Answer to Question #279909 in Statistics and Probability for Yttad

Question

Find the range the standard deviation and the variance for the given samples

Answer 1

Assume the data is: 48, 91, 87, 93, 59, 68, 92, 100, 81

Solution:

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

$Range=100-48=52$

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

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

$\approx79.9$

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

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

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

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

$\approx311.6$

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