# Mean Deviation Formula

The process of collecting and analyzing the data is known as statistics. The deviation in statistics is known as the variation between the other values of the variable and the observed value. Let’s study more about the deviation

### Mean deviation

The mean deviation of the standard distribution is the measurement of its central tendency. It can be calculated by using three methods. They are Arithmetic mean, Median, Mode.

Mean deviation is used to show how far the observations are situated from the average of the observed data. Each one of these is to be considered of an absolute value. In this, the negative signs are totally ignored. According to this, it is said that the deviation of both sides is of equivalent nature. The suitable average of the mean deviation could be of mean, median, or mode of data. For an instance the formula for the mean deviation of individual, discrete and continuous series.

**Types of mean deviation**

There are three types of mean deviation. They are individual series, discrete series, and continuous.

**Individual series**

When the data is provided individually in the form of a series then it is known as the individual series. It is basically a form of raw data in the form of a series that forms an arrangement individually. In individual series, items are represented in a single form.

For example, let us assume the following scores by the players in a cricket match:

56, 97, 46, 88, 67, 59, 62, 78, 90, 58, 77

In the above data, it is not given that how many players score 56 runs and more than 78 in a single sight.

**Discrete series**

The discrete series is the series that is used to reflect each specific value of the observed variable. One of the variables corresponds to an integer value. In discrete series exact measurement of the items in the data is visible. For example, the wages of 20 workers are below in the table as,

Wages

(number of workers)

Frequency 2000 6 2500 4 3000 2 3500 3 4000 3 4500 2 Conclusively 6 workers are getting RS 2000 wages paid, and 4 workers are getting RS 2500 paid, and so on.

**Continuous series**

A continuous series is a series that maintains the items in certain definite classes. The items in the class interval lose their individual identity, and those individual items are merged in one or another class intervals. Each class has continuity in which means that the end of one class should be the beginning for the other one. That’s why it is named a continuous series.

For example, the continuous series is depicted as follows

Age Frequency 10-15 4 15-20 12 20-25 16 25-30 22 30-35 10 35-40 8 40-45 6 45-50 4

**Mean deviation from the arithmetic mean (formula)**

**Individual series**

Mean Deviation (M.D) = ∑∣X – X̄∣ / N

Where,

∑ – Summation

x – Observation

X̄ – Mean

N – Number of observation

**Discrete series**

Mean Deviation (M.D) = ∑f∣X – X̄∣ / ∑f

Where,

∑ – Summation

x – Observation

X̄ – Mean

f – frequency of observation

**Continuous series**

Mean Deviation (M.D) = ∑f∣X – X̄∣ / ∑f

Where,

∑ – Summation

x – Mid-value of the class

X̄ – Mean

f – frequency of observation

**Mean deviation from the median (formula)**

**Individual series**

Mean Deviation (M.D) = ∑|X – M| / N

Where,

∑ – Summation

x – Observation

M – Median

N – Number of observation

**Discrete series**

Mean Deviation (M.D) = ∑ f|X – M| / ∑ f

Where,

∑ – Summation

x – Observation

M – Median

N – Frequency of observations

**Continuous series**

Mean Deviation (M.D) = ∑ f∣X – X̄∣ / ∑f

Where,

∑ – Summation

x – Observation

M – Median

N – Frequency of observations

**Mean deviation by mode (formula)**

**Individual series**

Mean Deviation (M.D) = ∑|X – Mode| / N

Where,

∑ – Summation

x – Observation

M – Mode

N – Number of observations

**Discrete series**

Mean Deviation (M.D) = ∑ f|X – Mode| / ∑ f

Where,

∑ – Summation

x – Observation

M – Mode

N – Frequency of observations

**Continuous series**

Mean Deviation (M.D) = ∑ f |X – Mode| / ∑ f

Where,

∑ – Summation

x – Observation

M – Mode

N – Frequency of observations

**Steps to calculate mean deviation**

- Firstly we have to calculate the arithmetic mean, median, or by mode of the given data
- Now we have to calculate the deviation from mean, median, or mode and have to ignore the negative items
- Now we have to multiply the deviations with the frequency of the data. This step can only be done while solving either the discrete or the continuous series this step doesn’t work in individual series.
- Now sum up all the deviations
- Apply the formula and solve the question.

### Similar Problems

**Question 1: Calculate the mean deviation from the median and the co-efficient of mean deviation from the following data:**

**Marks of the students: 88, 14, 78, 69, 44, 54, 18, 79, 40.**

**Solution:**

Arrange the data in ascending order: 14, 18, 40, 44, 54, 69, 78, 79, 88.

