Chapter 4, Problem 4.8P

$\boxed{SOC}$ Data on several variables measuring overall health and well-being for 11 nations are reported here for 2010, with projection to 2020. Are nations becoming more or less diverse on these variables? Calculate the mean, range, and standard deviation for each year for each variable. Summarize the results in a paragraph.

	Life Expectancy (years)		Infant Mortality Rate*		Fertility Rate#
Nation	2010	2020	2010	2020	2010	2020
Canada	81	82	5.0	4.4	1.6	1.6
China	75	76	16.5	12.6	1.5	1.5
Egypt	72	75	26.2	17.9	3.0	2.7
Germany	79	81	4.0	3.6	1.4	1.5
Japan	82	83	2.8	2.7	1.2	1.3
Mali	52	57	113.7	91.9	6.5	5.5
Mexico	76	78	17.8	13.2	2.3	2.1
Peru	71	74	27.7	20.2	2.3	2.0
Ukraine	69	70	8.7	7.3	1.3	1.4
U.S	78	80	6.1	5.4	2.1	2.1
Zambia	52	54	68.4	50.6	6.0	5.3

Source: U.S. Bureau of the Census, 2012. Statistical Abstract of the United States: 2012. p. 842.

Notes:

*Number of deaths of children under one year of age per 1000 live births.

#Average number of children per female.

To determine

To find:

The range, mean and standard deviation of male and female of the given variable

Answer to Problem 4.8P

Solution:

The range, mean and standard deviation of the given variable is given in the table below,

	Life Expectancy		Infant Mortality Rate		Fertility Rate
	2010	2020	2010	2020	2010	2020
Range	30	29	110.9	89.2	5.3	4.2
Mean	71.54	73.63	26.99	20.89	2.65	2.45
Standard Deviation	9.99	9.31	32.74	25.91	1.77	1.44

Explanation of Solution

Given:

The following table shows the data of 3 different variables of 11 nations.

	Life Expectancy (years)		Infant Mortality Rate		Fertility Rate
Nation	2010	2020	2010	2020	2010	2020
Canada	81	82	5.0	4.4	1.6	1.6
China	75	76	16.5	12.6	1.5	1.5
Egypt	72	75	26.2	17.9	3.0	2.7
Germany	79	81	4.0	3.6	1.4	1.5
Japan	82	83	2.8	2.7	1.2	1.3
Mali	52	57	113.7	91.9	6.5	5.5
Mexico	76	78	17.8	13.2	2.3	2.1
Peru	71	74	27.7	20.2	2.3	2.0
Ukraine	69	70	8.7	7.3	1.3	1.4
U.S	78	80	6.1	5.4	2.1	2.1
Zambia	52	54	68.4	50.6	6.0	5.3

Formula used:

Let the data values be $X_i$’s.

The formula to calculate range is given by,

$\text{Range}=\text{High Score}-\text{Low score}$

The formula to calculate mean is given by,

$\overline{X}=\frac{{\displaystyle \sum _{i=1}^N{X}_i}}{N}$

The formula to calculate standard deviation is given by,

$s=\sqrt{\frac{{\displaystyle \sum {\left(X_i-\overline{X}\right)}^2}}{N}}$

Where, $N$ is the population size.

Calculation:

Consider the data of life expectancy in the year 2010.

Arrange the data in the increasing order.

The data in increasing order is given by,

S. No	Life Expectancy
1	52
2	52
3	69
4	71
5	72
6	75
7	76
8	78
9	79
10	81
11	82

The highest value is 82 and the lowest value is 52.

The range is given by,

$\text{Range}=\text{Highest value}-\text{Lowest value}$

Substitute 82 for highest value and 52 for lowest value in the above mentioned formula,

$\begin{array}{rll}\text{Range}& =& 82-52\\ & =& 30\end{array}$

The size of the population is 11.

