Mean Deviation MCQ Quiz - Objective Question with Answer for Mean Deviation - Download Free PDF

Explanation:

The mean of 10 natural numbers 

⇒ \(\overline{x} = \frac{1+2+3+4+....10}{10}\) 

= \(\frac{10\times11}{2\times10}\) = 5.5

Mean deviation = \(\frac{[1- 5.5] [2-5.5] [3- 5.5] ... [10- 5.5]}{10} = 2.5\)

∴ Option (b) is correct.

Concept Used:

• Mean deviation about the mean = \(\frac{\sum |x_i - \bar{x}|}{n}\)

• Mean deviation about the median = \(\frac{\sum |x_i - \text{median}|}{n}\)

• Mean deviation of m and M = \(\frac{|m - \bar{X}| + |M - \bar{X}|}{2}\) where \(\bar{X}\) is the mean of m and M.

Calculation:

First, arrange the data in ascending order: 2, 3, 5, 7, 11, 15, 20

Mean \(\bar{x} = \frac{2+3+5+7+11+15+20}{7} = \frac{63}{7} = 9\)

Median = 7 (middle value)

Mean deviation about the mean (m):

\(m = \frac{|2-9| + |3-9| + |5-9| + |7-9| + |11-9| + |15-9| + |20-9|}{7}\)

\(m = \frac{7+6+4+2+2+6+11}{7} = \frac{38}{7}\)

Mean deviation about the median (M):

\(M = \frac{|2-7| + |3-7| + |5-7| + |7-7| + |11-7| + |15-7| + |20-7|}{7}\)

\(M = \frac{5+4+2+0+4+8+13}{7} = \frac{36}{7}\)

Mean of m and M:

\(\bar{X} = \frac{m+M}{2} = \frac{\frac{38}{7} + \frac{36}{7}}{2} = \frac{74}{14} = \frac{37}{7}\)

Mean deviation about the mean of m and M:

\(\frac{|m - \bar{X}| + |M - \bar{X}|}{2} = \frac{|\frac{38}{7} - \frac{37}{7}| + |\frac{36}{7} - \frac{37}{7}|}{2} = \frac{\frac{1}{7} + \frac{1}{7}}{2} = \frac{\frac{2}{7}}{2} = \frac{1}{7}\)

Hence option 1 is correct

Concept Used:

Mean deviation about the mean is calculated using the formula:

Mean Deviation = \(\frac{(Σfᵢ|xᵢ - x̄|) }{N}\)

where fᵢ is the frequency of the i-th class, xᵢ is the midpoint of the i-th class interval, x̄ is the mean of the data, and N is the total frequency.

Calculation:

First, we find the midpoints (xᵢ) of each class interval:

0-2: x₁ = 1

2-4: x₂ = 3

4-6: x₃ = 5

6-8: x₄ = 7

8-10: x₅ = 9

Next, we calculate the mean (x̄) of the data:

x̄ = \(\frac{(Σfᵢxᵢ) }{ N}\)

x̄ = \(\frac{(1×1 + 3×3 + 5×5 + 3×7 + 1×9) }{(1 + 3 + 5 + 3 + 1)}\)

x̄ = \(\frac{65 }{13}\) = 5

Now, we calculate the absolute deviations |xᵢ - x̄|:

|x₁ - x̄| = |1 - 5| = 4

|x₂ - x̄| = |3 - 5| = 2

|x₃ - x̄| = |5 - 5| = 0

|x₄ - x̄| = |7 - 5| = 2

|x₅ - x̄| = |9 - 5| = 4

Next, we calculate Σfᵢ|xᵢ - x̄|:

Σfᵢ|xᵢ - x̄| = 1×4 + 3×2 + 5×0 + 3×2 + 1×4 = 4 + 6 + 0 + 6 + 4 = 20

Mean Deviation = \(\frac{(Σfᵢ|xᵢ - x̄|) }{ N}\) = \(\frac{20 }{ 13}\)

Hence option 4 is correct

Explanation -

x + y = 10 .........(1)

Median = 18 = M

M.D. = \(\frac{\sum \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right|}{\sum \mathrm{f}_{\mathrm{i}}}\)

1.25 = \(\frac{36+x+2 y}{40}\)

x + 2y = 14 .........(1)

by (1) & (2)

x = 6, y = 4

⇒ 4x + 5y = 24 + 20 = 44 

Hence Option(2) is correct.

