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# Math Examples of Descriptive Statistics Review

• ## The following table of data shown marks (out of 10) obtains by 25 students. Marks 1 2 3 4 5 6 7 8 9 10 Frequency 0 3 2 5 1 5 4 3 1 1 Find the dispersion of given data in table?

Dispersion of any data is measure in the term of variance and Standard Deviation. For finding the standard deviation we use the formula -
Standard Deviation (σ) = √ VARIANCE and
VARIANCE (σ2) = (∑ fx2  / ∑f ) - (∑ fx  / ∑f )2
Where µ = mean = sum of data / no. of event
Step 1: find the fx and fx2
fx 0 6 6 20 5 30 28 24 9 10 Fx2 0 36 36 400 25 900 784 576 81 100
Step 2:  now we measure the variance of data
∑f = 0+3+2+5+1+5+4+3+1+1 = 25
∑fx = 0+6+6+20+5+30+28+24+9+10 = 138
∑fx2 = 0+36+36+400+25+900+784+576+81+100 = 2938
VARIANCE (σ2) = (∑ fx2  / ∑f ) - (∑ fx  / ∑f )2
=  (2938/25) – (138/25)2
= 117.52 – (5.52)2
= 117.52-30.47 = 87.05
Step 3:  Standard Deviation (σ) = √ VARIANCE
= √ 87.05= 9.33
Dispersion of the given data in the term of variance and standard are 87.05 and 9.33.

## Find the dispersion of the following data of percentages of student pass in different class 82, 70, 73, 78, 81, 84, 100?

Dispersion of any data is measure in the term of variance and Standard Deviation. For finding the standard deviation we use the formula -
Standard Deviation (σ) = √ VARIANCE and
VARIANCE (σ2) = ∑ (X- µ)2 / N
Where µ = mean = sum of data / no. of event
Step 1: First we find the Mean of data
µ = (82+70+73+78+81+84+100) / 7
= 568/7= 81.14
Step 2: Now we measure the variance of Set of given data
(X- µ) 0.86 -11.14 8.14 3.14 0.14 2.86 18.86 (X- µ)2 0.73 124.09 66.25 9.86      0.019 8.18 355.69
VARIANCE (σ2) = ∑ (X- µ)2 / N
= (0.73+124.09+66.25+9.86+.019+8.18+355.69) / 7
= 564.819/7 = 80.68
Step 3:  Standard Deviation (σ) = √ VARIANCE
= √ 80.68 = 8.98
Dispersion of the given data in the term of variance and standard are 80.68 and 8.98.

## Find the dispersion of the student study per week in hours 35, 38, 40, 45, 50, 30?

Dispersion of any data is measure in the term of variance and Standard Deviation. For finding the standard deviation we use the formula -
Standard Deviation (σ) = √ VARIANCE and
VARIANCE (σ2) = ∑ (X- µ)2 / N
Where µ = mean = sum of data / no. of event
Step 1: first we find the Mean of data
µ = (35+38+40+45+50+30/ 6
=238/6 = 39.67
Step 2: Now we measure the variance of Set of given data
(X- µ) -4.67 -1.67 0.33 5.33 10.33 -9.67 (X- µ)2 21.81 2.79 0.11 28.41 106.71 93.51
VARIANCE (σ2) = ∑ (X- µ)2 / N
= (21.81+2.79+.11+28.41+106.71+93.51) / 6
= 106.7089/6= 17.78
Step 3:  Standard Deviation (σ) = √ VARIANCE
= √ 17.78= 4.21
Dispersion of the given data in the term of variance and standard are 17.78 and 4.21.

## Find the dispersion of the following data 30, 50, 40, 60, 80, 70, 55, 65?

Dispersion of any data is measure in the term of variance and Standard Deviation. For find the standard deviation we used the formula -
Standard Deviation (σ) = √ VARIANCE and
VARIANCE (σ2) = ∑ (X- µ)2 / N
Where µ = mean = sum of data / no. of event
Step 1: first we find the Mean of data
µ = (30+50+40+60+80+70+55+65) / 8
= 450/8 = 56.25
Step 2: now we measure the variance of Set of given data
(X- µ) -26.25 6.25 16.25 3.75 23.75 13.75 -1.25 8.75 (X- µ)2 689.06 39.06 264.06 14.06 564.06 189.06 1.56 76.56
VARIANCE (σ2) = ∑ (X- µ)2 / N
= (689.06+39.06+264.06+14.06+564.06+189.06+1.56+76.56) / 8
= 1837.48/ 8 = 229.68
Step 3:  Standard Deviation (σ) = √ VARIANCE
= √ 229.68= 15.16.
Dispersion of the given data in the term of variance and standard are 229.68 and 15.16.