The mean is given by,

$\overline{X}=\frac{{\displaystyle \sum _{i=1}^N{X}_i}}{N}$

Substitute 11 for $N$, 81 for $X_1$, 75 for $X_2$ and so on in the above mentioned formula,

$\begin{array}{rll}\overline{X}& =& \frac{81+75+\mathrm{...}+52}{11}\\ & =& \frac{787}{11}\\ & =& 71.54\end{array}$ ……$\left(1\right)$

Consider the following table of sum of squares,

$\left(X_i\right)$	$\left(X_i-\overline{X}\right)$	${\left(X_i-\overline{X}\right)}^2$
81	9.46	89.4916
75	3.46	11.9716
72	0.46	0.2116
79	7.46	55.6516
82	10.46	109.4116
52	$-19.54$	381.8116
76	4.46	19.8916
71	$-0.54$	0.2916
69	$-2.54$	6.4516
78	6.46	41.7316
52	$-19.54$	381.8116
	${\displaystyle \sum \left(X_i-\overline{X}\right)=0}$	${\displaystyle \sum {\left(X_i-\overline{X}\right)}^2=1098.728}$

From equation $\left(1\right)$, substitute $81$ for $X_1$ and $71.54$ for $\overline{X}$ in $\left(X_1-\overline{X}\right)$.

$\begin{array}{rll}\left(X_1-\overline{X}\right)& =& \left(81-71.54\right)\\ \left(X_1-\overline{X}\right)& =& 9.46\end{array}$

Square the both sides of the equation.

$\begin{array}{rll}{\left(X_1-\overline{X}\right)}^2& =& {\left(9.46\right)}^2\\ {\left(X_1-\overline{X}\right)}^2& =& 89.4916\end{array}$

Proceed in the same manner to calculate ${\left(X_i-\overline{X}\right)}^2$ for all the $1\le i\le N$ for the rest data and refer table for the rest of the ${\left(X_i-\overline{X}\right)}^2$ values calculated. Then the value of ${{\displaystyle \sum \left(X_i-\overline{X}\right)}}^2$ is calculated as,

$\begin{array}{rll}{{\displaystyle \sum \left(X_i-\overline{X}\right)}}^2& =& 89.4916+11.9716+\mathrm{...}+381.8116\\ & =& 1098.728\end{array}$ ……$\left(2\right)$

The standard deviation is given by,

$s=\sqrt{\frac{{\displaystyle \sum {\left(X_i-\overline{X}\right)}^2}}{N}}$

From equation $\left(2\right)$, substitute $1098.728$ for ${{\displaystyle \sum \left(X_i-\overline{X}\right)}}^2$ and 11 for $N$ in the above mentioned formula,

$\begin{array}{rll}s& =& \sqrt{\frac{1098.728}{11}}\\ & =& \sqrt{99.88433}\\ & \approx & 9.99\end{array}$

Thus, standard deviation of life expectancy in the year 2010 is 9.99.

Consider the data of life expectancy in the year 2020.

Arrange the data in the increasing order.

The data in increasing order is given by,

S. No	Life Expectancy
1	54
2	57
3	70
4	74
5	75
6	76
7	78
8	80
9	81
10	82
11	83

The highest value is 83 and the lowest value is 54.

The range is given by,

$\text{Range}=\text{Highest value}-\text{Lowest value}$

Substitute 83 for highest value and 54 for lowest value in the above mentioned formula,

$\begin{array}{rll}\text{Range}& =& 83-54\\ & =& 29\end{array}$

The size of the population is 11.

The mean is given by,

$\overline{X}=\frac{{\displaystyle \sum _{i=1}^N{X}_i}}{N}$

Substitute 11 for $N$, 82 for $X_1$, 76 for $X_2$ and so on in the above mentioned formula,

$\begin{array}{rll}\overline{X}& =& \frac{82+76+\mathrm{...}+54}{11}\\ & =& \frac{810}{11}\\ & =& 73.63\end{array}$ ……$\left(3\right)$

Consider the following table of sum of squares,

$\left(X_i\right)$	$\left(X_i-\overline{X}\right)$	${\left(X_i-\overline{X}\right)}^2$
82	8.37	70.0569
76	2.37	5.6169
75	1.37	1.8769
81	7.37	54.3169
83	9.37	87.7969
57	$-16.63$	276.5569
78	4.37	19.0969
74	0.37	0.1369
70	$-3.63$	13.1769
80	6.37	40.5769
54	$-19.63$	385.3369
	${\displaystyle \sum \left(X_i-\overline{X}\right)=0}$	${\displaystyle \sum {\left(X_i-\overline{X}\right)}^2=954.5459}$

From equation $\left(3\right)$, substitute $82$ for $X_1$ and $73.63$ for $\overline{X}$ in $\left(X_1-\overline{X}\right)$.