Concept:

Mean deviation about the mean = \(\frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{{\rm{i}} = {\rm{n}}} \left| {{{\rm{x}}_{\rm{i}}}-{\rm{\;\bar x}}} \right|}}{{\rm{n}}}\), Where \({\rm{\bar x}}\) is mean.

Calculation:

Given numbers are 10, 9, 21, 16, 24

Total numbers = 5

\({\rm{Mean}} = {\rm{\bar x}} = \;\frac{{10 + 9 + 21 + 16 + 24}}{5} = \frac{{80}}{5} = 16\)

We know that mean deviation about the mean = \(\frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{{\rm{i}} = {\rm{n}}} \left| {{{\rm{x}}_{\rm{i}}}-{\rm{\;\bar x}}} \right|}}{{\rm{n}}}\)

Mean deviation from the mean = \(\frac{{\left| {10-{\rm{\;}}16} \right| + \left| {9-{\rm{\;}}16} \right| + \left| {21-{\rm{\;}}16} \right| + \left| {16{\rm{\;}}-{\rm{\;}}16} \right| + \left| {24-{\rm{\;}}16} \right|\;}}{5}\)

Formula

Median = (n + 1)/2

Calculation

Median = value of (n + 1)/2^th item

⇒ (48 + 1)/2 = 24.5^th item

24.5 lies in cf of 36

∴ Median = 12

Mean deviation about median = (1/48)(36)

⇒ 0.75

∴ Mean deviation about median is 0.75

Concept:

Mean deviation about the mean = \(\frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{{\rm{i}} = {\rm{n}}} \left| {{{\rm{x}}_{\rm{i}}}-{\rm{\;\bar x}}} \right|}}{{\rm{n}}}\), Where \({\rm{\bar x}}\) is mean.

Calculation:

Given numbers are 10, 9, 21, 16, 24

Total numbers = 5

\({\rm{Mean}} = {\rm{\bar x}} = \;\frac{{10 + 9 + 21 + 16 + 24}}{5} = \frac{{80}}{5} = 16\)

We know that mean deviation about the mean = \(\frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{{\rm{i}} = {\rm{n}}} \left| {{{\rm{x}}_{\rm{i}}}-{\rm{\;\bar x}}} \right|}}{{\rm{n}}}\)

Mean deviation from the mean = \(\frac{{\left| {10-{\rm{\;}}16} \right| + \left| {9-{\rm{\;}}16} \right| + \left| {21-{\rm{\;}}16} \right| + \left| {16{\rm{\;}}-{\rm{\;}}16} \right| + \left| {24-{\rm{\;}}16} \right|\;}}{5}\)

CONCEPT:

Mean Deviation for ungrouped data about the median:

For ‘n’ observation x₁, x₂………….. x_n, the mean deviation about their mean \(\bar x\) is given by \(M.D = \frac{{\mathop \sum \nolimits_{i = 1}^n \left| {{x_i} - M} \right|}}{N}\). where N is the number of observations and M is the median.

CALCULATIONS:

Arrange data in ascending order –

2, 3, 4, 6, 7, 9, 11

No. of observation n = 7 (odd)

Median will be \(\left( {\frac{{n + 1}}{2}} \right)th\) term i.e. 4^th term = 6.

