Q 3.62. - Elementary Statistics

Question

 Consider the following four data sets. 

a. Compute the mean of each data set.

b. Although the four data sets have the same means, in what respect are they quite different?

c. Which data set appears to have the least variation? the greatest variation?

d. Compute the range of each data set.

e. Use the defining formula to compute the sample standard deviation of each data set.

f. From your answers to parts (d) and (e), which measure of variation better distinguishes the spread in the four data sets: the range or the standard deviation? Explain your answer.

g. Are your answers from parts (c) and (e) consistent?

Step-by-Step Solution

Answer

Part a)All mean data is $5$

Part b) they are quite different in terms of variation. 

Part c)the deviation from the mean is the highest in that case. 

Part d)Range of 4 Datas is $8,8,0,8$

Part e)Sample S,D of 4 dates are $3.33,4.22,0,2.71$

Part f) does not give any information about the spread in the data. 

Part g)Dta 2 highest and data 3 least

Step 1:Given information

Step 2:Explaination Part a)

$\text{ Mean of data set }\mathrm{I}=\frac{(1+1+2+2+5+5+8+8+9+9)}{10}$

$=5$

$\text{ Mean of data set II }=\frac{(1+1+1+1+1+9+9+9+9+9)}{10}$

$=5$

$\text{ Mean of data set III }=\frac{(5+5+5+5+5+5+5+5+5+5)}{10}$

$=5$

$\text{ Mean of data set IV }=\frac{(2+4+4+4+4+4+4+4+10+10)}{10}$

$=5$

Step 2:Explaination Part b)

From the above, we observe that,

Although the four data sets have same means, they are quite different in terms of variation.

Step 2:Explaination Part c)

Data set III appears to have the last variation, since all the values in that data set are the same.

Data set II appears to have the greatest variation, since the deviation from the mean is the highest in that case.

Step 2:Explaination Part d)

$\text{ Range of data set }l=9-1=8$

$\text{ Range of data set II }=9-1=8$

$\text{ Range of data set III }=5-5=0$

$\text{ Range of data set IV }=10-2=8$

Step 2:Explaination Part e)

 For data set-l, 

$\sum {\left(x_i-\overline{x}\right)}^2\text{ of data set }\mathrm{I}=100$

$\therefore \text{ Sample standard deviation }=\sqrt{\frac{100}{9}}\left[\because \sqrt{\frac{\sum {\left(x_i-\overline{x}\right)}^2}{n-1}}\right]$

$=3.33333$

$\text{ Sample standard deviation of data set-l is, }3.33$

$\text{ For data set-ll }$

$\sum {\left(x_i-\overline{x}\right)}^2\text{ of data set }\Vert =160$

$\therefore \text{ Sample standard deviation }=\sqrt{\frac{160}{9}}$

$=4.21637$

$\text{ Sample standard deviation of data set-ll is, }4.22$

 For data set-III,

$\sum {\left(x_i-\overline{x}\right)}^2\text{ of data set III }=0$

$\text{ Sample standard deviation of data set-III is, }0$

For data set-IV, 

$\sum {\left(x_i-\overline{x}\right)}^2\text{ of data set }\mathrm{IV}=66$

$\therefore \text{ Sample standard deviation }=\sqrt{\frac{66}{9}}$

$=2.70801$

$\text{ Sample standard deviation of data set-IV is, }2.71$

Step 2:Explaination Part f)

Standard deviation is a better measure of variation which distinguishes the spread in the four data sets, since the range of the data sets I,II and IV is 8 and set III is zero which does not give any information about the spread in the data.

Step 2:Explaination Part g)

Yes, the answers from part (c) and part (e) are consistent, since in part (e) data set II has the highest standard deviation of$4.22$and data set III has the least standard deviation of 0