$\begin{array}{rll}\left(X_1-\overline{X}\right)& =& \left(82-73.63\right)\\ \left(X_1-\overline{X}\right)& =& 8.37\end{array}$

Square the both sides of the equation.

$\begin{array}{rll}{\left(X_1-\overline{X}\right)}^2& =& {\left(8.37\right)}^2\\ {\left(X_1-\overline{X}\right)}^2& =& 70.0569\end{array}$

Proceed in the same manner to calculate ${\left(X_i-\overline{X}\right)}^2$ for all the $1\le i\le N$ for the rest data and refer table for the rest of the ${\left(X_i-\overline{X}\right)}^2$ values calculated. Then the value of ${{\displaystyle \sum \left(X_i-\overline{X}\right)}}^2$ is calculated as,

$\begin{array}{rll}{{\displaystyle \sum \left(X_i-\overline{X}\right)}}^2& =& 70.0569+5.6169+\mathrm{...}+385.3369\\ & =& 954.5459\end{array}$ ……$\left(4\right)$

The standard deviation is given by,

$s=\sqrt{\frac{{\displaystyle \sum {\left(X_i-\overline{X}\right)}^2}}{N}}$

From equation $\left(4\right)$, substitute $954.5459$ for ${{\displaystyle \sum \left(X_i-\overline{X}\right)}}^2$ and 11 for $N$ in the above mentioned formula,

$\begin{array}{rll}s& =& \sqrt{\frac{954.5459}{11}}\\ & =& \sqrt{86.7769}\\ & \approx & 9.31\end{array}$

Thus, standard deviation of life expectancy in the year 2020 is 9.31.

Consider the data of infant mortality rate in the year 2010.

Arrange the data in the increasing order.

The data in increasing order is given by,

S. No	Infant Mortality Rate
1	2.8
2	4
3	5
4	6.1
5	8.7
6	16.5
7	17.8
8	26.2
9	27.7
10	68.4
11	113.7

The highest value is 113.7 and the lowest value is 2.8.

The range is given by,

$\text{Range}=\text{Highest value}-\text{Lowest value}$

Substitute 113.7 for highest value and 2.8 for lowest value in the above mentioned formula,

$\begin{array}{rll}\text{Range}& =& 113.7-2.8\\ & =& 110.9\end{array}$

The size of the population is 11.

The mean is given by,

$\overline{X}=\frac{{\displaystyle \sum _{i=1}^N{X}_i}}{N}$

Substitute 11 for $N$, 5 for $X_1$, 16.5 for $X_2$ and so on in the above mentioned formula,

$\begin{array}{rll}\overline{X}& =& \frac{5+16.5+\mathrm{...}+26.2}{11}\\ & =& \frac{296.9}{11}\\ & =& 26.99\end{array}$ ……$\left(5\right)$

Consider the following table of sum of squares,

$\left(X_i\right)$	$\left(X_i-\overline{X}\right)$	${\left(X_i-\overline{X}\right)}^2$
5	$-21.99$	483.5601
16.5	$-10.49$	110.0401
26.2	$-0.79$	0.6241
4	$-22.99$	528.5401
2.8	$-24.19$	585.1561
113.7	86.71	7518.624
17.8	$-9.19$	84.4561
27.7	0.71	0.5041
8.7	$-18.29$	334.5241
6.1	$-20.89$	436.3921
68.4	41.41	1714.788
	${\displaystyle \sum \left(X_i-\overline{X}\right)=0}$	${\displaystyle \sum {\left(X_i-\overline{X}\right)}^2=11797.21}$

WFrom equation $\left(5\right)$, substitute $5$ for $X_1$ and $26.99$ for $\overline{X}$ in $\left(X_1-\overline{X}\right)$.

$\begin{array}{rll}\left(X_1-\overline{X}\right)& =& \left(5-26.99\right)\\ \left(X_1-\overline{X}\right)& =& -21.99\end{array}$

Square the both sides of the equation.

