Chapter 3.2, Problem 11E

Multiple Births The numbers of various multiple births in the United States for the past 10 years are listed. Find the range, variance, and standard deviation of the data sets. Which set of data is the most variable?

Source: World Almanac 2012.

To determine

The range, variance and standard deviation of the given data; the set of data that is the most variable.

Answer to Problem 11E

The range of triplets birth, quadruplets birth quintuplets birth are 1,233, 167, and 45 respectively.

The variance of triplet birth, quadruplet birth, and, quintuplet birth are 291845.4, 3,268.44, and 162.69.

The standard deviation of triplets birth, quadruplets birth, and quintuplet birth are 540.22, 57.22, and 12.75.

The triplet birth data is most variable and quintuplet birth data is least variable.

Explanation of Solution

Given info:

The data for the Triplets birth are shown below,

5,877	7,110	5937
6,898	6,118	6,885
6,208	6,742	6,750
	6,742

The data for the Quadruplets birth is,

345	468	369
434	355	501
418	506	439
	512

The data for the quintuplets is,

46	85	91
69	67	85
68	77	86
	67

Calculation:

Formula to calculate the range is,

$\text{Range}=\text{highest}\text{value}-\text{lowest}\text{value}$

Substitute 7110 for highest value and 5877 for lowest value in the above formula to calculate the range of triplets birth.

$\begin{array}{c}\text{Range}=\text{7110-5877}\\ \text{=1233}\end{array}$

Substitute 512 for highest value and 345 for lowest value in the above formula to calculate the range of quadruplet’s birth.

$\begin{array}{c}\text{Range}=512-345\\ =167\end{array}$

Substitute 91 for the highest value and 46 for lowest value in the above formula to calculate the range of quintuplet’s birth.

$\begin{array}{c}\text{Range}=\text{91-46}\\ =\text{45}\end{array}$

Formula to calculate the variance is,

$\text{Variance}=\frac{{\displaystyle \sum {\left(X-\mu \right)}^2}}{N}$ (1)

Where,

• $X$ is individual value.
• $\mu$ is the population mean.
• $N$ is the population size.

Step to calculate the population variance of triplet’s birth,

Determine the mean of the data.

$\mu =\frac{{\displaystyle \sum X}}{N}$ (2)

Substitute the values of $X$ in the equation (2) to get the mean of the data.

$\begin{array}{c}\mu =\frac{5877+6898+6208+7110+6118+6742+6742+5937+6885+7650}{10}\\ =\frac{66167}{10}\\ =6616.7\end{array}$

Subtract the mean from each data value $\left(X-\mu \right)$.

Determine square each result ${\left(X-\mu \right)}^2$.

Determine sum of squares ${\displaystyle \sum {\left(X-\mu \right)}^2}$.

Divide the sum by $N$ to get the variance $\frac{{\displaystyle \sum {\left(X-\mu \right)}^2}}{N}$.

Table for calculating variance given below,

Values $X$	$(X-\mu )$	${\left(X-\mu \right)}^2$
5877	$5877-6616.7=-739.7$	547156.1
6898	$6898-6616.7=-282.3$	79129.69
6208	$6208-6616.7=-408.7$	167035.7
7110	$7110-6616.7=493.3$	243344.9
6118	$6118-6616.7=-498.7$	248701.7
6742	$6742-6616.7=125.3$	15700.09
6742	$6742-6616.7=125.3$	15700.09
5937	$5937-6616.7=-679.7$	461992.1
6885	$6885-6616.7=268.3$	71984.89
7650	$7650-6616.7=1033.3$	1067709
TOTAL		2918454

Substitute 2918454 for ${\displaystyle \sum {\left(X-\mu \right)}^2}$ and 10 for $N$ in the equation (1) to get the population variance,

$\begin{array}{c}\text{Variance}=\frac{2918454}{10}\\ =291845.4\end{array}$

Take the square root of the 2918454 to get the population standard deviation.