## Find the dispersion of the following data 14, 10, 15, 7, 40, 16?

Dispersion of any data is measure in the term of variance and Standard Deviation. For finding the standard deviation we use the formula -
Standard Deviation (σ) = √ VARIANCE and
VARIANCE (σ2) = ∑ (X- µ)2 / N
Where µ = mean = sum of data / no. of event
Step 1: First we find the Mean of data
µ = (14+10+15+7+40+16) / 6
= 102/6 = 17
Step 2: Now we measure the variance of Set of given data
(X- µ) -3 -7 -2 -10 -23 -1 (X- µ)2 9 49 4 100 529 1
VARIANCE (σ2) = ∑ (X- µ)2 / N
= (9+49+4+100+529+1) / 6
= 692 / 6 = 115.33
Step 3:  Standard Deviation (σ) = √ VARIANCE
= √115.33= 10.74
Dispersion of the given data in the term of variance and standard are 115.33 and 10.74.

## Find the dispersion of the following data 2, 4, 5, 8, 3, 6, 12, 8?

Dispersion of any data is measure in the term of variance and Standard Deviation. For finding the standard deviation we use the formula -
Standard Deviation (σ) = √ VARIANCE and
VARIANCE (σ2) = ∑ (X- µ)2 / N
Where µ = mean = sum of data / no. of event
Step 1: First we find the Mean of data
µ = (2+4+5+8+3+6+12+8) / 8
= 48/8 = 6
Step 2: now we measure the variance of Set of given data
(X- µ) -4 -2 -1 2 3 0 6 2 (X- µ)2 16 4 1 4 9 0 36 4
VARIANCE (σ2) = ∑ (X- µ)2 / N
= (16+4+1+4+9+0+36+4) / 8
= 74 / 8 = 9.25
Step 3:  Standard Deviation (σ) =  √ VARIANCE
= √ 9.25 = 3.04.
Dispersion of the given data in the term of variance and standard are 9.25 and 3.04.

## Find Standard deviation of the given sample data (4, 6, 8, 10, 12, 14) using mean method?

In this question the total number of observations or data points in the sample is given as n = 6.
Y (Y – Y’)2 4 (4 – 9)2 = 25 6 (6 – 9)2 = 9 8 (8 – 9)2 = 1 10 (10 -9)2 = 1 12 (12 – 9)2 = 9 14 (14 - 9)2 = 25 ∑ Y = 54 ∑ (Y – Y’)2 = 70
So the Mean value is Y’ = Sum of all observations / total number of observation,
Y’= ∑Y / n,
Y’ = 54/6,
Y’= 9.
Standard Deviation (S) = √ [∑ (Y – Y’)2]/n,
S = √(70 / 6),
S = 3.42.
So the Standard Deviation of this sample is 3.42.

## Given a data set ‘S’ having many data values which are: S = (1, 2, 3, 2, 2, 4, 4, 4, 5, 3). Find the standard deviation of the given set?

This is one of the Standard Deviation Examples. In this Given data Set is S = (1, 2, 3, 2, 2, 4, 4, 4, 5, 3) and total number of data values N = 10, now to find standard deviation of this we first need to calculate the average or Mean value of this data set as:
Mean (Y’) = Sum of all observations/total number of observation
Mean (Y’) = ∑Y / n,
Mean (Y’) = (1 + 2 + 3 + 2 + 2 + 4 + 4 + 4 + 5 + 3) / 10,
Mean (Y’) = 30/10,
Mean (Y’) = 3,
Now to calculate standard deviation we need to calculate the variance of each observation as: (Y – Y’)
(1 - 3)2 = (-2)2 = 4,
(2 - 3)2 = (-1)2 = 1,
(3 - 3)2 = (0)2 = 0,
(2 - 3)2 = (-1)2 = 1,
(2 - 3)2= (-1)2 = 1,
(4 - 3)2 = (1)2 = 1,
(4 - 3)2 = (1)2 = 1,
(4 - 3)2 = (1)2 = 1,
(5 - 3)2 = (2)2 = 4,
(3 - 3)2 = (0)2 = 0,
So here we found the variance of each observation now we need to calculate the average of these variances as:
Average of variance =∑ (Y –Y’)2/n,
= (4 + 1 + 0 + 1 + 1 + 1 + 1 + 1 + 4 + 0) / 10,
= 14/10.
Now we will take Square root of the average value of the above:
Standard Deviation (S) = Square root of the average of variance = sqrt [∑ (Y –Y’)2/n],
Standard Deviation   (S) = √ (14/10),
Standard Deviation   (S) = √ (1.4),
So here standard deviation is the √1.4.