$\begin{array}{rll}{\left(X_1-\overline{X}\right)}^2& =& {\left(-21.99\right)}^2\\ {\left(X_1-\overline{X}\right)}^2& =& 483.5601\end{array}$

Proceed in the same manner to calculate ${\left(X_i-\overline{X}\right)}^2$ for all the $1\le i\le N$ for the rest data and refer table for the rest of the ${\left(X_i-\overline{X}\right)}^2$ values calculated. Then the value of ${{\displaystyle \sum \left(X_i-\overline{X}\right)}}^2$ is calculated as,

$\begin{array}{rll}{{\displaystyle \sum \left(X_i-\overline{X}\right)}}^2& =& 483.5601+110.0401+\mathrm{...}+1714.788\\ & =& 11797.21\end{array}$ ……$\left(6\right)$

The standard deviation is given by,

$s=\sqrt{\frac{{\displaystyle \sum {\left(X_i-\overline{X}\right)}^2}}{N}}$

From equation $\left(6\right)$, substitute $11797.21$ for ${{\displaystyle \sum \left(X_i-\overline{X}\right)}}^2$ and 11 for $N$ in the above mentioned formula,

$\begin{array}{rll}s& =& \sqrt{\frac{11797.21}{11}}\\ & =& \sqrt{1072.474}\\ & \approx & 32.74\end{array}$

Thus, standard deviation of infant mortality rate in the year 2010 is 32.74.

Consider the data of infant mortality rate in the year 2020.

Arrange the data in the increasing order.

The data in increasing order is given by,

S. No	Infant Mortality Rate
1	2.7
2	3.6
3	4.4
4	5.4
5	7.3
6	12.6
7	13.2
8	17.9
9	20.2
10	50.6
11	91.9

The highest value is 91.9 and the lowest value is 2.7.

The range is given by,

$\text{Range}=\text{Highest value}-\text{Lowest value}$

Substitute 91.9 for highest value and 2.7 for lowest value in the above mentioned formula,

$\begin{array}{rll}\text{Range}& =& 91.9-2.7\\ & =& 89.2\end{array}$

The size of the population is 11.

The mean is given by,

$\overline{X}=\frac{{\displaystyle \sum _{i=1}^N{X}_i}}{N}$

Substitute 11 for $N$, 4.4 for $X_1$, 12.6 for $X_2$ and so on in the above mentioned formula,

$\begin{array}{rll}\overline{X}& =& \frac{4.4+12.6+\mathrm{...}+50.6}{11}\\ & =& \frac{229.8}{11}\\ & =& 20.89\end{array}$ ……$\left(7\right)$

Consider the following table of sum of squares,

$\left(X_i\right)$	$\left(X_i-\overline{X}\right)$	${\left(X_i-\overline{X}\right)}^2$
4.4	$-16.49$	271.9201
12.6	$-8.29$	68.7241
17.9	$-2.99$	8.9401
3.6	$-17.29$	298.9441
2.7	$-18.19$	330.8761
91.9	71.01	5042.42
13.2	$-7.69$	59.1361
20.2	$-0.69$	0.4761
7.3	$-13.59$	184.6881
5.4	$-15.49$	239.9401
50.6	29.71	882.6841
	${\displaystyle \sum \left(X_i-\overline{X}\right)=0}$	${\displaystyle \sum {\left(X_i-\overline{X}\right)}^2=7388.749}$

From equation $\left(7\right)$, substitute $4.4$ for $X_1$ and $20.89$ for $\overline{X}$ in $\left(X_1-\overline{X}\right)$.

$\begin{array}{rll}\left(X_1-\overline{X}\right)& =& \left(4.4-20.89\right)\\ \left(X_1-\overline{X}\right)& =& -16.49\end{array}$

Square the both sides of the equation.

$\begin{array}{rll}{\left(X_1-\overline{X}\right)}^2& =& {\left(-16.49\right)}^2\\ {\left(X_1-\overline{X}\right)}^2& =& 271.9201\end{array}$

Proceed in the same manner to calculate ${\left(X_i-\overline{X}\right)}^2$ for all the $1\le i\le N$ for the rest data and refer table for the rest of the ${\left(X_i-\overline{X}\right)}^2$ values calculated. Then the value of ${{\displaystyle \sum \left(X_i-\overline{X}\right)}}^2$ is calculated as,

$\begin{array}{rll}{{\displaystyle \sum \left(X_i-\overline{X}\right)}}^2& =& 271.9201+68.7241+\mathrm{...}+882.6841\\ & =& 7388.749\end{array}$ ……$\left(8\right)$

The standard deviation is given by,

$s=\sqrt{\frac{{\displaystyle \sum {\left(X_i-\overline{X}\right)}^2}}{N}}$

From equation $\left(8\right)$, substitute $7388.749$ for ${{\displaystyle \sum \left(X_i-\overline{X}\right)}}^2$ and 11 for $N$ in the above mentioned formula,

$\begin{array}{rll}s& =& \sqrt{\frac{7388.749}{11}}\\ & =& \sqrt{671.7045}\\ & \approx & 25.91\end{array}$

Thus, standard deviation of infant mortality in the year 2020 is 25.91.

