Answer to Question #277691 in Statistics and Probability for DAVID

Question

Find the Range, Standard Deviation, and Variance, for the given data samples. 1. 93, 67, 49, 55, 92, 87, 77, 66, 73, 96, 54 2. 48, 91, 87, 93, 59, 68, 92, 100, 81 3. 8, 6, 8, 6, 8, 6, 8, 6, 8, 6, 8, 6, 8 4. -8, -5, -12, -1, 4, 7, 11 5. -23, -17, -19, -5, -4, -11, -31

Answer 1

1.

$$49, 54, 55, 66, 67, 73, 77, 87, 92, 93, 96$$

$n=11$

$$Range=96-49=47$$

$$mean=\dfrac{1}{n}\sum _i x_i=\dfrac{1}{11}(49+54+55+66+67$$ $$+73+77+87+92+93+96)=\dfrac{809}{11}$$

$$Var(X)=s^2=\dfrac{1}{n-1}\sum (x_i-\bar{x})^2$$

$$=\dfrac{1}{11-1}((49-\dfrac{809}{11})^2+(54-\dfrac{809}{11})^2$$

$$+(55-\dfrac{809}{11})^2+(66-\dfrac{809}{11})^2+(67-\dfrac{809}{11})^2$$

$$+(73-\dfrac{809}{11})^2+(77-\dfrac{809}{11})^2+(87-\dfrac{809}{11})^2$$

$$+(92-\dfrac{809}{11})^2+(93-\dfrac{809}{11})^2+(96-\dfrac{809}{11})^2)$$

$$\approx284.472727$$

$$s=\sqrt{s^2}\approx\sqrt{284.472727}\approx16.866319$$

2.

$$48, 59, 68, 81, 87, 91, 92, 93, 100$$

$n=9$

$$Range=100-48=52$$

$$mean=\dfrac{1}{n}\sum _i x_i=\dfrac{1}{9}(48+59+68+81+87$$ $$+91+92+93+100)=\dfrac{719}{9}$$

$$Var(X)=s^2=\dfrac{1}{n-1}\sum (x_i-\bar{x})^2$$

$$=\dfrac{1}{9-1}((48-\dfrac{719}{9})^2+(59-\dfrac{719}{9})^2$$

$$+(68-\dfrac{719}{9})^2+(81-\dfrac{719}{9})^2+(87-\dfrac{719}{9})^2$$

$$+(91-\dfrac{719}{9})^2+(92-\dfrac{719}{9})^2+(93-\dfrac{719}{9})^2)$$

$$+(100-\dfrac{719}{9})^2)\approx 311.611111$$

$$s=\sqrt{s^2}\approx\sqrt{ 311.611111}\approx17.652510$$

3.

$$6, 6, 6, 6, 6, 6, 8, 8, 8, 8, 8, 8, 8$$

$n=13$

$$Range=8-6=2$$

$$mean=\dfrac{1}{n}\sum _i x_i=\dfrac{1}{13}(6+6+6+6+6$$ $$+6+8+8+8+8+8+8+8)=\dfrac{92}{13}$$

$$Var(X)=s^2=\dfrac{1}{n-1}\sum (x_i-\bar{x})^2$$

$$=\dfrac{1}{13-1}((6-\dfrac{92}{13})^2+(6-\dfrac{92}{13})^2$$

$$+(6-\dfrac{92}{13})^2+(6-\dfrac{92}{13})^2+(6-\dfrac{92}{13})^2$$

$$+(6-\dfrac{92}{13})^2+(8-\dfrac{92}{13})^2+(8-\dfrac{92}{13})^2$$

$$+(8-\dfrac{92}{13})^2+(8-\dfrac{92}{13})^2+(8-\dfrac{92}{13})^2$$

$$+(8-\dfrac{92}{13})^2+(8-\dfrac{92}{13})^2\approx 1.076923$$

$$s=\sqrt{s^2}\approx\sqrt{1.076923}\approx1.037749$$

4.

$$-12, -8, -5, -1, 4, 7, 11$$

$n=7$

$$Range=11-(-12)=23$$

$$mean=\dfrac{1}{n}\sum _i x_i=\dfrac{1}{7}(-12-8-5-1+4$$ $$Var(X)=s^2=\dfrac{1}{n-1}\sum (x_i-\bar{x})^2$$

$$=\dfrac{1}{7-1}((-12+\dfrac{4}{7})^2+(-8+\dfrac{4}{7})^2$$

$$+(-5+\dfrac{4}{7})^2+(-1+\dfrac{4}{7})^2+(4+\dfrac{4}{7})^2$$

$$+(7+\dfrac{4}{7})^2+(11+\dfrac{4}{7})^2)\approx 69.619048$$

$$s=\sqrt{s^2}\approx\sqrt{69.619048}\approx8.343803$$

5.

$$-31, -23, -19, -17, -11, -5, -4$$

$n=7$

$$Range=-4-(-31)=27$$

$$mean=\dfrac{1}{n}\sum _i x_i=\dfrac{1}{7}(-31-23-19-17-11$$ $$-5-4)=-\dfrac{110}{7}$$

$$Var(X)=s^2=\dfrac{1}{n-1}\sum (x_i-\bar{x})^2$$

$$=\dfrac{1}{7-1}((-31+\dfrac{110}{7})^2+(-23+\dfrac{110}{7})^2$$

$$+(-19+\dfrac{110}{7})^2+(-17+\dfrac{110}{7})^2+(-11+\dfrac{110}{7})^2$$

$$+(-5+\dfrac{110}{7})^2+(-4+\dfrac{110}{7})^2)\approx 95.571429$$

$$s=\sqrt{s^2}\approx\sqrt{ 95.571429}\approx 9.776064$$