Mean deviation about their median M is given by \(M.D\left( M \right) = \frac{{\left| {{x_i} - M} \right|}}{n}\)

Here, M = 6

\(M.D\left( M \right) = \frac{{\left| {2 - 6} \right| + \left| {3 - 6} \right| + \left| {4 - 6} \right| + \left| {6 - 6} \right| + \left| {7 - 6} \right| + \left| {9 - 6} \right| + \left| {11 - 6} \right|}}{7}\)

\(M.D\left( M \right) = \frac{{18}}{7} = 2.57\)

Concept:

Mean: It is the average of given observation. Let x₁, x₂, …, x_n be n observations, then

Mean \( = {\rm{\bar X}} = \frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} {{\rm{x}}_{\rm{i}}}}}{{\rm{n}}}\)

Mean deviation: Let x₁, x₂, …, x_n be n observations, then:

Mean deviation \( = \frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} \left| {{{\rm{x}}_{\rm{i}}} - {\rm{\bar x}}} \right|}}{{\rm{n}}}\)

Calculation:

Given: Data: 4, 7, 8, 9, 10, 12, 13, 17.

Mean \({\rm{\bar X}} = \frac{{4 + 7 + 8 + 9 + 10 + 12 + 13 + 17}}{8}\)

\( \Rightarrow {\rm{\bar X}} = \frac{{80}}{8}\)

⇒ X̅ = 10

Mean deviation \( = \frac{{\mathop \sum \nolimits_{\rm{i}}^{\rm{n}} \left| {{{\rm{X}}_{\rm{i}}} - {\rm{\bar X}}} \right|}}{{\rm{n}}}\)

\(= \frac{{6 + 3 + 2 + 1 + 0 + 2 + 3 + 7}}{8}\)

= 3

Concept:

Mean deviation from mean = \(\displaystyle \frac{Σ∣x_i​ − x̅ ∣​}{N}\) 

Coefficient of mean deviation = \(\displaystyle \frac{Mean\ Deviation}{\bar x}\)

where N is the number of observation

x̅ is the mean of the data

x is the value of the observation.

Calculation:

\(\displaystyle Mean=\bar x=\frac{Sum\ of \ observation\ }{Number\ of \ observation}\)

\(\displaystyle Mean=\bar x=\frac{21+34+23+39+26+37+40+20+33+27}{10} \)

⇒ x̅ = \(\displaystyle \frac{300}{10}=30\)

 Sum of deviation from mean is given by Σ∣xi​ − x̅∣,

⇒ Σ∣x_i​ − x̅∣ = ∣21 − 30∣ + ∣34 − 30∣ + ∣23 − 30∣ + ∣39 − 30∣ + ∣26 − 30∣ + ∣37 − 30∣ + ∣40 − 30∣ + ∣20 − 30∣ + ∣33 − 30∣ + ∣27 − 30∣

⇒ Σ∣xi​ − x̅∣ = 9 + 4 + 7 + 9 + 4 + 7 + 10 + 10 + 3 + 3

⇒ Σ∣xi​ − x̅∣ = 66
Mean deviation from mean = \(\displaystyle \frac{Σ∣x_i​ − x̅ ∣​}{10}\) = \(\displaystyle \frac{66}{10}\) ​= 6.6
Coefficient of mean deviation = \(\displaystyle \frac{6.6}{30}\) = 0.22

∴ Coefficient of mean deviation = 0.22

Concept:

To find the mean deviation, use the following steps

Step 1: Find the mean value for the given data values

Step 2: Now, subtract mean value form each of the data value given (Note: Ignore the minus symbol)

Step 3: Now, find the mean of those values obtained in step 2.