Consider the data of fertility rate in the year 2010.

Arrange the data in the increasing order.

The data in increasing order is given by,

S. No	Fertility Rate
1	1.2
2	1.3
3	1.4
4	1.5
5	1.6
6	2.1
7	2.3
8	2.3
9	3
10	6
11	6.5

The highest value is 6.5 and the lowest value is 1.2.

The range is given by,

$\text{Range}=\text{Highest value}-\text{Lowest value}$

Substitute 6.5 for highest value and 1.2 for lowest value in the above mentioned formula,

$\begin{array}{rll}\text{Range}& =& 6.5-1.2\\ & =& 5.3\end{array}$

The size of the population is 11.

The mean is given by,

$\overline{X}=\frac{{\displaystyle \sum _{i=1}^N{X}_i}}{N}$

Substitute 11 for $N$, 1.6 for $X_1$, 1.5 for $X_2$ and so on in the above mentioned formula,

$\begin{array}{rll}\overline{X}& =& \frac{1.6+1.5+\mathrm{...}+6}{11}\\ & =& \frac{29.2}{11}\\ & =& 2.65\end{array}$ ……$\left(9\right)$

Consider the following table of sum of squares,

$\left(X_i\right)$	$\left(X_i-\overline{X}\right)$	${\left(X_i-\overline{X}\right)}^2$
1.6	$-1.05$	1.1025
1.5	$-1.15$	1.3225
3	0.35	0.1225
1.4	$-1.25$	1.5625
1.2	$-1.45$	2.1025
6.5	3.85	14.8225
2.3	$-0.35$	0.1225
2.3	$-0.35$	0.1225
1.3	$-1.35$	1.8225
2.1	$-0.55$	0.3025
6	3.35	11.2225
	${\displaystyle \sum \left(X_i-\overline{X}\right)=0}$	${\displaystyle \sum {\left(X_i-\overline{X}\right)}^2=34.6275}$

From equation $\left(9\right)$, substitute $1.6$ for $X_1$ and $2.65$ for $\overline{X}$ in $\left(X_1-\overline{X}\right)$.

$\begin{array}{rll}\left(X_1-\overline{X}\right)& =& \left(1.6-2.65\right)\\ \left(X_1-\overline{X}\right)& =& -1.05\end{array}$

Square the both sides of the equation.

$\begin{array}{rll}{\left(X_1-\overline{X}\right)}^2& =& {\left(-1.05\right)}^2\\ {\left(X_1-\overline{X}\right)}^2& =& 1.1025\end{array}$

Proceed in the same manner to calculate ${\left(X_i-\overline{X}\right)}^2$ for all the $1\le i\le N$ for the rest data and refer table for the rest of the ${\left(X_i-\overline{X}\right)}^2$ values calculated. Then the value of ${{\displaystyle \sum \left(X_i-\overline{X}\right)}}^2$ is calculated as,

$\begin{array}{rll}{{\displaystyle \sum \left(X_i-\overline{X}\right)}}^2& =& 1.1025+1.3225+\mathrm{...}+11.2225\\ & =& 34.6275\end{array}$ ……$\left(10\right)$

The standard deviation is given by,

$s=\sqrt{\frac{{\displaystyle \sum {\left(X_i-\overline{X}\right)}^2}}{N}}$

From equation $\left(10\right)$, substitute $34.6275$ for ${{\displaystyle \sum \left(X_i-\overline{X}\right)}}^2$ and 11 for $N$ in the above mentioned formula,

$\begin{array}{rll}s& =& \sqrt{\frac{34.6275}{11}}\\ & =& \sqrt{3.147955}\\ & \approx & 1.77\end{array}$

Thus, standard deviation of fertility rate in the year 2010 is 1.77.

Consider the data of fertility rate in the year 2020.

Arrange the data in the increasing order.

The data in increasing order is given by,

S. No	Fertility Rate
1	1.3
2	1.4
3	1.5
4	1.5
5	1.6
6	2
7	2.1
8	2.1
9	2.7
10	5.3
11	5.5

The highest value is 5.5 and the lowest value is 1.3.