## Calculate the standard deviation for the given data: 15, 9, 18, 3, 8, 11, 14, 16, 4, 10 and 2?

We know that the formula for Standard Deviation is:
S = √∑(x – x’)2
N
where‘s’ is the standard deviation,
x is value in the data Set;
x’ is the Mean of the values;
N is the number of the values.
Now we can Calculate the standard deviation step by step:
For finding the standard deviation it is necessary to find the mean to the given data.
The formula for the finding the mean is:
X = ∑x,
N
Here in this above formula we can also solve the value of sigma or we can say the sum of all the given data.
= x1+ x2+ x3 + x4 …. + xN
N
= 15 + 9 + 18 + 3 + 8 + 11 + 14 + 16 + 4 + 10 + 2;
11
On further solving we get:
= 110,
11
= 10;
So the mean value is 10;
Now we calculate x – x’ from the given data:
X1 – x = 15 – 10 = 5;
X2 – x = 9 – 10 = -1;
X3 – x = 18 – 10 = -8;
X4 – x = 3 – 10 = -7;
X5 – x = 8 – 10 = -2;
X6 – x = 11 – 10 = 1;
X7 – x = 14 – 10 = 4;
X8 – x = 16 – 10 = 6;
X9 – x = 4 – 10 = -6;
X10 – x = 10 – 10 = 0;
X11 – x = 2 – 10 = -8;

Now we have to calculate ∑(X – x’)2;
∑(X – x’)2 = (X1 – x’)2 + (X2 – x’)2 +… (Xn – x’)2

= (5)2 + (-1)2 + (-8)2 + (-7)2 + (-2)2 + (1)2 + (4)2 + (6)2 + (-6)2 + (0)2 +
(-8)2;
= 25 + 1 + 64 + 49 + 4 + 1 + 16 + 36 + 36 + 0 + 64 = 296;
Now put all the values in the formula:
S = √∑(x – x’)2,
N
= √ 296,
11
= √296
11
On further solving we get √32.88
On further solving the value we get:
= 5.73
So the value of standard deviation is 5.73.

## Calculate the standard divination for the given data: 5, 9, 8, 3, 7, 1 and 2?

We know that the formula for Standard Deviation is:
S = √∑(x – x’)2
N
In the given standard deviation formula‘s’ is the standard deviation,
x is value in the data Set;
x’ is the Mean of the values;
‘N’ is the number of the values.
Now we can Calculate the standard deviation step by step:
For finding the standard deviation it is necessary to find the mean to the given data.
The formula for the finding the mean is:
X = ∑x,
N
Here in this above formula we can also solve the value of sigma or we can say the sum of all the given data.
= x1+ x2+ x3 + x4 …. + xN,
N
= 5 + 9 + 8 + 3 + 7 + 1 + 2;
7
On further solving we get:
= 35
7
= 5;
So the mean value is 5;
Now we calculate x – x’ from the given data:
X1 – x = 5 – 5 = 0;
X2 – x = 9 – 5 = 4;
X3 – x = 8 – 5 = 3;
X4 – x = 3 – 5 = -2;
X5 – x = 7 – 5 = 2;
X6 – x = 1 – 5 = -4;
X7 – x = 2 – 5 = -3;
Now we have to calculate ∑(X – x’)2;
∑(X – x’)2 = (X1 – x’)2 + (X2 – x’)2 +… (Xn – x’)2,

= (0)2 + (4)2 + (3)2 + (-2)2 + (2)2 + (-4)2 + (-3)2;
= 0 + 16 + 9 + 4 + 4 + 16 + 9 = 58;
Now put all the values in the standard deviation formula:
S = √∑(x – x)2,
N
= √ 58,
7
= √58,
7
On further solving we get:
= √8.2
On further solving the value we get:
= 2.86
So the value of standard deviation is 2.86.

## Calculate the standard deviation for values 5, 4, 8, 10, 9, 4 and 2?

We know that the formula for Standard Deviation is:
S = √∑(x – x’)2
N
Where, ‘s’ is the standard deviation,
‘x’ is value in the data Set;
x’ is the Mean of the values;
‘N’ is the number of the values.
Now we will Calculate the standard deviation step by step:
For finding the standard deviation, it is necessary to find the mean of the given data.
The formula for the finding the mean is:
X’ = ∑x,
N
or
x’ = x1+ x2+ x3 + x4 …. + xN
N
= 5 + 4 + 8 + 10 + 9 + 2 + 4;
7
On further solving we get:
= 42
7
= 6;
So the mean value is 6;
Now we calculate x – x’ from the given data:
X1 – x = 5 – 6 = -1;
X2 – x = 4 – 6 = -2;
X3 – x = 8 – 6 = 2;
X4 – x = 10 – 6 = 4;
X5 – x = 9 – 6 = 3;
X6 – x = 2 – 6 = -4;
X7 – x = 4 – 6 = -2;
Now we have to calculate ∑(X – x’)2;
∑(X – x’)2 = (X1 – x’)2 + (X2 – x’)2 +… (Xn – x’)2,

= (-1)2 + (-2)2 + (2)2 + (4)2 + (3)2 + (-3)2 + (-2)2;
= 1 + 4 + 4 + 16 + 9 + 16 + 4 = 54;
Now put all the values in the standard deviation formula:
S = √∑(x – x’)2,
N
= √ 54,
7 – 1
= √54,
6
On further solving the value we get:
= √9
So the value of standard deviation is 3.

