Standard Frequency Distribution
A Frequency Distribution shows us a summarized grouping of data divided into mutually exclusive classes and the number of occurrences in a class
. It is a way of showing unorganized data e.g. to show results of an election, income of people for a certain region, sales of a product within a certain period, student loan amounts of graduates, etc. Source: Wikipedia.
Types of Standard Frequency Distribution:
Different types of standard frequency distribution are,
Univariate frequency distribution
Joint frequency distribution tables
Univariate frequency distribution:
It is a list of values that can be ordered by the quantity. It can show the values for each value appears for number of times.
Joint frequency distribution:
It is used as two-way tables. It is also called as bivariate joint frequency distribution.
Example Problem for Univariate Standard Frequency Distribution:
Example 1:
Construct the Univariate frequency distribution table for the given data. For the following salary has given to the workers in the office.
Salary for the workers in an office is 3001-4000 is 15
4001-5000 is 22
5001-6000 is 19
6001-7000 is 25
7001-8000 is 17
8001-9000 is 14
9001-10000 are 12
Solution:
Determine the range:
10000 3001 = 7000
Determine the intervals:
Choose the interval as 1000
Construct the Univariate frequency distribution table.
SalaryNo of employeesCumulative frequency
3001-4000 15 15
4001-5000 22 37
5001-6000 19 56
6001-7000 25 81
7001-8000 17 98
8001-9000 14 112
9001-10000 12 124
Example Problem for Joint Standard Frequency Distribution:
Example 2:
In a college, boys and girls are select different departments to study on the department that can be given. Using that set of values construct the joint standard frequency distribution table.
Boys and girls are select different departments in college are Computer science, Information technology, Civil, and Mechanical.
In Computer science 64 boys and 36 girls
In Information technology 43 boys and 57 girls
In Civil 25 boys and 5 girls
In Mechanical 65 boys and 12 girls are selected their departments.
Solution:
Construct a joint frequency distribution table for the given set of data.
Joint FrequencyComputer ScienceInformation TechnologyCivilMechanicalTotal
Boys 64 43 25 65 197
Girls 36 57 5 12 110
Total 100 90 30 77 297
In probability theory and statistics, the standard deviation of a statistical population, a data set, or a probability distribution is the square root of its variance. Simply, standard deviation is the positive square root of variance whereas variance is the mean of the square of the deviations from the mean. Standard deviation is a widely used measure of the variability or dispersion.
The example and practice problems are given below percent standard deviation which helps you for doing percentage problems in standard deviation.
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Example Problems for Percent Standard Deviation:
Example problem1 on percent standard deviation:
Coefficient of variation of two distribution are 65 % and 80%, and their standard deviations are 20 and 16 respectively. Find their arithmetic means.
Solution:
Step 1: Assign variables
Coefficient of variation for 1st distribution, cv1 = 65 s1 = 20
Coefficient of variation for 2nd distribution, cv2 = 80 s2 = 16
Step 2: Consider barx1 and barx 2 be the means of 1st and 2nd distribution respectively.
Step 3: Calculate the barx1 from cv1 percent
cv1 = s1 / barx1 * 100
Therefore,
barx1 = (s1 * 100) / cv1
= (20 * 100) / 65
= 30.77
Step 4: Calculate barx2 from cv2 percent
cv2 = s2 / barx2 * 100
Therefore,
barx2 = (s2 * 100) / cv2
= (16 * 100) / 80
= 20
step 5: Solution
Arithmetic means, barx1 = 30.77
barx2 = 20
Example problem 2 on percent standard deviation:
The mean and variance of the heights and weights of the students of a class are given below. Show that the weights are more variable than heights in percentage.
Solution:
For the height, we have
variance, s2 = 116.64 cm2
Therefore, the standard deviation is SD, s = sqrt116.64 cm = 10.8cm
Also we have, mean height = 160cm
Therefore,
Coefficient of variation for 1st distribution, cv1 = ( (SD)/(mean) * 100)%
= ( 10.8/160 * 100)%
= 4.75 % = 4.75 percentage
For variance, s2 = 17.64 kg2 , the standard deviation is SD, s = sqrt17.64 kg = 4.2 kg
And mean weight = 50.4kg
Coefficient of variation for 2nd distribution, cv2 = ( (SD)/(mean) * 100)%
= ( 4.2/50.4 * 100)%
= 25/3 %
= 8 1/3 %
From this it is clear that Coefficient of variation in weights is grater than Coefficient of variation in heights.
Hence, weights are more variable than heights.
Practice Problems for Percent Standard Deviation:
1) Coefficient of variation of two distribution are 60 % and 75%, and their standard deviations are 18 and 15 respectively. Find their arithmetic means.
2) The mean and variance of the heights and weights of the students of a class are given below. Which shows more variability, heights or weights?
Solutions:
1) 30, 20
2) Heights
by:mathqa
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