Median = Value of the (N + 1)

^{TH}/ 2 term= Value of the (9 + 1)

^{TH}/ 2 term = 54Calculation of mean deviation:

X |X – M| 14 40 18 36 40 14 44 10 54 0 69 15 78 24 79 25 88 34 N = 9 ∑|X–M|=198 M.D. = ∑|X – M| / N

= 198/9

= 22

Co-efficient of Mean Deviation from Median = M.D./M

= 22/54

= 0.4074

**Question 2: Calculate the mean deviation about the mean using the following data**

**5, 8, 14, 16, 20, 6, 8, 19.**

**Solution:**

First, we have to find the mean of the data that we are provided with

Mean of the given data = Sum of all the terms total number of terms

X̄ = 5 + 8 + 14 + 16 + 20 + 6 + 8 + 19

= 96/8

= 12

Next, find the mean deviation

X _{i}X _{i}– x̄|X _{i }– x̄|5 5 – 12 = -7 |-7| = 7 8 8 – 12 = -4 |-4| = 4 14 14 – 12 = 2 |2| = 2 16 16 – 12 = 4 |4| = 4 20 20 – 12 = 8 |8| = 8 6 6 – 12 = -6 |-6| = 6 8 8 – 12 = -4 |-4| = 4 19 19 – 12 = 7 |7| = 7 ∑|X _{i }− x̄| = 42Mean deviation about mean = ∑|X

_{i }− X̄| / 8= 42/8

= 5.25

**Question 3: Find the mean deviation about the median for the following data.**

Class | Frequency (f) |

5-15 | 16 |

15-25 | 5 |

25-35 | 8 |

35-45 | 6 |

45-55 | 3 |

**Solution:**

Class f cf x _{i}|x – x̄| f. |x – x̄| 5-15 16 16 10 11 176 15-25 5 21 20 1 5 25-35 8 29 30 9 72 35-45 6 35 40 19 114 45-55 3 38 50 29 87 Total N = 38 To find the median class,

N/2 = 38/2 = 19

thus cf is nearest to 20

Thus, median class is 15 – 25.

l = 15, i = 10, f = 5, cf = 16, ∑

^{5}_{1 }f_{i}/2= 19Substituting these values in the formula,

M = l+(∑

^{5}_{1 }f_{i}/2 − cf)/f × h= 21

Mean deviation about median = ∑

^{5}_{1 }f_{i}|x_{i }− M| / ∑^{5}_{1}f_{i}= 295.6 / 38 = 7.778

Answer: Mean deviation about median = 7.778

**Question 4: Find the mean deviation about the mean for {17, 24, 37, 18, 4}**

**Solution: **

The data is ungrouped, thus mean = (17 + 24 + 37 + 18 + 4) / 5 = 20

x |x – x̄| 17 3 24 4 37 17 18 2 4 16 Total 42 Using the formula,

∑

^{5}_{1}|x_{i}− μ|/5= 42 / 5 = 8.4

Mean deviation about mean = 8.4

**Question 5: Determine the mean deviation for the data values 4, 2, 9, 7, 3, 5.**

**Solution:**

Given data values are 4, 2, 9, 7, 3, 5.

We know that the procedure to calculate the mean deviation.

First, find the mean for the given data:

Mean, µ = (4 + 2 + 9 + 7 + 3 + 5)/6

µ = 30/6

µ = 5

Therefore, the mean value is 5.

Now, subtract each mean from the data value, and ignore the minus symbol if any

(Ignore”-”)

4 – 5 = 1

2 – 5 = 3

9 – 5 = 4

7 – 5 = 2

3 – 5 = 2

5 – 5 = 0

Now, the obtained data set is 1, 3, 4, 2, 2, 0.

Finally, find the mean value for the obtained data set

Therefore, the mean deviation is

= (1 + 3 + 4 + 2 + 2 + 0) /6

= 12/6

= 2

Hence, the mean deviation for 4, 2, 9, 7, 3, 5 is 2.

**Question 6: Find the mean deviation from the mean of given series also calculate its coefficient,**

C.I. | F |

0-2 | 3 |

2-4 | 5 |

4-6 | 6 |

6-8 | 4 |

8-10 | 2 |

**Solution:**

C-I f x fx x – x̄ |x – x̄| f|x – x̄| 0-2 3 1 3 1-4.7=-3.7 3.7 11.1 2-4 5 3 15 3-4.7=-1.7 1.7 8.5 4-6 6 5 30 5-4.7=0.3 0.3 1.8 6-8 4 7 28 7-4.7=2.3 2.3 9.2 8-10 2 9 18 9-4.7=4.3 4.3 8.6 ∑f = 20 ∑fx = 94 ∑f|x – x̄| = 39.2 x̄= ∑fx/∑f = 94/20 = 4.7

M,D = ∑f|x – x̄|/∑f = 39.2/20 = 1.96

Coefficient of M,D, = M.D./mean = 1.96/4.7 = 0.417