Answer to Question #216849 in Statistics and Probability for Yalou

Question

Find the range, the standard deviation, and the variance for the given samples. Round noninteger to the nearest tenth.

1. 48, 91, 87, 93, 59, 68, 92, 100, 81
2. 93, 67, 49, 55, 92, 87, 77, 66, 73, 96, 54
3. 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
4. 8, 6, 8, 6, 8, 6, 8, 6, 8, 6, 8, 6, 8
5. -8, -5, -12, -1, 4, 7, 11
6. -23, -17, -19, -5, -4, -11, -31

Answer 1

1.

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

$$Range=100-48=52$$

$$mean=\bar{x}=\dfrac{\displaystyle\sum_{i=1}^nx_i}{n}=\dfrac{1}{11}(48+59+68+81+87$$

$$+91+92+93+100)=\dfrac{719}{9}$$

$$\approx79.9$$ $$Variance=s^2=\dfrac{\displaystyle\sum_{i=1}^n(x_i-\bar{x})^2}{n-1}$$

$$=\dfrac{1}{8}((48-\dfrac{719}{9})^2+(59-\dfrac{719}{9})^2+(68-\dfrac{719}{9})^2$$

$$+(81-\dfrac{719}{9})^2+(87-\dfrac{719}{9})^2+(91-\dfrac{719}{9})^2$$

$$+(92-\dfrac{719}{9})^2+(93-\dfrac{719}{9})^2+(100-\dfrac{719}{9})^2)$$

$$\approx311.6$$

$$s=\sqrt{s^2}\approx17.7$$

2,

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

$$Range=96-49=47$$ $$mean=\bar{x}=\dfrac{\displaystyle\sum_{i=1}^nx_i}{n}=\dfrac{1}{11}(49+54+55+66+67$$

$$+73+77+87+92+93+96)=\dfrac{809}{11}$$

$$\approx73.5$$ $$Variance=s^2=\dfrac{\displaystyle\sum_{i=1}^n(x_i-\bar{x})^2}{n-1}$$

$$=\dfrac{1}{10}((49-\dfrac{809}{11})^2+(54-\dfrac{809}{11})^2+(55-\dfrac{809}{11})^2$$

$$+(66-\dfrac{809}{11})^2+(67-\dfrac{809}{11})^2+(73-\dfrac{809}{11})^2$$

$$+(77-\dfrac{809}{11})^2+(87-\dfrac{809}{11})^2+(92-\dfrac{809}{11})^2$$

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

$$\approx284.5$$

$$s=\sqrt{s^2}\approx16.9$$

3.

$$4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4$$

$$Range=4-4=0$$

$$mean=\bar{x}=\dfrac{\displaystyle\sum_{i=1}^nx_i}{n}=\dfrac{1}{17}(4+4+4+4+4$$

$$+4+4+4+4+4+4+4+4+4+4+4+4)$$

$$=4$$ $$Variance=s^2=\dfrac{\displaystyle\sum_{i=1}^n(x_i-\bar{x})^2}{n-1}$$

$$=\dfrac{1}{16}((4-4)^2+(4-4)^2+(4-4)^2+(4-4)^2$$

$$+(4-4)^2+(4-4)^2+(4-4)^2+(4-4)^2+(4-4)^2$$

$$+(4-4)^2+(4-4)^2+(4-4)^2+(4-4)^2+$$

$$+(4-4)^2+(4-4)^2+(4-4)^2+(4-4)^2)=0$$

$$s=\sqrt{s^2}=0$$

4.

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

$$Range=8-6=2$$

$$mean=\bar{x}=\dfrac{\displaystyle\sum_{i=1}^nx_i}{n}=\dfrac{1}{13}(6+6+6+6+6+6$$

$$+8+8+8+8+8+8+8)=\dfrac{92}{13}$$

$$\approx7.1$$ $$Variance=s^2=\dfrac{\displaystyle\sum_{i=1}^n(x_i-\bar{x})^2}{n-1}$$

$$=\dfrac{1}{12}((6-\dfrac{92}{13})^2+(6-\dfrac{92}{13})^2+(6-\dfrac{92}{13})^2$$

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

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

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

$$+(8-\dfrac{92}{13})^2)\approx1.1$$

$$s=\sqrt{s^2}\approx1.0$$

5.

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

$$Range=11-(-12)=23$$ $$mean=\bar{x}=\dfrac{\displaystyle\sum_{i=1}^nx_i}{n}=\dfrac{1}{7}(4-12-8-5-1+4$$

$$+7+11)=\dfrac{-4}{7}\approx-0.6$$

$$Variance=s^2=\dfrac{\displaystyle\sum_{i=1}^n(x_i-\bar{x})^2}{n-1}$$

$$=\dfrac{1}{6}((-12-\dfrac{-4}{7})^2+(-8-\dfrac{-4}{7})^2+(-5-\dfrac{-4}{7})^2$$

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

$$+(11-\dfrac{-4}{7})^2)\approx69.6$$

$$s=\sqrt{s^2}\approx8.3$$

6.

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

$$Range=-4-(-31)=27$$ $$mean=\bar{x}=\dfrac{\displaystyle\sum_{i=1}^nx_i}{n}=\dfrac{1}{7}(-31-23-19-17$$

$$-11-5-4)=\dfrac{-110}{7}\approx-15.7$$

$$Variance=s^2=\dfrac{\displaystyle\sum_{i=1}^n(x_i-\bar{x})^2}{n-1}$$

$$=\dfrac{1}{6}((-31-\dfrac{-110}{7})^2+(-23-\dfrac{-110}{7})^2$$

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

$$(-11-\dfrac{-110}{7})^2+(-5-\dfrac{-110}{7})^2$$

$$+(-4-\dfrac{-110}{7})^2\approx95.6$$

$$s=\sqrt{s^2}\approx9.8$$