  Welcome to Live Chat

Welcome to LiveWebTutors Services, World's leading Academic solutions provider with Millions of Happy Students.

Call Back 24x7 Support Available

chat now

In A Hurry? Get A Callback Table of Content

# Mean Deviation

Mean deviation can be defined as the mean of the absolute deviations and it is denoted by Dx. The formula of the mean deviation can be given as Dx = 1/N sigma 1 to n |xi – x bar| where x bar is the mean of the given deviations and N is the sum of the frequency or number of deviations. The value of N depends upon the type of the statics data given and xi is the given deviations.

Let us discuss the process of finding the mean deviation of the given data. First we need to find out the mean of the given data then we need to find out the deviations about the mean and find out the sum of deviations now find the mean of the sum of deviations. Then we will get the required mean deviation of the given data.

Let us discuss mean deviation with some of the examples.

Consider the first example to be: A student marks in five subjects are given as follows find out the mean deviation of the marks of the student. The marks are 92, 75, 95, 90 and 98.

The given marks of the student are 92, 75, 95, 90 and 98. Now let us find out the mean of the given marks then we will get it as mean = (92 + 75 + 95 + 90 + 98) / 5 which is equal to 90. Now we got the mean of the marks as 90. Now let us find out the mean deviations, we will get the mean deviations if we subtract given observations from the mean. The mean deviations are (92 – 90) = 2, (75 – 90) = -15, (95 – 90) = 5, (90 – 90) = 0 and (98 – 90) = 8. Now if we find the modulus of these deviations then we will get the positive observations that is the observations which we will get after finding the modulus are 2, 15, 5, 0 and 8. Now let us find out the mean of these deviations, first we need to find out the sum of the deviations then we will get it as (2 + 15 + 0 + 5 + 8) which is equal to 30. If we find out the mean then we will get it as 30/5 which is equal to 6. Now the mean deviation is 6.

Consider the second example be: find out the mean deviation of observations 9, 3, 6.

The given observations are 9, 3, 6. First let us find out the mean of the given observations then we will get it as (9 + 3 + 6) / 3 which is equal to 18/3 = 6. Now let us find out the mean deviations then we will get it as (9 – 6) = 3, (3 – 6) = -3, (6 – 6) = 0. Now let us find out the modulus of the observations then we will get it as 3, 3, 0. Now let us find out the mean of these deviations for that we need to find out the sum of deviations which is equal to (3 + 3 + 0) = 6. The mean now is 6/3 = 2. Now the required mean deviation is 2.

Our Amazing Features
• On Time Delivery

There is no deadline that can stop our writers from delivering quality assignments to the students.

• Plagiarism Free Work

Get authentic and unique assignments by using our 100% plagiarism-free services.

• 24 X 7 Live Help

The experienced team of Live web tutors has got your back at all times of the day. Get connected with our customer support executives to receive instant and live solutions for your assignment problems.

• Services For All Subjects

We can build quality assignments in the subjects you're passionate about. Be it Engineering and Literature or Law and Marketing we have an expert writer for all.

• Best Price Guarantee

Get premium service at a pocket-friendly rate. At live web tutors, we understand the tight budget of students and thus offer our services at highly affordable prices.

FREE RESOURCES
FREE SAMPLE FILE
live review Our Mission Client Satisfaction
• I ordered my HRM assignment from this platform. The solution helped me to pass my assessment with flying colors. Well done! Keep up the good work.

27 Oct 2020

Arnisa

• Recently, I ordered my history case study from this platform. The professionals did great work with the content. I am yet to be graded for it but as far as I reviewed the text I am definitely impressed with the quality provided to me at this price. Thank you for your support.

27 Oct 2020

Tammy

• I got good grades with my nursing assignment, all thanks to the experts of this platform. My writing skills are not that great hence the experts of this portal helped me to pass my program with good grades.

27 Oct 2020

Sidorela

• I ordered my English assignment from this platform. The service was very quick and impressive. The customer support executives were very polite and helpful. I am grateful for their assistance.

27 Oct 2020

Jenny

• Overall, it was a pleasant experience. I ordered my management thesis from this platform. The experts did thorough and detailed work on my assignment. All my instructions were implemented precisely.

27 Oct 2020

Fiona

View All Review