Chapter 11.3, Problem 3Q

To determine

To find the Range and Standard Deviation

Answer to Problem 3Q

Range = Male > Female

Standard Deviation = Male > Female

Explanation of Solution

Given information:Female: $6,2,4,3,4$ and Male: $5,3,6,6,9$

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:

Female: $6,2,4,3,4$ and Male: $5,3,6,6,9$

So, In Ascending Order

Female: $2,3,4,4,6$ and Male: $3,5,6,6,9$

1) Female

• Range

Now,

Range = Highest − Lowest

So,

Range = $6-2$

Hence,

Range = $4$

• Mean

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

So,

Mean = $\frac{2+3+4+4+6}{5}$

Now,

Mean = $\frac{19}{5}$

Hence,

Mean $\mu =3.8$

• Standard Deviation

  $x_i=2,3,4,4,6$

and

Mean $\mu =3.8$

So,

  $\left|x_i-\mu \right|=1.8,0.8,0.2,0.2,2.2$

Now,

  ${\left(\left|x_i-\mu \right|\right)}^2=3.24,0.64,0.04,0.04,4.84$

So,

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

Now,

Standard Deviation = $\sqrt{\frac{3.24+0.64+0.04+0.04+4.84}{5}}$

So,

Standard Deviation = $\sqrt{\frac{8.8}{5}}$

Now,

Standard Deviation = $\sqrt{1.76}$

Hence,

Standard Deviation = $1.33$

2) Male

• Range

Now,

Range = Highest − Lowest

So,

Range = $9-3$

Hence,

Range = $6$

• Mean

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

So,

Mean = $\frac{3+5+6+6+9}{5}$

Now,

Mean = $\frac{29}{5}$

Hence,

Mean $\mu =5.8$

• Standard Deviation

  $x_i=3,5,6,6,9$

and

Mean $\mu =5.8$

So,

  $\left|x_i-\mu \right|=2.8,0.8,0.2,0.2,3.2$

Now,

  ${\left(\left|x_i-\mu \right|\right)}^2=7.84,0.64,0.04,0.04,10.24$

So,

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

Now,

Standard Deviation = $\sqrt{\frac{7.84+0.64+0.04+0.04+10.24}{5}}$

So,

Standard Deviation = $\sqrt{\frac{18.8}{5}}$

Now,

Standard Deviation = $\sqrt{3.76}$

Hence,

Standard Deviation = $1.94$