Mean deviation \(\rm = \dfrac {1}{n} \sum |x - \bar x|\)

Calculations:

Given data is 2, 9, 9, 3, 6, 9, 4 

⇒ n = 7

Mean \(\rm \bar x = \dfrac {2+9+9+3+6+9+4}{7}\)

⇒  \(\rm \bar x = 6\)

Mean deviation 

\(\rm = \dfrac {1}{n} \sum |x - \bar x|\;\)

⇒ Mean deviation 

\(\rm = \dfrac {1}{7} {(|2-6|+|9-6|+|9-6|+|3-6|+ |6-6|+|9-6|+|4-6|)}\)

⇒ Mean deviation =  2.57

Hence, the mean deviation of the data 2, 9, 9, 3, 6, 9, 4 is 2.57

Formula used

Mean Deviation about median= \(∑\rm \frac{|x_{i} - M|}{n}\)

Where, 

x_i = individual term 

M = Median

n = total number of terms

Calculation:

52, 56, 66, 70, 75, 80, 82

Median = 70

⇒  M = 70

⇒ The value of (x_i - M) are

⇒ -18, -14, -4, 0, 5, 10, 12

⇒ Mean Deviation = \(\rm \frac{18 + 14 + 4 + 0 + 5 + 10 + 12}{7}\)

∴ The mean deviation about median is 9

Important Points

The mean is the average of a data set.

The median is the middle of the set of numbers.

Mean deviation about mean 

\(∑\rm \frac{|x_{i} - x̅|}{n}\) where x̅ = mean 

x_i = individual term 

n = total number of terms

Concept:

Deviation = |data value - mean|

Calculation:

Let, data value be \(\rm a_1, a_{2},a_{3},....,a_n\)

\(\left(a_{1}-2.5\right)+\left(a_{2}-2.5\right)+\left(a_{3}-2.5\right)+\ldots+\left(a_{n}-2.5\right) =50\)

\(\left(a_1+a_{2}+a_{3}+.....+a_n\right)-2.5(n)=50\)

\( \left(a_1+a_{2}+a_{3}+.....+a_n\right) = 50+2.5n \) and 

\(\left(a_{1}-3.5\right)+\left(a_{2}-3.5\right)+\left(a_{n}-3.5\right)=-50\)

\(\left(a_1+a_{2}+a_{3}+.....+a_n\right)-3.5(n)=-50\)

\(\left(a_1+a_{2}+a_{3}+.....+a_n\right) = -50+3.5n\)

Now,

-50 + 3.5n = 50 + 2.5n

⇒ n = 100

⇒ n = 100

Hence, option (4) is correct. 

Concept:

• Mean of a distribution = \(∑ f_i x_i \over ∑ f_i\) 
• Mean deviation about mean x̅ = \({∑ f_i|x_i -\overline x | \over ∑ f_i}\)

where fi are frequencies and xi are the mid points of class

Calculation:

Calculating mean

⇒ Mean = \(∑ f_i x_i \over ∑ f_i\) = \(690 \over 20\) = 34.5

CONCEPT:

Mean Deviation for ungrouped data:

For ‘n’ observation x₁, x₂………….. x_n, the mean deviation about their mean \(\bar x\) is given by \(M.D= \frac{{\mathop \sum \nolimits_{i = 1}^n \left| {{x_i} - \bar x} \right|}}{N}\).where N is the number of observations

CALCULATIONS:

Given data of numbers are p, 6, 6, 7, 8, 11, 15, and 16.

Mean \(\rm\bar x = \frac{{Sum\;of\;all\;the\;observations}}{{Total\;number\;of\;observations}} = \frac{{p + 6 + 6 + 7 + 8 + 11 + 15 + 16}}{8} = 3p\)

⇒ p + 69 = 24p

⇒  23p = 69                  

∴  p = 3

So given data is 3, 6, 6, 7, 8, 11, 15, 16 and mean is 3p, i.e. 9.

∴ Mean deviation \(= \frac{{\left| {3 - 9} \right| + \left| {6 - 9} \right| + \left| {6 - 9} \right| + \left| {7 - 9} \right| + \left| {8 - 9} \right| + \left| {11 - 9} \right| + \left| {15 - 9} \right| + \left| {16 - 9} \right|}}{8}\)