons, the median is the average of the eighth(16/2) and ninth(16/2+1) salaries when the salaries are arranged in increasing order from left to right. The eighth and ninth salaries are each $30,000. Therefore, the median is $\tilde{x}$=Rs 30,000.

(b) The new mean salary is

$\tilde{x}=\frac{608,000+80,000}{16}=\frac{688,000}{16}$= Rs 40,000

The median is still Rs 30,000, the average of the eighth and ninth salaries, which did not change. Hence, the mean moves in the direction of the increased salary, but the median does not change.

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