## Find line of regression of ‘y’ on ‘x’. ‘y’ where we have 3 observation on price ‘X’ and supply and ‘y’ following data which is obtained ∑ x = 2 , ∑ y = 1, ∑ x2 = 20 , ∑ y2 = 50 , ∑ xy = 30?

For finding line of regression we need to follow the below steps.
Step 1: In first step we write the equation of regression of ‘y’ on ‘x’.
Y = a + bx.  Here ‘a’ and ‘b’ are regression values which are dependent on Linear Regression.
Step 2: Now we write the norms equation,
∑ y = a + b ∑x   , this is first equation of linear regression.
∑ xy = a∑x + b ∑x2  ,this is second equation of linear regression.
3a + 2b = 1      eq (1),
2a + 20b = 30      eq(2),
Step 3: In this step we solve the both equation (1) and (2).
Here for solving these equation first we are equal both coefficient value of a then we will solve. Now we subtract both equations.
a= 0.666,
b= 1.57,
Now we get final equation value Y = 0.666 + 1.57x.

## Find the regression value of the given set of data: (-1, 0), (2, 2), (5, 3), (6, 1) (1, 2)?

In the first step we write ‘x’ and ‘y’ value which are given in problem.
We write
x = -1, 2, 5, 6, 1,
And y = 0, 2, 3, 1, 2
Step 2: In this step we find the all summation value.
Summation of x = ∑x = 13,
Summation of y= ∑y = 8,
Summation of xy =∑xy = 27,
Summation of x2 = ∑ x2 = 67,
Step 3: Now we write Linear Regression formula and then we putting value.
a = (n∑ xy - ∑x ∑y)/n∑x2 – (∑x)2 , this is first equation of linear regression in this equation we will put summation value.
= (5*8 – 104)/5*67 – 169,
= 40 -104 / 166 = 64 / 166 = 0.385,
b = 1/n(∑y – a ∑x) this is second equation linear regression in this equation we will put the summation value.
= 1/5(8 - 0.385 *13) =1/5(8+ 5.01) = 13.01/5 = 2.602,
Now we get the value of a = 0.385, b = 2.602.

## Find the regression value of the given set of data: (2, 1), (3, 2), (3, -1)?

For solving regression value we need to follow the below steps.
Step 1: In the first step we write ‘x’ and ‘y’ value which are given in problem.
We write
x = 2, 3, 3,
And y = 1, 2, -1.
Step 2: In this step we find the all summation value.
Summation of x =∑x = 8,
Summation of y = ∑y = 2,
Summation of xy =∑xy = 2,
Summation of x2 = ∑ x2 = 6.
Step 3: Now we write the formula of regression and then we put the all summation value in regression equation.
a = (n∑ xy - ∑x ∑y)/n∑x2 – (∑x)2 , this is first equation of Linear Regression in this equation we will put summation value.
= (3*2 – 16)/3*6 – 64,
= -10 / 18 - 64 = -10/-46 = 4.6,
b = 1/n (∑y – a ∑x) this is second equation of linear regression in this equation we will put summation value.
= 1/3 (2 + 4.6*8) = 1/3 (2+ 36.8) = 38.8/3 = 12.9,
Now we get the value of a = 4.6, b = 12.9.

## Find the regression value of the given set of data: (-1, 0), (2, -1), (1, 2), (2, 1)?

For solving regression value we need to follow the below steps.
Step 1: In the first step we write the ‘x’ and ‘y’ value which are given in problem.
We write
x = -1, 2, 1, 2
And y = 0, -1, 2, 1,
Step 2: In this step we find the all summation value.
Summation of x = ∑x = 4,
Summation of y = ∑y = 2,
Summation of xy = ∑xy = 0,
Summation of x2 = ∑ x2 = 10,
Step 3: Now we write the formula of regression and then we put the all summation value in regression equation.
a = (n∑ xy - ∑x ∑y)/n∑x2 – (∑x)2, this is first equation of Linear Regression in this equation we will put summation value.
(4*0 – 8)/4*10 -16,
= -8 / 40-16 = -8 / 24 = -1/3 = -.33
b = 1/n (∑y – a ∑x), this is second equation of linear regression in this equation we will put summation value.
= ¼ (2 + .33*4) = (2+ 1.32) = 3.32 / 4 = .83,
Now we get the value of a = -.33, b = .83.
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