$\begin{array}{c}\text{Standard}\text{deviation}=\sqrt{291845.4}\\ \approx 540.22\end{array}$

Steps to calculate the population variance of quadruplet’s birth,

Determine the mean of the data,

Substitute the value $X$ in the equation (2) to get the mean of the data.

$\begin{array}{c}\mu =\frac{345+434+418+468+355+506+512+396+501+439}{10}\\ =\frac{4374}{10}\\ =437.4\end{array}$

Subtract the mean from each data value $\left(X-\mu \right)$.

Determine square each result ${\left(X-\mu \right)}^2$.

Determine sum of squares ${\displaystyle \sum {\left(X-\mu \right)}^2}$.

Divide the sum by $N$ to get the variance $\frac{{\displaystyle \sum {\left(X-\mu \right)}^2}}{N}$.

Table for calculating variance given below,

Values $X$	$(X-\mu )$	${\left(X-\mu \right)}^2$
345	$345-437.4=-92.4$	8537.76
434	$434-437.4=-3.4$	11.56
418	$418-437.4=-19.4$	376.36
468	$468-437.4=30.6$	936.36
355	$355-437.4=-82.4$	6789.76
506	$506-437.4=68.6$	4705.96
512	$512-437.4=74.6$	5565.16
396	$396-437.4=-41.4$	1713.96
501	$501-437.4=63.6$	4044.96
439	$439-437.4=1.6$	2.56
TOTAL		32684.4

Substitute 32684.4 for ${\displaystyle \sum {\left(X-\mu \right)}^2}$ and 10 for $N$ in equation (1) to get the population variance of the data,

$\begin{array}{c}\text{Variance}=\frac{32684.4}{10}\\ =3268.44\end{array}$

Take the square root of the variance which is 3268.44 to get the population standard deviation of the data.

$\begin{array}{c}\text{Standard}\text{deviation}=\sqrt{3268.44}\\ \approx 57.17\end{array}$

Steps to calculate the population variance of the quintuplet birth,

Determine the mean of the data

Substitute the values of the $X$ in the equation (2) to get the mean of the quintuplet birth,

$\begin{array}{c}\mu =\frac{46+69+68+85+67+77+67+91+85+86}{10}\\ =\frac{741}{10}\\ =74.1\end{array}$

Subtract the mean from each data value $\left(X-\mu \right)$.

Determine square each result ${\left(X-\mu \right)}^2$.

Determine sum of squares ${\displaystyle \sum {\left(X-\mu \right)}^2}$.

Divide the sum by $N$ to get the variance $\frac{{\displaystyle \sum {\left(X-\mu \right)}^2}}{N}$.

Table for calculating variance given below,

Values $X$	$(X-\mu )$	${\left(X-\mu \right)}^2$
46	$46-74.1=-28.1$	789.61
69	$69-74.1=-5.1$	26.01
68	$68-74.1=-6.1$	37.21
85	$85-74.1=10.9$	118.81
67	$67-74.1=-7.1$	50.41
77	$77-74.1=2.9$	8.41
67	$67-74.1=-7.1$	50.41
91	$91-74.1=16.9$	285.61
85	$85-74.1=10.9$	118.81
86	$86-74.1=11.9$	141.61
TOTAL		1626.9

Substitute 1626.9 for ${\displaystyle \sum {\left(X-\mu \right)}^2}$ and 10 for $N$ in the equation (1) to get the population, variance of the data.

$\begin{array}{c}\text{Variance}=\frac{1626.9}{10}\\ =162.69\end{array}$

Take the square root of the 162.69 to get the population standard deviation for the quintuplet birth data.

$\begin{array}{c}\text{Standard}\text{deviation}=\sqrt{162.69}\\ \approx 12.75\end{array}$

Thus, the population variance and population standard deviation for the quintuplets birth data is 162.69 and 12.75 respectively.

Standard deviation for triplet birth is 540.22 for quadruplets birth is 57.17 and for quintuplet birth is 12.75.

The maximum standard deviation gives more variability. So, triplet birth data is most variable and quintuplet birth data is least variable.