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# Math Examples of Numerical Descriptive Statistics

• ## 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.

## 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.
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