Chapter 11.3, Problem 4Q

To determine

To find the Range and Standard Deviation

Answer to Problem 4Q

Range = Seniors>Juniors

Standard Deviation = Seniors > Juniors

Explanation of Solution

Given information: Juniors: $19,15,20,10,14,21,18,15$ and Seniors: $22,19,29,32,15,26,30,19$

Formula Used: Range = Highest − Lowest, Standard Deviation = $\sqrt{\frac{\sum {\left(\left|x_i-\mu \right|\right)}^2}{N}}$ and Mean $\mu =\frac{Sumofallobservations}{Totalobservations}$

Calculation:

Juniors: $19,15,20,10,14,21,18,15$ and Seniors: $22,19,29,32,15,26,30,19$

So, In Ascending Order

Juniors: $10,14,15,15,18,19,20,21$ and Seniors: $15,19,19,22,26,29,30,32$

1) Juniors

• Range

Now,

Range = Highest − Lowest

So,

Range = $21-10$

Hence,

Range = $11$

• Mean

Mean = $\frac{Sumofallobservations}{Totalobservations}$

So,

Mean = $\frac{10+14+15+15+18+19+20+21}{8}$

Now,

Mean = $\frac{132}{8}$

Hence,

Mean $\mu =16.5$

• Standard Deviation

  $x_i=10,14,15,15,18,19,20,21$

and

Mean $\mu =16.5$

So,

  $\left|x_i-\mu \right|=6.5,2.5,1.5,1.5,1.5,2.5,3.5,4.5$

So,

Standard Deviation = $\sqrt{\frac{\sum {\left(\left|x_i-\mu \right|\right)}^2}{N}}$

Now,

Standard Deviation = $\sqrt{\frac{42.25+6.25+2.25+2.25+2.25+6.25+12.25+20.25}{8}}$

So,

Standard Deviation = $\sqrt{\frac{94}{8}}$

Now,

Standard Deviation = $\sqrt{11.75}$

Hence,

Standard Deviation = $3.43$

2) Seniors

• Range

Now,

Range = Highest − Lowest

So,

Range = $32-15$

Hence,

Range = $17$

• Mean

Mean = $\frac{Sumofallobservations}{Totalobservations}$

So,

Mean = $\frac{15+19+19+22+26+29+30+32}{8}$

Now,

Mean = $\frac{192}{8}$

Hence,

Mean $\mu =24$

• Standard Deviation

  $x_i=15,19,19,22,26,29,30,32$

and

Mean $\mu =24$

So,

  $\left|x_i-\mu \right|=9,5,5,2,2,5,6,8$

So,

Standard Deviation = $\sqrt{\frac{\sum {\left(\left|x_i-\mu \right|\right)}^2}{N}}$

Now,

Standard Deviation = $\sqrt{\frac{81+25+25+4+4+25+36+64}{8}}$

So,

Standard Deviation = $\sqrt{\frac{264}{8}}$

Now,

Standard Deviation = $\sqrt{33}$

Hence,

Standard Deviation = $5.74$