The range is given by,

$\text{Range}=\text{Highest value}-\text{Lowest value}$

Substitute 5.5 for highest value and 1.3 for lowest value in the above mentioned formula,

$\begin{array}{rll}\text{Range}& =& 5.5-1.3\\ & =& 4.2\end{array}$

The size of the population is 11.

The mean is given by,

$\overline{X}=\frac{{\displaystyle \sum _{i=1}^N{X}_i}}{N}$

Substitute 11 for $N$, 1.6 for $X_1$, 1.5 for $X_2$ and so on in the above mentioned formula,

$\begin{array}{rll}\overline{X}& =& \frac{1.6+1.5+\mathrm{...}+5.3}{11}\\ & =& \frac{27}{11}\\ & =& 2.45\end{array}$ ……$\left(11\right)$

Consider the following table of sum of squares,

$\left(X_i\right)$	$\left(X_i-\overline{X}\right)$	${\left(X_i-\overline{X}\right)}^2$
1.6	$-0.85$	0.7225
1.5	$-0.95$	0.9025
2.7	0.25	0.0625
1.5	$-0.95$	0.9025
1.3	$-1.15$	1.3225
5.5	3.05	9.3025
2.1	$-0.35$	0.1225
2	$-0.45$	0.2025
1.4	$-1.05$	1.1025
2.1	$-0.35$	0.1225
5.3	2.85	8.1225
	${\displaystyle \sum \left(X_i-\overline{X}\right)=0}$	${\displaystyle \sum {\left(X_i-\overline{X}\right)}^2=22.8875}$

From equation $\left(11\right)$, substitute $1.6$ for $X_1$ and $2.45$ for $\overline{X}$ in $\left(X_1-\overline{X}\right)$.

$\begin{array}{rll}\left(X_1-\overline{X}\right)& =& \left(1.6-2.45\right)\\ \left(X_1-\overline{X}\right)& =& -0.85\end{array}$

Square the both sides of the equation.

$\begin{array}{rll}{\left(X_1-\overline{X}\right)}^2& =& {\left(-0.85\right)}^2\\ {\left(X_1-\overline{X}\right)}^2& =& 0.7225\end{array}$

Proceed in the same manner to calculate ${\left(X_i-\overline{X}\right)}^2$ for all the $1\le i\le N$ for the rest data and refer table for the rest of the ${\left(X_i-\overline{X}\right)}^2$ values calculated. Then the value of ${{\displaystyle \sum \left(X_i-\overline{X}\right)}}^2$ is calculated as,

$\begin{array}{rll}{{\displaystyle \sum \left(X_i-\overline{X}\right)}}^2& =& 0.7225+0.9025+\mathrm{...}+8.1225\\ & =& 22.8875\end{array}$ ……$\left(12\right)$

The standard deviation is given by,

$s=\sqrt{\frac{{\displaystyle \sum {\left(X_i-\overline{X}\right)}^2}}{N}}$

From equation $\left(12\right)$, substitute $22.8875$ for ${{\displaystyle \sum \left(X_i-\overline{X}\right)}}^2$ and 11 for $N$ in the above mentioned formula,

$\begin{array}{rll}s& =& \sqrt{\frac{22.8875}{11}}\\ & =& \sqrt{2.080682}\\ & \approx & 1.44\end{array}$

Thus, standard deviation of fertility rate in the year 2020 is 1.44.

The mean life expectancy in the year 2010 is 71.54 and in the year 2020 it is 73.63, on an average the life expectancy of the nations has increased over the ten year period. The standard deviation of life expectancy decreased in the ten year duration, there is less variability in 2020 as compared to year 2010. The mean infant mortality rate of year 2010 is more than the year 2020, the number of infant deaths has decreased during the time period. The mean fertility rate of year 2010 is 2.65 and in year 2020 it is 2.45, the number of children born has decreased between the years. The nations are becoming more diverse.

Conclusion:

Therefore, the range, mean and standard deviation of the given variable is given in the table below,

	Life Expectancy		Infant Mortality Rate		Fertility Rate
	2010	2020	2010	2020	2010	2020
Range	30	29	110.9	89.2	5.3	4.2
Mean	71.54	73.63	26.99	20.89	2.65	2.45
Standard Deviation	9.99	9.31	32.74	25.91	1.77	1.44